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Let's analyze the problem step-by-step: **Given:** - A real object is placed in front of a converging lens. - The question asks for the inequality involving the object distance \( p \) and the focal length \( f \) that must hold for the image to be real. --- ### Step 1: Recall the lens formula The lens formula relates the object distance \( p \), the image distance \( q \), and the focal length \( f \): \[ \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \] - \( p \): object distance (positive if object is on the same side as the incoming light) - \( q \): image distance (positive if the image is real, i.e., on the opposite side of the lens) ### Step 2: Conditions for a real image - For a **real image** formed by a converging lens (positive focal length), the image distance \( q \) must be **positive**: \[ q > 0 \] ### Step 3: Express \( q \) in terms of \( p \) and \( f \) From the lens formula: \[ q = \frac{pf}{p - f} \] ### Step 4: Determine the inequality involving \( p \) and \( f \) Since \( q > 0 \): \[ \frac{pf}{p - f} > 0 \] - \( p > 0 \) (object in front of the lens, typical assumption) - \( f > 0 \) (converging lens) Now, analyze the inequality: \[ \frac{pf}{p - f} > 0 \] Because \( p > 0 \) and \( f > 0 \), the numerator \( pf \) is positive. The sign of the entire fraction depends on the denominator \( p - f \): - For the fraction to be positive, numerator and denominator must have the same sign: \[ p - f > 0 \quad \Rightarrow \quad p > f \] ### **Final answer:** \[ \boxed{ \text{For the image to be real, the object distance } p \text{ must satisfy:} \quad p > f } \] **In words:** *The object must be placed beyond the focal length of the converging lens for the image to be real.*

Let's analyze the problem step-by-step: --- ### Given Data: - Mass of neutron star: \( M_{NS} = 2.5 \text{ solar masses} \) - Orbital period: \( T = 0.40 \text{ years} \) - Orbital radius (distance between the two objects): \( R = 2.0 \text{ AU} \) - Mass of the blue star: \( M_{blue} = ? \) --- ### Step 1: Understand the system The neutron star and the blue star form a binary system orbiting their common center of mass. Since the neutron star is much less massive than the blue star (assuming, but in this case, the neutron star's mass is quite large), the problem simplifies if we consider the entire orbit of the neutron star around the blue star. But more precisely, both objects orbit their common center of mass at distances inversely proportional to their masses. ### Step 2: Use Kepler's third law For a two-body system, Kepler's third law states: \[ T^2 = \frac{4\pi^2}{G M_{total}} R^3 \] Where: - \( T \) = orbital period - \( R \) = orbital radius of the neutron star (distance from the blue star) - \( M_{total} = M_{blue} + M_{NS} \) Rearranged to find total mass: \[ M_{total} = \frac{4 \pi^2 R^3}{G T^2} \] --- ### Step 3: Convert units - \( R = 2.0 \text{ AU} \) - \( T = 0.40 \text{ years} \) Use units where: - \( 1 \text{ AU} \) and \( 1 \text{ year} \) are standard units in astronomical calculations. - Gravitational constant in these units: \[ G = 4\pi^2 \text{ AU}^3 / (\text{Solar mass} \times \text{ year}^2) \] This is a common choice because: \[ T^2 = \frac{a^3}{M_{total}} \] where \( a \) is in AU, \( T \) in years, and \( M_{total} \) in solar masses. Thus, in these units: \[ M_{total} = \frac{a^3}{T^2} \] --- ### Step 4: Calculate total mass \[ M_{total} = \frac{(2.0)^3}{(0.40)^2} = \frac{8}{0.16} = 50 \text{ solar masses} \] ### Step 5: Find the blue star's mass Since the neutron star (mass \( 2.5 M_\odot \)) orbits the common center of mass, the distance from the center of mass to the neutron star, \( r_{NS} \), relates to the total orbital radius: \[ r_{NS} = R_{NS} = \frac{M_{blue}}{M_{total}} \times R \] Similarly, the distance from the blue star to the center of mass: \[ r_{blue} = R_{blue} = \frac{M_{NS}}{M_{total}} \times R \] Given that the neutron star is observed at a distance of \( R_{NS} = 2.0 \text{ AU} \), but the problem seems to treat \( R \) as the orbital separation, representing the orbit of the neutron star around the blue star, we interpret this as: \[ R_{NS} = R \left( \frac{M_{blue}}{M_{total}} \right) \] Since the neutron star is at a distance \( R_{NS} \) from the blue star: \[ R_{NS} = \frac{M_{blue}}{M_{total}} \times R \] Rearranged: \[ M_{blue} = M_{total} \times \frac{R_{NS}}{R} \] But in the problem, \( R = 2.0 \text{ AU} \) is the orbital radius of the neutron star around the blue star (assuming the neutron star is orbiting at 2 AU from the blue star). The total mass: \[ M_{total} = 50 M_\odot \] The radius of the neutron star's orbit about the center of mass: \[ r_{NS} = R \times \left( \frac{M_{blue}}{M_{total}} \right) \] But because the neutron star's orbital radius is \( R_{NS} \), and the total separation is \( R \): \[ R_{NS} = R \times \left( \frac{M_{blue}}{M_{total}} \right) \] Rearranged: \[ M_{blue} = M_{total} \times \frac{R_{NS}}{R} \] Given the problem states the neutron star is **orbiting at a distance of 2.0 AU**, i.e., \( R_{NS} = 2.0 \text{ AU} \). Since the total separation is \( R \), the orbital radius of the neutron star is \( R_{NS} \): \[ \Rightarrow R_{NS} = 2.0 \text{ AU} \] But the total orbital separation \( R_{total} \) (distance between the two objects) is the sum of their distances from the center of mass: \[ R_{total} = R_{blue} + R_{NS} \] Assuming the neutron star is at 2 AU from the blue star, and the system's total separation is \( R_{total} \). Alternatively, the problem simplifies by considering the neutron star's orbit at 2 AU, and the total mass as found. The ratios: \[ \frac{R_{NS}}{R_{total}} = \frac{M_{blue}}{M_{total}} \] Since the neutron star is at 2 AU: \[ \frac{M_{blue}}{M_{total}} = \frac{R_{NS}}{R_{total}} \] If we assume the total orbital separation is approximately the same as the orbital radius of the neutron star (which is typical if the mass of the neutron star is small compared to the blue star), then: \[ M_{blue} \approx M_{total} - M_{NS} = 50 - 2.5 = 47.5 \text{ solar masses} \] But this would be inconsistent unless the neutron star's orbital radius is small compared to the total separation. **Alternatively**, since the orbital period is given for the neutron star orbiting the blue star, and the orbital radius is 2 AU, then: \[ M_{total} = \frac{a^3}{T^2} = 50 M_\odot \] And the ratio: \[ \frac{R_{NS}}{R_{total}} = \frac{M_{blue}}{M_{total}} \] Given that the neutron star's orbital radius is 2 AU, and the total separation is \( R \), which is not specified explicitly but seems to be 2 AU from the problem statement, then: \[ \Rightarrow M_{blue} = M_{total} - M_{NS} = 50 - 2.5 = 47.5 M_\odot \] --- ### **Final answer:** \[ \boxed{ \text{The mass of the blue star is approximately } \boxed{47.5 \text{ solar masses}} } \] --- ### **Summary:** - Kepler's third law yields a total mass of about 50 solar masses. - The neutron star's orbit at 2 AU suggests the blue star's mass is about 47.5 solar masses, since the neutron star is much less massive and orbits close to the blue star's position in the system. --- **Note:** The key assumption here is that the orbital period and radius correspond to the neutron star orbiting the blue star, and the total mass is derived accordingly.

Let's approach this problem step-by-step, applying conservation of energy to find the proton's speed at different distances and the stopping point. --- ### Given Data: - Proton mass, \( m = 1.67 \times 10^{-27} \, \text{kg} \) - Proton charge, \( q_p = +1.60 \times 10^{-19} \, \text{C} \) - Point charge, \( Q = +1.50 \, \mu\text{C} = +1.50 \times 10^{-6} \, \text{C} \) - Initial distance, \( r_i = 6.50 \, \text{m} \) - Initial speed, \( v_i = 2.50 \times 10^{6} \, \text{m/s} \) - Final distance for part (a), \( r_f = 3.50 \, \text{m} \) - For part (b), the proton comes to a stop (\( v_f = 0 \)), at some distance \( r_f \). --- ## Part (a): Find the proton's speed when it is 3.50 meters from the charge ### Step 1: Write conservation of energy Total energy at initial position = total energy at final position: \[ K_i + U_i = K_f + U_f \] where: - \( K = \frac{1}{2} m v^2 \) (kinetic energy) - \( U = \frac{k Q q_p}{r} \) (electric potential energy) - \( k = \frac{1}{4 \pi \varepsilon_0} \approx 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2 \) ### Step 2: Write the energy equation \[ \frac{1}{2} m v_i^2 + \frac{k Q q_p}{r_i} = \frac{1}{2} m v_f^2 + \frac{k Q q_p}{r_f} \] Solve for \( v_f \): \[ \frac{1}{2} m v_f^2 = \frac{1}{2} m v_i^2 + \frac{k Q q_p}{r_i} - \frac{k Q q_p}{r_f} \] \[ v_f = \sqrt{ v_i^2 + \frac{2}{m} \left( \frac{k Q q_p}{r_i} - \frac{k Q q_p}{r_f} \right) } \] --- ### Step 3: Calculate the terms Compute the potential energies: - \( k Q q_p \): \[ k Q q_p = (8.99 \times 10^9) \times (1.50 \times 10^{-6}) \times (1.60 \times 10^{-19}) \] \[ = 8.99 \times 10^9 \times 2.40 \times 10^{-25} = (8.99 \times 2.40) \times 10^{9 - 25} = 21.58 \times 10^{-16} = 2.158 \times 10^{-15} \, \text{J} \quad (\text{approximate}) \] ### Step 4: Plug in the values \[ v_i = 2.50 \times 10^6 \, \text{m/s} \] \[ r_i = 6.50 \, \text{m} \] \[ r_f = 3.50 \, \text{m} \] \[ m = 1.67 \times 10^{-27} \, \text{kg} \] Calculate the energy difference: \[ \frac{2}{m} \left( \frac{k Q q_p}{r_i} - \frac{k Q q_p}{r_f} \right) = \frac{2 \times 2.158 \times 10^{-15}}{1.67 \times 10^{-27}} \left( \frac{1}{6.50} - \frac{1}{3.50} \right) \] Compute numerator: \[ 2 \times 2.158 \times 10^{-15} = 4.316 \times 10^{-15} \] Divide by mass: \[ \frac{4.316 \times 10^{-15}}{1.67 \times 10^{-27}} \approx 2.586 \times 10^{12} \] Compute the difference in reciprocals: \[ \frac{1}{6.50} \approx 0.1538 \] \[ \frac{1}{3.50} \approx 0.2857 \] \[ 0.1538 - 0.2857 = -0.1319 \] Multiply: \[ 2.586 \times 10^{12} \times (-0.1319) \approx -3.414 \times 10^{11} \] Now, the total inside the square root: \[ v_f = \sqrt{ v_i^2 + (-3.414 \times 10^{11}) } \] Calculate \( v_i^2 \): \[ (2.50 \times 10^6)^2 = 6.25 \times 10^{12} \] Thus: \[ v_f = \sqrt{ 6.25 \times 10^{12} - 3.414 \times 10^{11} } = \sqrt{ 6.25 \times 10^{12} - 0.3414 \times 10^{12} } = \sqrt{ (6.25 - 0.3414) \times 10^{12} } = \sqrt{ 5.9086 \times 10^{12} } \] \[ v_f \approx \sqrt{ 5.9086 \times 10^{12} } \approx 2.43 \times 10^{6} \, \text{m/s} \] --- ### **Final answer for (a):** \[ \boxed{ v_f \approx \mathbf{2.43 \times 10^{6} \, \text{m/s}} } \] --- ## Part (b): How far from the charge does the proton come to rest? ### Step 1: Use energy conservation At the turning point, the proton's kinetic energy is zero: \[ 0 + \frac{k Q q_p}{r_{stop}} = \text{initial total energy} \] Initial total energy: \[ E_{initial} = \frac{1}{2} m v_i^2 + \frac{k Q q_p}{r_i} \] At the stopping point: \[ E_{final} = \frac{k Q q_p}{r_{stop}} \] Set equal: \[ \frac{1}{2} m v_i^2 + \frac{k Q q_p}{r_i} = \frac{k Q q_p}{r_{stop}} \] Solve for \( r_{stop} \): \[ \frac{k Q q_p}{r_{stop}} = \frac{1}{2} m v_i^2 + \frac{k Q q_p}{r_i} \] \[ r_{stop} = \frac{k Q q_p}{\displaystyle \frac{1}{2} m v_i^2 + \frac{k Q q_p}{r_i}} \] --- ### Step 2: Calculate each term - \( \frac{1}{2} m v_i^2 \): \[ = 0.5 \times 1.67 \times 10^{-27} \times (2.50 \times 10^{6})^2 \] \[ = 0.5 \times 1.67 \times 10^{-27} \times 6.25 \times 10^{12} = 0.5 \times 1.67 \times 6.25 \times 10^{-15} \] \[ = 0.5 \times 10.43 \times 10^{-15} = 5.215 \times 10^{-15} \, \text{J} \] - \( \frac{k Q q_p}{r_i} \), previously calculated as approximately \( 2.158 \times 10^{-15} \, \text{J} \) Sum: \[ 5.215 \times 10^{-15} + 2.158 \times 10^{-15} = 7.373 \times 10^{-15} \, \text{J} \] - \( k Q q_p = 2.158 \times 10^{-15} \, \text{J} \) ### Step 3: Compute \( r_{stop} \): \[ r_{stop} = \frac{2.158 \times 10^{-15}}{7.373 \times 10^{-15}} \approx 0.293 \, \text{m} \] Convert to centimeters: \[ 0.293 \, \text{m} \times 100 = 29.3 \, \text{cm} \] --- ### **Final answer for (b):** \[ \boxed{ r_{stop} \approx \mathbf{29.3 \, \text{cm}} } \] --- ## **Summary:** - **(a)** Proton speed at 3.50 m: **~2.43×10^6 m/s** - **(b)** Distance from the charge when it stops: **~29.3 cm** Let me know if you'd like further clarifications!

Let's analyze both parts step-by-step, calculating impulse, average force, acceleration, and pressure experienced during the crash. --- ## **Part a) Not Wearing Seat Belt** ### Given Data: - Mass of passenger, \( m = 90\, \text{kg} \) - Initial speed, \( v_i = 65\, \text{mi/hr} \) - Final speed, \( v_f = 0\, \text{m/s} \) - Time to stop, \( \Delta t = 0.01\, \text{s} \) - Area of impact surface, \( A = 0.5\, \text{m}^2 \) --- ### Step 1: Convert initial velocity to m/s \[ 1\, \text{mi/hr} = \frac{1609\, \text{m}}{3600\, \text{s}} \approx 0.447\, \text{m/s} \] \[ v_i = 65\, \text{mi/hr} \times 0.447\, \text{m/s per mi/hr} \approx 29.055\, \text{m/s} \] --- ### Step 2: Calculate impulse \( J \) Impulse is the change in momentum: \[ J = \Delta p = m (v_f - v_i) \] \[ J = 90\, \text{kg} \times (0 - 29.055\, \text{m/s}) = - 2614.95\, \text{kg·m/s} \] **Magnitude of impulse:** \[ |J| \approx 2615\, \text{kg·m/s} \] --- ### Step 3: Calculate average force \( F_{avg} \) \[ F_{avg} = \frac{J}{\Delta t} = \frac{2615\, \text{kg·m/s}}{0.01\, \text{s}} = 261,500\, \text{N} \] **Direction:** Opposite to initial motion (a decelerating force). --- ### Step 4: Calculate acceleration \( a \) Using \( F = m a \): \[ a = \frac{F_{avg}}{m} = \frac{261,500\, \text{N}}{90\, \text{kg}} \approx 2905.56\, \text{m/s}^2 \] This is a high deceleration. --- ### Step 5: Calculate pressure \( P \) Pressure is force per unit area: \[ P = \frac{F_{avg}}{A} = \frac{261,500\, \text{N}}{0.5\, \text{m}^2} = 523,000\, \text{Pa} \] or **about 523 kPa**. --- ## **Part b) Wearing Seat Belt** ### Given Data: - Same as above, but **crash duration = 0.18 s** ### Step 1: Impulse \( J \) Same change in momentum: \[ J = -2615\, \text{kg·m/s} \] --- ### Step 2: Calculate average force \[ F_{avg} = \frac{J}{\Delta t} = \frac{2615}{0.18} \approx 14,528\, \text{N} \] Much lower force due to longer stopping time. --- ### Step 3: Calculate acceleration: \[ a = \frac{F_{avg}}{m} = \frac{14,528}{90} \approx 161.4\, \text{m/s}^2 \] Less severe deceleration. --- ### Step 4: Calculate pressure: \[ P = \frac{F_{avg}}{A} = \frac{14,528}{0.5} = 29,056\, \text{Pa} \] or **about 29 kPa**. --- ## **Summary:** | Quantity | Part a (No Seat Belt) | Part b (With Seat Belt) | |------------|------------------------|-------------------------| | Impulse \( J \) | **~2615 kg·m/s** | **~2615 kg·m/s** | | Force \( F_{avg} \) | **~261,500 N** | **~14,528 N** | | Acceleration \( a \) | **~2905.56 m/s²** | **~161.4 m/s²** | | Pressure \( P \) | **~523 kPa** | **~29 kPa** | This demonstrates how wearing a seat belt significantly reduces the force, acceleration, and pressure experienced during a crash, increasing safety. --- Let me know if you need further details!

Let's analyze each part step-by-step based on the given SHM equation: \[ x(t) = 3.5 \cos (4.5 t - 2.0) \] --- ### Part (a): Find the amplitude **Amplitude** \(A\) is the maximum displacement from equilibrium. From the equation: \[ x(t) = A \cos(\omega t + \phi) \] The coefficient of cosine is the amplitude: \[ \boxed{ A = 3.5 \text{ meters} } \] --- ### Part (b): Find the frequency in Hz Given the angular frequency: \[ \omega = 4.5 \text{ rad/sec} \] The frequency \(f\) is: \[ f = \frac{\omega}{2\pi} = \frac{4.5}{2\pi} \approx \frac{4.5}{6.2832} \approx 0.716 \text{ Hz} \] **Answer:** \[ \boxed{ f \approx 0.716 \text{ Hz} } \] --- ### Part (c): Position at \( t = 1.2\, \text{s} \) Plug into the original equation: \[ x(1.2) = 3.5 \cos (4.5 \times 1.2 - 2.0) \] Calculate the argument: \[ 4.5 \times 1.2 = 5.4 \] \[ \text{Argument} = 5.4 - 2.0 = 3.4 \text{ radians} \] Calculate the cosine: \[ \cos(3.4) \approx -0.968 \] Therefore, \[ x(1.2) = 3.5 \times (-0.968) \approx -3.39\, \text{meters} \] **Answer:** \[ \boxed{ x(1.2) \approx -3.39\, \text{m} } \] --- ### Part (d): Velocity at \( t = 1.2\, \text{s} \) Velocity is the derivative of position: \[ v(t) = \frac{dx}{dt} \] Differentiate: \[ x(t) = A \cos(\omega t + \phi) \] \[ v(t) = -A \omega \sin(\omega t + \phi) \] Plug in values: \[ v(1.2) = -3.5 \times 4.5 \times \sin(3.4) \] Calculate: \[ 3.5 \times 4.5 = 15.75 \] \[ \sin(3.4) \approx 0.233 \] Thus: \[ v(1.2) = -15.75 \times 0.233 \approx -3.67\, \text{m/s} \] **Answer:** \[ \boxed{ v(1.2) \approx -3.67\, \text{m/s} } \] --- ### Part (e): Acceleration at \( t = 1.2\, \text{s} \) Acceleration is the derivative of velocity: \[ a(t) = \frac{dv}{dt} = -A \omega^2 \cos(\omega t + \phi) \] Or directly from the position: \[ a(t) = -A \omega^2 \cos(\omega t + \phi) \] Calculate: \[ a(1.2) = -3.5 \times (4.5)^2 \times \cos(3.4) \] \[ (4.5)^2 = 20.25 \] \[ a(1.2) = -3.5 \times 20.25 \times (-0.968) \] Note the double negative: \[ a(1.2) = -3.5 \times 20.25 \times (-0.968) \approx +3.5 \times 20.25 \times 0.968 \] Calculate: \[ 3.5 \times 20.25 = 70.875 \] \[ 70.875 \times 0.968 \approx 68.7 \] **Answer:** \[ \boxed{ a(1.2) \approx +68.7\, \text{m/s}^2 } \] --- ### Part (f): Maximum acceleration Maximum magnitude occurs when \(\cos(\omega t + \phi) = \pm 1\): \[ a_{max} = A \omega^2 \] Calculate: \[ a_{max} = 3.5 \times (4.5)^2 = 3.5 \times 20.25 = 70.9\, \text{m/s}^2 \] **Answer:** \[ \boxed{ a_{max} \approx 70.9\, \text{m/s}^2 } \] --- ### Part (g): Maximum velocity Maximum velocity occurs when \(\sin(\omega t + \phi) = \pm 1\): \[ v_{max} = A \omega \] Calculate: \[ v_{max} = 3.5 \times 4.5 = 15.75\, \text{m/s} \] **Answer:** \[ \boxed{ v_{max} \approx 15.75\, \text{m/s} } \] --- ## **Summary:** | Part | Result | |--------|---------| | (a) | Amplitude \(A = 3.5\, \text{m}\) | | (b) | Frequency \(f \approx 0.716\, \text{Hz}\) | | (c) | Position at \(t=1.2\, \text{s} \): \(\approx -3.39\, \text{m}\) | | (d) | Velocity at \(t=1.2\, \text{s}\): \(\approx -3.67\, \text{m/s}\) | | (e) | Acceleration at \(t=1.2\, \text{s}\): \(\approx 68.7\, \text{m/s}^2\) | | (f) | Max acceleration: \(\approx 70.9\, \text{m/s}^2\) | | (g) | Max velocity: \(\approx 15.75\, \text{m/s}\) | Let me know if you'd like any further clarification!

Let's carefully evaluate each part step-by-step. --- ### Given sets: - \( U = \{10, 20, 30, 40, 50, 60, 70, 80\} \) - \( A = \{10, 30, 50, 70\} \) - \( B = \{40, 50, 60, 70\} \) - \( C = \{20, 40, 60\} \) --- ### Part (a): Find \((A \cap B)^c\) **Step 1:** Find \(A \cap B\): \[ A \cap B = \{10, 30, 50, 70\} \cap \{40, 50, 60, 70\} = \{50, 70\} \] **Step 2:** Find the complement of \(A \cap B\) with respect to \(U\): \[ (A \cap B)^c = U - (A \cap B) = \{10, 20, 30, 40, 50, 60, 70, 80\} - \{50, 70\} \] \[ = \{10, 20, 30, 40, 60, 80\} \] **Answer (a):** \[ \boxed{ (A \cap B)^c = \{10, 20, 30, 40, 60, 80\} } \] --- ### Part (b): Find \( B \cup (A^c) \) **Step 1:** Find \(A^c\): \[ A^c = U - A = \{10, 20, 30, 40, 50, 60, 70, 80\} - \{10, 30, 50, 70\} = \{20, 40, 60, 80\} \] **Step 2:** Find \( B \cup A^c \): \[ B \cup A^c = \{40, 50, 60, 70\} \cup \{20, 40, 60, 80\} \] \[ = \{20, 40, 50, 60, 70, 80\} \] **Answer (b):** \[ \boxed{ B \cup A^c = \{20, 40, 50, 60, 70, 80\} } \] --- ### Part (c): Find \(A^c \cup C\) **Step 1:** Recall \(A^c = \{20, 40, 60, 80\}\) (from above). **Step 2:** Find the union: \[ A^c \cup C = \{20, 40, 60, 80\} \cup \{20, 40, 60\} = \{20, 40, 60, 80\} \] (since all elements of \(C\) are already in \(A^c\), the union is just \(A^c\)). **Answer (c):** \[ \boxed{ A^c \cup C = \{20, 40, 60, 80\} } \] --- **Final summarized answers:** | Part | Result | |---------|----------------------------------------------------| | (a) | \(\boxed{\{10, 20, 30, 40, 60, 80\}}\) | | (b) | \(\boxed{\{20, 40, 50, 60, 70, 80\}}\) | | (c) | \(\boxed{\{20, 40, 60, 80\}}\) | Let me know if you'd like further explanations!

Let's carefully analyze and solve the system of equations using Cramer's rule. --- ### Given equations: 1) \( 2x + 3y - 5x = 1 \) 2) \( x + y - z = 2 \) 3) \( 2y + z = 8 \) --- ### Step 1: Simplify equation (1) \[ 2x - 5x + 3y = 1 \] \[ -3x + 3y = 1 \] Divide through by 3: \[ - x + y = \frac{1}{3} \] Rearranged: \[ x - y = -\frac{1}{3} \] Now, the system is: \[ \boxed{ \begin{cases} x - y = -\frac{1}{3} \quad (1') \\ x + y - z = 2 \quad (2) \\ 2y + z = 8 \quad (3) \end{cases} } \] --- ### Step 2: Write the equations in standard form Expressed as: \[ \begin{cases} x - y + 0 \cdot z = -\frac{1}{3} \quad (1') \\ x + y - z = 2 \quad (2) \\ 0 \cdot x + 2 y + z = 8 \quad (3) \end{cases} \] The coefficients matrix \(A\): \[ A = \begin{bmatrix} 1 & -1 & 0 \\ 1 & 1 & -1 \\ 0 & 2 & 1 \end{bmatrix} \] --- ### Step 3: Compute the determinant of \(A\) \[ \det(A) = \begin{vmatrix} 1 & -1 & 0 \\ 1 & 1 & -1 \\ 0 & 2 & 1 \end{vmatrix} \] Using expansion along the first row: \[ \det(A) = 1 \times \begin{vmatrix} 1 & -1 \\ 2 & 1 \end{vmatrix} - (-1) \times \begin{vmatrix} 1 & -1 \\ 0 & 1 \end{vmatrix} + 0 \times (\text{something}) \] Calculate minors: - \( M_{11} = (1)(1) - (-1)(2) = 1 + 2 = 3 \) - \( M_{12} = (1)(1) - (-1)(0) = 1 - 0 = 1 \) Now, substitute: \[ \det(A) = 1 \times 3 - (-1) \times 1 + 0 = 3 + 1 = 4 \] **Determinant \(\det(A) = 4 \neq 0\)**, so the system has a unique solution. --- ### Step 4: Find determinants for \(x, y, z\) **a) \( \det(A_x) \)**: replace first column of \(A\) with constants \[ A_x = \begin{bmatrix} -\frac{1}{3} & -1 & 0 \\ 2 & 1 & -1 \\ 8 & 2 & 1 \end{bmatrix} \] Calculate \(\det(A_x)\): \[ \det(A_x) = -\frac{1}{3} \times \begin{vmatrix} 1 & -1 \\ 2 & 1 \end{vmatrix} - (-1) \times \begin{vmatrix} 2 & -1 \\ 8 & 1 \end{vmatrix} + 0 \times (\text{something}) \] Compute minors: - \( M_{11} = 1 \times 1 - (-1) \times 2 = 1 + 2 = 3 \) - \( M_{12} = 2 \times 1 - (-1) \times 8 = 2 + 8 = 10 \) Substitute: \[ \det(A_x) = -\frac{1}{3} \times 3 - (-1) \times 10 + 0 = -1 + 10 = 9 \] --- **b) \( \det(A_y) \)**: replace second column with constants \[ A_y = \begin{bmatrix} 1 & -\frac{1}{3} & 0 \\ 1 & 2 & -1 \\ 0 & 8 & 1 \end{bmatrix} \] Compute \(\det(A_y)\): \[ \det(A_y) = 1 \times \begin{vmatrix} 2 & -1 \\ 8 & 1 \end{vmatrix} - (-\frac{1}{3}) \times \begin{vmatrix} 1 & -1 \\ 0 & 1 \end{vmatrix} + 0 \times (\text{something}) \] Minors: - \( M_{11} = 2 \times 1 - (-1) \times 8 = 2 + 8 = 10 \) - \( M_{12} = 1 \times 1 - (-1) \times 0 = 1 - 0 = 1 \) Calculate: \[ \det(A_y) = 1 \times 10 - (-\frac{1}{3}) \times 1 = 10 + \frac{1}{3} = \frac{30}{3} + \frac{1}{3} = \frac{31}{3} \] --- **c) \( \det(A_z) \)**: replace third column with constants \[ A_z = \begin{bmatrix} 1 & -1 & -\frac{1}{3} \\ 1 & 1 & 2 \\ 0 & 2 & 8 \end{bmatrix} \] Compute \(\det(A_z)\): \[ \det(A_z) = 1 \times \begin{vmatrix} 1 & 2 \\ 2 & 8 \end{vmatrix} - (-1) \times \begin{vmatrix} 1 & 2 \\ 0 & 8 \end{vmatrix} + (-\frac{1}{3}) \times \begin{vmatrix} 1 & 1 \\ 0 & 2 \end{vmatrix} \] Calculate minors: - \( M_{11} = 1 \times 8 - 2 \times 2 = 8 - 4 = 4 \) - \( M_{12} = 1 \times 8 - 2 \times 0 = 8 - 0 = 8 \) - \( M_{13} = 1 \times 2 - 1 \times 0 = 2 - 0 = 2 \) Substitute back: \[ \det(A_z) = 1 \times 4 - (-1) \times 8 + (-\frac{1}{3}) \times 2 = 4 + 8 - \frac{2}{3} = 12 - \frac{2}{3} \] Express as improper fraction: \[ 12 - \frac{2}{3} = \frac{36}{3} - \frac{2}{3} = \frac{34}{3} \] --- ### Step 5: Find \(x, y, z\) \[ x = \frac{\det(A_x)}{\det(A)} = \frac{9}{4} = 2.25 \] \[ y = \frac{\det(A_y)}{\det(A)} = \frac{31/3}{4} = \frac{31/3}{4} = \frac{31}{3} \times \frac{1}{4} = \frac{31}{12} \approx 2.58 \] \[ z = \frac{\det(A_z)}{\det(A)} = \frac{34/3}{4} = \frac{34}{3} \times \frac{1}{4} = \frac{34}{12} = \frac{17}{6} \approx 2.83 \] --- ## **Final solutions:** \[ \boxed{ \begin{cases} x \approx 2.25 \\ y \approx 2.58 \\ z \approx 2.83 \end{cases} } \] --- Let me know if you'd like any further clarification!

Let's analyze each factor step-by-step and explain how it influences the magnitude of the magnetic force, including the reasoning. --- ### a) Strength of the magnetic field (\(B\)) **Effect:** **Directly proportional** to the magnetic force. **Reason:** The magnetic force on a moving charge or current-carrying conductor depends on the magnetic field strength. According to the Lorentz force law: \[ \vec{F} = q \vec{v} \times \vec{B} \] or for a current-carrying wire: \[ \vec{F} = I \vec{L} \times \vec{B} \] **Implication:** A stronger magnetic field (\(B\)) results in a larger force because the magnetic field exerts a greater force on moving charges or currents. --- ### b) Speed of the magnet passing through the coil (\(v\)) **Effect:** **Directly proportional** to the induced emf and hence to the force, in context of electromagnetic induction. **Reason:** When a magnet moves relative to a coil, it changes the magnetic flux linkage over time, inducing an emf (Faraday’s Law): \[ \mathcal{E} = -N \frac{d\Phi}{dt} \] - The faster the magnet moves (\(v\)), the quicker the magnetic flux (\(\Phi\)) changes, increasing \(\frac{d\Phi}{dt}\). - A larger emf induces a greater current in the coil, which, through its own magnetic field, interacts with the magnet, producing a force. **Implication:** Higher speed increases the magnitude of the induced current and the resulting magnetic force during the interaction. --- ### c) The number of loops wrapped around a cylinder (\(N\)) **Effect:** **Directly proportional** to the magnetic force. **Reason:** The emf induced in the coil (Faraday’s Law) is: \[ \mathcal{E} = -N \frac{d\Phi}{dt} \] - Increasing the number of loops (\(N\)) amplifies the emf for a given change in flux. - A larger emf produces a bigger current in the coil, which in turn generates a stronger magnetic field that interacts with the magnet. **Implication:** More loops mean a stronger induced magnetic field and a larger force exerted on the magnet. --- ### d) Current (\(I\)) **Effect:** **Directly proportional** to the magnetic force. **Reason:** The current in the coil creates its own magnetic field (according to Ampère’s Law). The interaction between this magnetic field and the magnet produces a force. - The induced emf leads to a current: \[ I = \frac{\mathcal{E}}{R} \] - A larger emf (from higher \(N\), faster \(v\), or stronger \(B\)) causes a larger current. - The magnetic force between two magnetic fields depends on the magnitude of the currents and the magnetic field strength. **Implication:** Higher current results in a stronger magnetic field generated by the coil, which increases the force exerted on the magnet. --- ### **Summary:** | Factor | Influence on Magnetic Force | Reason | |---------|------------------------------|---------| | **Strength of magnetic field (\(B\))** | **Increases** | The force depends on \(B\); stronger \(B\) exerts a larger force. | | **Speed of the magnet (\(v\))** | **Increases** | Faster change in flux induces a larger emf and current, boosting force. | | **Number of loops (\(N\))** | **Increases** | More loops induce higher emf and current, strengthening magnetic interaction. | | **Current (\(I\))** | **Increases** | Larger current produces a stronger magnetic field, resulting in a greater force. | Let me know if you'd like further clarification!

Let's analyze the problem carefully and find the rate at which the distance between the two bikes is changing. --- ### Given Data: - Bike A approaches the intersection at **12 km/h**. - Bike B leaves the intersection at **6 km/h**. - When: - Bike A is **8 km** from the intersection. - Bike B is **15 km** from the intersection. - The bikes are moving **perpendicular** to each other (i.e., at right angles). --- ### Step 1: **Set up coordinate system** - Place the intersection at the origin \((0,0)\). - Assume: - Bike A moves **along the x-axis** toward the intersection: - Since approaching, its position at time \(t\) is: \[ x_A(t) = 8 - v_A t \] where \(v_A = 12\, \text{km/h}\). - Bike B moves **along the y-axis** away from the intersection: \[ y_B(t) = 15 + v_B t \] where \(v_B = 6\, \text{km/h}\). (Note: Since B is leaving, its distance increases, so position increases with time.) --- ### Step 2: **Write the positions at the given moment** At the instant of interest: - \(x_A = 8\, \text{km}\), so: \[ 8 = 8 - v_A t \Rightarrow t = 0 \] - \(y_B = 15\, \text{km}\), so: \[ 15 = 15 + v_B t \Rightarrow t=0 \] This confirms we're measuring the rate at that instant \(t=0\). ### Step 3: **Express the positions as functions of time** - \(x_A(t) = 8 - 12 t\) - \(y_B(t) = 15 + 6 t\) ### Step 4: **Write the distance between the two bikes** Since they are moving perpendicular (along x and y axes), the distance \(D(t)\): \[ D(t) = \sqrt{[x_A(t)]^2 + [y_B(t)]^2} \] At \(t=0\): \[ x_A(0) = 8\, \text{km} \] \[ y_B(0) = 15\, \text{km} \] The rate of change of \(D(t)\): \[ \frac{dD}{dt} = \frac{1}{2 D} \times 2 \left( x_A \frac{dx_A}{dt} + y_B \frac{dy_B}{dt} \right) = \frac{ x_A \frac{dx_A}{dt} + y_B \frac{dy_B}{dt} }{ D } \] --- ### Step 5: **Calculate derivatives** \[ \frac{dx_A}{dt} = -12\, \text{km/h} \] \[ \frac{dy_B}{dt} = 6\, \text{km/h} \] --- ### Step 6: **Compute \(D\) at \(t=0\)** \[ D(0) = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17\, \text{km} \] --- ### Step 7: **Compute numerator of \(\frac{dD}{dt}\)** \[ x_A \frac{dx_A}{dt} + y_B \frac{dy_B}{dt} = (8)(-12) + (15)(6) = -96 + 90 = -6\, \text{km}^2/\text{h} \] --- ### **Step 8: **Calculate \(\frac{dD}{dt}\) at \(t=0\)** \[ \boxed{ \frac{dD}{dt} = \frac{-6}{17} \approx -0.353\, \text{km/h} } \] --- ### **Final answer:** \[ \boxed{ \text{The rate at which the distance between the bikes is changing} \approx \mathbf{-0.353\, \text{km/h}} } \] - The negative sign indicates **the distance is decreasing** at this instant. --- ### **Is the rate increasing or decreasing?** - Since the magnitude of the rate is negative and the bikes are moving **perpendicular** with different speeds, the *rate of change of the distance* depends on their relative positions. - To check whether the rate is increasing or decreasing, consider the second derivative, but based on the problem, at this instant, the *distance is decreasing* at about 0.353 km/h. --- **Summary:** - **The distance between the bikes is decreasing at approximately 0.353 km/h at the moment when the distances are 8 km and 15 km from the intersection.**

Let's analyze each part carefully. --- ## Given: - The fundamental frequency of a standing wave on a string: \[ f = \frac{v}{2L} \] where: - \(v\) = wave speed on the string - \(L\) = length of the string The wave speed: \[ v = \sqrt{\frac{T}{u}} \] where: - \(T\) = tension in the string - \(u\) = mass per unit length (\(\mu\)) --- ## Part (a): Find the length \(L_2\) of the second string for frequency 5.0 Hz, given the first string's parameters. ### Step 1: Understand the proportionality From the fundamental frequency: \[ f = \frac{v}{2L} \] If the tension \(T\) and mass per unit length \(\mu\) are the same, then the wave speed \(v\) remains constant: \[ v = \text{constant} \] So, the frequency is inversely proportional to the length: \[ f \propto \frac{1}{L} \] ### Step 2: Set up ratio from the given data For the first string: \[ f_1 = 4.0\, \text{Hz} \] \[ L_1 = 2.0\, \text{m} \] For the second string: \[ f_2 = 5.0\, \text{Hz} \] \[ L_2 = ? \] Assuming the same tension and mass per unit length, the ratio: \[ \frac{f_2}{f_1} = \frac{L_1}{L_2} \] Rearranged: \[ L_2 = L_1 \times \frac{f_1}{f_2} \] Plug in the values: \[ L_2 = 2.0\, \text{m} \times \frac{4.0}{5.0} = 2.0 \times 0.8 = 1.6\, \text{m} \] --- ### **Answer (a):** \[ \boxed{ L_2 = \mathbf{1.6\, \text{m}} } \] --- ## Part (b): Which quantities are assumed to remain constant, and how could you practically achieve that? ### Quantities kept constant: - **Tension \(T\)**: The tension in the string affects wave speed \(v\), so holding \(T\) constant keeps \(v\) unchanged. - **Mass per unit length \(\mu\)**: The density and mass distribution along the string affect \(v\). Keeping \(\mu\) constant ensures the wave speed remains the same. ### How to keep them constant practically: - Use the **same string material and diameter** for both strings to keep \(\mu\) constant. - Use the **same tension** in both strings, which can be achieved by applying the same pulling force or tensioning mechanism. --- ## Part (c): Keep the same length \(L=2.0\, \text{m}\) but change the frequency from 4.0 Hz to 5.0 Hz. ### What quantity must change? From the wave speed relation: \[ f = \frac{v}{2L} \] - If length \(L\) remains constant, then to increase frequency from 4.0 Hz to 5.0 Hz, the wave speed \(v\) must increase proportionally: \[ v \propto f \] - Since \(v = \sqrt{\frac{T}{\mu}}\), increasing \(v\) implies increasing the tension \(T\), assuming \(\mu\) remains constant. ### How to achieve it practically: - Increase the tension \(T\) in the string (e.g., tighten the string further with a tuning peg or weight). - Keep the same string material (constant \(\mu\)) and length. - By increasing \(T\), you increase \(v\), which raises the frequency at the same length. --- ## **Summary:** | Part | Explanation | |--------|--------------| | (a) | **Length \(L_2 = 1.6\, \text{m}\)**, found via inverse proportionality of frequency and length assuming constant tension and \(\mu\). | | (b) | **Quantities held constant:** tension \(T\) and mass per unit length \(\mu\). **Practically:** use the same string material and tensioning method. | | (c) | **Quantity changed:** tension \(T\). **How:** increase tension to raise wave speed and thus increase frequency, keeping the length the same. | --- Let me know if you'd like further clarification!

Let's analyze the difference in critical angles for orange (610 nm) and blue (470 nm) light in water surrounded by air. --- ### Step 1: Recall the formula for critical angle The critical angle \(\theta_c\) for total internal reflection when light goes from a medium with refractive index \(n_1\) (water) to a lower refractive index \(n_2\) (air) is: \[ \sin \theta_c = \frac{n_2}{n_1} \] - Since air has \(n_2 \approx 1.00\), - Water has \(n_1 \approx 1.33\) (approximately constant for visible wavelengths). **Note:** The refractive index depends slightly on wavelength, but for the visible spectrum, the variation is small. For simplicity, we'll assume: \[ n_{water} \approx 1.33 \quad \text{for both wavelengths} \] ### Step 2: Calculate \(\sin \theta_c\) \[ \sin \theta_c = \frac{1.00}{1.33} \approx 0.75 \] ### Step 3: Find the critical angle \[ \theta_c = \arcsin(0.75) \approx 48.6^\circ \] This is approximately the same for both wavelengths because the refractive index variation is very small in this range. --- ### Step 4: Considering wavelength dependence **Refractive index variation:** - For **610 nm (orange)**, \(n_{orange} \approx 1.33\) - For **470 nm (blue)**, \(n_{blue} \approx 1.34\) This slight difference causes the critical angles to differ slightly. Calculate for each: \[ \sin \theta_{c, orange} = \frac{1.00}{1.33} \approx 0.75 \] \[ \theta_{c, orange} \approx 48.6^\circ \] \[ \sin \theta_{c, blue} = \frac{1.00}{1.34} \approx 0.746 \] \[ \theta_{c, blue} \approx \arcsin(0.746) \approx 48.0^\circ \] ### **Difference in critical angles:** \[ \Delta \theta_c \approx 48.6^\circ - 48.0^\circ = 0.6^\circ \] --- ### **Final answer:** The critical angles for orange and blue light differ by approximately **0.6 degrees** in water surrounded by air. **Note:** This small difference arises due to the wavelength dependence of the refractive index (dispersion).

Let's carefully work through the problem step-by-step to find the resistivity (\(\rho\)) of the wire. --- ### Given Data: - Voltage, \( V = 110\, \text{V} \) - Radius of wire, \( r = 0.30\, \text{mm} = 0.30 \times 10^{-3}\, \text{m} = 3.0 \times 10^{-4}\, \text{m} \) - Length of wire, \( L = 3.5\, \text{m} \) - Total electrons flowing, \( N_e = 9.22 \times 10^{20} \) - Time, \( t = 2\, \text{min} = 120\, \text{s} \) --- ### Step 1: Find the total charge \(Q\) passing through the wire Number of electrons times charge per electron: \[ Q = N_e \times e \] where \( e = 1.60 \times 10^{-19}\, \text{C} \). Calculate: \[ Q = 9.22 \times 10^{20} \times 1.60 \times 10^{-19} = (9.22 \times 1.60) \times 10^{20 - 19} = 14.75 \times 10^{1} = 147.5\, \text{C} \] --- ### Step 2: Find the current \( I \) Current is charge per unit time: \[ I = \frac{Q}{t} = \frac{147.5\, \text{C}}{120\, \text{s}} \approx 1.229\, \text{A} \] --- ### Step 3: Find the cross-sectional area \(A\) Area of the circular cross-section: \[ A = \pi r^2 = \pi \times (3.0 \times 10^{-4})^2 \] \[ A = \pi \times 9.0 \times 10^{-8} \approx 3.1416 \times 9.0 \times 10^{-8} \approx 2.827 \times 10^{-7}\, \text{m}^2 \] --- ### Step 4: Find the resistance \( R \) Using Ohm's law: \[ V = IR \Rightarrow R = \frac{V}{I} = \frac{110\, \text{V}}{1.229\, \text{A}} \approx 89.5\, \Omega \] --- ### Step 5: Calculate resistivity \(\rho\) Resistivity is related to resistance: \[ R = \frac{\rho L}{A} \] Rearranged: \[ \rho = \frac{R A}{L} \] Substitute known values: \[ \rho = \frac{89.5\, \Omega \times 2.827 \times 10^{-7}\, \text{m}^2}{3.5\, \text{m}} \] Calculate numerator: \[ 89.5 \times 2.827 \times 10^{-7} \approx (89.5 \times 2.827) \times 10^{-7} \] \[ 89.5 \times 2.827 \approx 253.2 \] Now: \[ \rho = \frac{253.2 \times 10^{-7}}{3.5} \approx 72.34 \times 10^{-7} \] Express in standard form: \[ \rho \approx 7.23 \times 10^{-6}\, \Omega \cdot \text{m} \] --- ### **Final answer:** \[ \boxed{ \rho \approx \mathbf{7.23 \times 10^{-6}\, \Omega \cdot \text{m}} } \] **(rounded to 3 significant figures)** --- **Summary:** The resistivity of the wire is approximately **7.23 μΩ·m**.

Let's analyze each question carefully. --- ### 1. **Converging lens: condition for a real image** **Question:** For a real object and converging lens, what inequality involving \( p \) and \( f \) must hold for the image to be real? --- **Step-by-step reasoning:** - The lens formula: \[ \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \] - For a **converging lens**, \(f > 0\). - **Real images** are formed when the image distance \(q > 0\). - From the lens formula: \[ q = \frac{pf}{p - f} \] - For \(q > 0\): \[ \frac{pf}{p - f} > 0 \] - Since \(p > 0\) and \(f > 0\), the numerator \(pf > 0\). To keep the entire fraction positive, the denominator must also be positive: \[ p - f > 0 \] \[ p > f \] --- **Final answer:** \[ \boxed{ \text{For the image to be real, } p > f } \] --- ### 2. **Mass of blue star from neutron star orbit** **Given:** - Neutron star mass, \( M_{NS} = 2.5 M_\odot \) - Orbital period, \( T = 0.40 \text{ yr} \) - Distance from blue star, \( R = 2.0 \text{ AU} \) --- **Step-by-step:** - Use Kepler's third law (in units where \(T\) in years, \(a\) in AU, \(M_{total}\) in solar masses): \[ T^2 = \frac{a^3}{M_{total}} \] \[ M_{total} = \frac{a^3}{T^2} \] - Total mass: \[ M_{total} = \frac{(2.0)^3}{(0.40)^2} = \frac{8}{0.16} = 50\, M_\odot \] - The total mass is: \[ M_{total} = M_{blue} + M_{NS} \] - So, \[ M_{blue} = M_{total} - M_{NS} = 50 - 2.5 = \boxed{47.5\, M_\odot} \] --- **Final answer:** \[ \boxed{ \text{Mass of blue star} \approx 47.5\, M_\odot } \] --- ### 3. **Proton approaching a point charge** **Given:** - Proton: \( m = 1.67 \times 10^{-27} \text{kg} \) - \( q_{p} = +1.60 \times 10^{-19} \text{C} \) - Point charge Q: \( +1.50 \mu C = 1.50 \times 10^{-6} \text{C} \) - Initial speed \( v_i = 2.50 \times 10^{6} \text{m/s} \) - Initial distance \( r_i = 6.50 \text{m} \) --- **Part (a): How fast when at 3.50 m?** **Method:** Use conservation of energy: \[ \text{Initial total energy} = \text{Final total energy} \] \[ \frac{1}{2} m v_i^2 + \frac{k Q q_p}{r_i} = \frac{1}{2} m v_f^2 + \frac{k Q q_p}{r_f} \] Solve for \(v_f\): \[ v_f = \sqrt{ v_i^2 + \frac{2}{m} \left( \frac{k Q q_p}{r_i} - \frac{k Q q_p}{r_f} \right) } \] Calculate \(k Q q_p\): \[ k = 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2 \] \[ k Q q_p = 8.99 \times 10^9 \times 1.50 \times 10^{-6} \times 1.60 \times 10^{-19} \approx 2.16 \times 10^{-15} \text{J} \] Calculate the energy difference: \[ \Delta U = \frac{2.16 \times 10^{-15}}{6.50} - \frac{2.16 \times 10^{-15}}{3.50} \approx (3.32 \times 10^{-16}) - (6.17 \times 10^{-16}) = -2.85 \times 10^{-16} \text{J} \] Now, kinetic energy: \[ \frac{1}{2} m v_i^2 = 0.5 \times 1.67 \times 10^{-27} \times (2.50 \times 10^{6})^2 \approx 5.22 \times 10^{-15} \text{J} \] Calculate: \[ v_f = \sqrt{ 5.22 \times 10^{-15} - 2 \times 2.85 \times 10^{-16} / 1.67 \times 10^{-27} } \] But to keep it straightforward, plugging into the earlier formula: \[ v_f = \sqrt{ 6.25 \times 10^{12} + \frac{2 \times (-2.85 \times 10^{-16})}{1.67 \times 10^{-27}} } \] \[ = \sqrt{ 6.25 \times 10^{12} - \frac{5.70 \times 10^{-16}}{1.67 \times 10^{-27}} } \] \[ = \sqrt{ 6.25 \times 10^{12} - 3.41 \times 10^{11} } \approx \sqrt{ 5.91 \times 10^{12} } \approx 2.43 \times 10^{6} \text{ m/s} \] **Answer:** \[ \boxed{ v_f \approx 2.43 \times 10^{6} \text{ m/s} } \] --- **Part (b): How far from the charge when the proton stops?** Set final kinetic energy zero: \[ 0 = \frac{1}{2} m v_i^2 + \frac{k Q q_p}{r_i} - \frac{k Q q_p}{r_{stop}} \] Rearranged: \[ \frac{k Q q_p}{r_{stop}} = \frac{1}{2} m v_i^2 + \frac{k Q q_p}{r_i} \] Calculate: \[ r_{stop} = \frac{k Q q_p}{ \frac{1}{2} m v_i^2 + \frac{k Q q_p}{r_i} } \] Numerator: \(2.16 \times 10^{-15} \text{J}\) Kinetic energy: \(5.22 \times 10^{-15} \text{J}\) Potential energy at initial: \[ \frac{2.16 \times 10^{-15}}{6.50} \approx 3.32 \times 10^{-16} \text{J} \] Total numerator: \[ 3.32 \times 10^{-16} + 5.22 \times 10^{-15} \approx 5.55 \times 10^{-15} \text{J} \] Calculate: \[ r_{stop} = \frac{2.16 \times 10^{-15}}{5.55 \times 10^{-15}} \approx 0.389\, \text{m} \] In centimeters: \[ 0.389\, \text{m} \times 100 = 38.9\, \text{cm} \] --- ### **Summary:** - **(a)** Speed at 3.5 m: **~2.43×10^6 m/s** - **(b)** Distance when stops: **~38.9 cm** --- ### 4. **Critical angles in water** **Question:** How much do the critical angles for orange (610 nm) and blue (470 nm) light differ in water surrounded by air? --- **Step-by-step:** - Critical angle: \[ \theta_c = \arcsin \left( \frac{n_2}{n_1} \right) \] - \(n_2 \approx 1.00\) (air) - \(n_1\) depends on wavelength due to dispersion: | Wavelength | Approximate \(n_1\) in water | |--------------|------------------------------| | 610 nm (orange) | 1.33 | | 470 nm (blue) | 1.34 | - Calculate \(\theta_c\): \[ \theta_{c, \text{orange}} \approx \arcsin \left( \frac{1.00}{1.33} \right) \approx 48.6^\circ \] \[ \theta_{c, \text{blue}} \approx \arcsin \left( \frac{1.00}{1.34} \right) \approx 48.0^\circ \] **Difference:** \[ \approx 48.6^\circ - 48.0^\circ = 0.6^\circ \] **Final answer:** The critical angles differ by approximately **0.6 degrees**. --- ### 5. **Resistivity of the wire** **Given:** - Voltage \(V=110\, \text{V}\) - Radius \(r=0.30\, \text{mm} = 3.0 \times 10^{-4}\, \text{m}\) - Length \(L=3.5\, \text{m}\) - Electrons \(N_e=9.22 \times 10^{20}\) flow in \(t=120\, \text{s}\) --- **Step 1:** Total charge: \[ Q = N_e \times e = 9.22 \times 10^{20} \times 1.60 \times 10^{-19} \approx 147.5\, \text{C} \] **Step 2:** Current: \[ I = \frac{Q}{t} = \frac{147.5}{120} \approx 1.229\, \text{A} \] **Step 3:** Cross-sectional area: \[ A = \pi r^2 = \pi (3.0 \times 10^{-4})^2 \approx 2.827 \times 10^{-7}\, \text{m}^2 \] **Step 4:** Resistance: \[ R = \frac{V}{I} = \frac{110}{1.229} \approx 89.5\, \Omega \] **Step 5:** Resistivity: \[ \rho = \frac{R A}{L} = \frac{89.5 \times 2.827 \times 10^{-7}}{3.5} \approx 7.23 \times 10^{-6}\, \Omega \cdot \text{m} \] --- ## **Final answer:** \[ \boxed{ \rho \approx \mathbf{7.23 \times 10^{-6}\, \Omega \cdot \text{m}} } \] --- ### 6. **Motion of a charged object in a uniform electric field** **Question:** Which statement is correct? a) Remain motionless b) Accelerate to some constant speed in direction of the field c) Accelerate to some constant speed in opposite direction d) Constant acceleration in direction of the field e) Constant acceleration opposite to the field --- **Answer:** - The positive charge experiences a force: \[ \vec{F} = q \vec{E} \] - It will accelerate continuously in the direction of the electric field (since force is constant). - It will **not** reach a constant speed unless there are other forces (like friction), which are not mentioned. **Correct statement:** \[ \boxed{ \text{(d)} \quad \text{The object will experience a constant acceleration in the direction of the electric field} } \] --- ### 7. **Electric field strength** **Question:** At 1 cm from an electron, the electric field is \(E\). At what distance is the field \(E/2\)? **Solution:** - Electric field due to point charge: \[ E = \frac{k |q|}{r^2} \] - To find radius where \(E' = E/2\): \[ \frac{k |q|}{r'^2} = \frac{1}{2} \times \frac{k |q|}{r^2} \] \[ \Rightarrow \frac{1}{r'^2} = \frac{1}{2} \times \frac{1}{r^2} \] \[ r'^2 = 2 r^2 \] \[ r' = r \sqrt{2} \] Given \(r = 1\, \text{cm}\): \[ r' = 1\, \text{cm} \times \sqrt{2} \approx 1.414\, \text{cm} \] --- ### **Final answer:** \[ \boxed{ r \approx 1.41\, \text{cm} } \] --- **Summary:** - The critical angles differ by about **0.6°**. - Resistivity

False. **Explanation:** A typical fire burns in a conical or roughly spherical pattern, not specifically in an "X" pattern. The flames tend to rise upward due to hot air rising (convection), creating a flame that moves upward in search of oxygen. The movement is primarily vertical, not lateral in an "X" shape.

Let's analyze the problem carefully. Given the position vector as a function of \( t \): \[ \mathbf{r}(t) = (2t, -3t^5, -3t) \] and asked to find the **tangential** and **normal** components of the acceleration vector at \( t=1 \). --- ### Step 1: Find the velocity vector \(\mathbf{v}(t)\) \[ \mathbf{v}(t) = \frac{d\mathbf{r}}{dt} \] Differentiate each component: \[ v_x(t) = \frac{d}{dt}(2t) = 2 \] \[ v_y(t) = \frac{d}{dt}(-3t^5) = -15 t^4 \] \[ v_z(t) = \frac{d}{dt}(-3t) = -3 \] At \( t=1 \): \[ \mathbf{v}(1) = (2, -15, -3) \] --- ### Step 2: Find the acceleration vector \(\mathbf{a}(t)\) \[ \mathbf{a}(t) = \frac{d\mathbf{v}}{dt} \] Differentiate each component: \[ a_x(t) = 0 \] \[ a_y(t) = \frac{d}{dt}(-15 t^4) = -60 t^3 \] \[ a_z(t) = 0 \] At \( t=1 \): \[ \mathbf{a}(1) = (0, -60, 0) \] --- ### Step 3: Compute the magnitude of velocity \(\|\mathbf{v}\|\) \[ \|\mathbf{v}\| = \sqrt{v_x^2 + v_y^2 + v_z^2} = \sqrt{2^2 + (-15)^2 + (-3)^2} = \sqrt{4 + 225 + 9} = \sqrt{238} \approx 15.43 \] --- ### Step 4: Find the **tangential component** of acceleration \(a_t\) \[ a_t = \frac{\mathbf{a} \cdot \mathbf{v}}{\|\mathbf{v}\|} \] Compute the dot product: \[ \mathbf{a} \cdot \mathbf{v} = (0)(2) + (-60)(-15) + (0)(-3) = 0 + 900 + 0 = 900 \] Now: \[ a_t = \frac{900}{15.43} \approx 58.33 \] --- ### **Answer for the tangential component:** \[ \boxed{ a_t \approx 58.33 } \] --- ### Step 5: Find the **normal component** \(a_n\) The magnitude of the normal component is: \[ a_n = \sqrt{\|\mathbf{a}\|^2 - a_t^2} \] Calculate \(\|\mathbf{a}\|\): \[ \|\mathbf{a}\| = \sqrt{0^2 + (-60)^2 + 0^2} = 60 \] Then, \[ a_n = \sqrt{60^2 - 58.33^2} = \sqrt{3600 - 3399.9} \approx \sqrt{200.1} \approx 14.14 \] --- ### **Final answers:** \[ \boxed{ \text{Tangential acceleration } a_t \approx 58.33 \\ \text{Normal acceleration } a_n \approx 14.14 } \] These components describe how the acceleration vector decomposes into the component along the current velocity (tangential) and perpendicular to it (normal) at \( t=1 \).

Let's analyze the problem carefully. Given the position vector: \[ \mathbf{r}(t) = (2t, -3t^5, -3t) \] and asked to find the **tangential** and **normal** components of the acceleration vector at \( t=1 \). --- ### Step 1: Find the velocity vector \(\mathbf{v}(t)\) Differentiate each component: \[ v_x(t) = \frac{d}{dt}(2t) = 2 \] \[ v_y(t) = \frac{d}{dt}(-3t^5) = -15 t^4 \] \[ v_z(t) = \frac{d}{dt}(-3t) = -3 \] At \( t=1 \): \[ \mathbf{v}(1) = (2, -15, -3) \] --- ### Step 2: Find the acceleration vector \(\mathbf{a}(t)\) Differentiate each component of \(\mathbf{v}(t)\): \[ a_x(t) = \frac{d}{dt}(2) = 0 \] \[ a_y(t) = \frac{d}{dt}(-15 t^4) = -60 t^3 \] \[ a_z(t) = \frac{d}{dt}(-3) = 0 \] At \( t=1 \): \[ \mathbf{a}(1) = (0, -60, 0) \] --- ### Step 3: Calculate the magnitude of the velocity \(\|\mathbf{v}\|\) \[ \|\mathbf{v}\| = \sqrt{2^2 + (-15)^2 + (-3)^2} = \sqrt{4 + 225 + 9} = \sqrt{238} \approx 15.43 \] --- ### Step 4: Find the **tangential acceleration** \(a_t\) The tangential acceleration is the component of \(\mathbf{a}\) along \(\mathbf{v}\): \[ a_t = \frac{\mathbf{a} \cdot \mathbf{v}}{\|\mathbf{v}\|} \] Calculate the dot product: \[ \mathbf{a} \cdot \mathbf{v} = (0)(2) + (-60)(-15) + (0)(-3) = 0 + 900 + 0 = 900 \] Now: \[ a_t = \frac{900}{15.43} \approx 58.33 \] --- ### Step 5: Find the **normal acceleration** \(a_n\) The magnitude of the normal component: \[ a_n = \sqrt{\|\mathbf{a}\|^2 - a_t^2} \] Calculate \(\|\mathbf{a}\|\): \[ \|\mathbf{a}\| = \sqrt{0^2 + (-60)^2 + 0^2} = 60 \] Then: \[ a_n = \sqrt{60^2 - 58.33^2} = \sqrt{3600 - 3399.9} \approx \sqrt{200.1} \approx 14.14 \] --- ### **Final answers:** \[ \boxed{ \text{Tangential acceleration } a_t \approx 58.33\, \text{units} \\ \text{Normal acceleration } a_n \approx 14.14\, \text{units} } \] These components describe how the acceleration vector decomposes into the component along the current velocity (tangential) and perpendicular to it (normal) at \( t=1 \).

Let's analyze and find the critical points of the function: \[ z = (x^2 - 8x)(y^2 - 6y) \] and classify each critical point as a local maximum, minimum, or saddle point. --- ### Step 1: Find the partial derivatives Given: \[ z = (x^2 - 8x)(y^2 - 6y) \] which is a product of functions of \(x\) and \(y\). --- ### Step 2: Compute \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\) Use the product rule: \[ \frac{\partial z}{\partial x} = \frac{\partial}{\partial x} (x^2 - 8x) \times (y^2 - 6y) + (x^2 - 8x) \times \frac{\partial}{\partial x} (y^2 - 6y) \] Since \(y^2 - 6y\) is independent of \(x\): \[ \frac{\partial z}{\partial x} = (2x - 8)(y^2 - 6y) \] Similarly, \[ \frac{\partial z}{\partial y} = (x^2 - 8x) \times (2y - 6) \] --- ### Step 3: Find critical points Critical points occur where: \[ \frac{\partial z}{\partial x} = 0 \quad \text{and} \quad \frac{\partial z}{\partial y} = 0 \] which implies: \[ (2x - 8)(y^2 - 6y) = 0 \quad (1) \] \[ (x^2 - 8x)(2y - 6) = 0 \quad (2) \] --- ### Step 4: Solve for critical points From (1): \[ (2x - 8) = 0 \quad \Rightarrow \quad x = 4 \] or \[ y^2 - 6y = 0 \quad \Rightarrow \quad y(y - 6) = 0 \quad \Rightarrow \quad y=0 \text{ or } y=6 \] From (2): \[ x^2 - 8x = 0 \quad \Rightarrow \quad x(x - 8) = 0 \quad \Rightarrow \quad x=0 \text{ or } x=8 \] or \[ 2y - 6 = 0 \quad \Rightarrow \quad y=3 \] --- ### Step 5: Find all critical points by combining solutions **Case A:** \(x=4\) - From (2): - \(x=4\) does **not** satisfy either \(x=0\) or \(x=8\), so check the other factor: \[ 2y - 6 = 0 \Rightarrow y=3 \] At \(x=4, y=3\): \[ \frac{\partial z}{\partial x} = (2 \times 4 -8)(3^2 - 6 \times 3) = (8-8)(9 - 18) = 0 \times (-9) = 0 \] \[ \frac{\partial z}{\partial y} = (4^2 - 8 \times 4)(2 \times 3 - 6) = (16 - 32)(6 - 6) = (-16) \times 0 = 0 \] So, **\((4,3)\)** is a critical point. --- **Case B:** \(x=0\) - From (1): \[ (2 \times 0 -8)(y^2 - 6 y) = (-8)(y^2 - 6 y) = 0 \] which gives: \[ y^2 - 6 y = 0 \Rightarrow y=0 \text{ or } y=6 \] - From (2): \[ x=0 \Rightarrow (0^2 - 8 \times 0)(2 y - 6) = 0 \times (2 y - 6) = 0 \] which is always zero, so these are critical points. **Critical points:** - \((0,0)\) - \((0,6)\) --- **Case C:** \(x=8\) - From (1): \[ (2 \times 8 -8)(y^2 - 6 y) = (16-8)(y^2 - 6 y) = 8(y^2 - 6 y) = 0 \] which gives: \[ y^2 - 6 y=0 \Rightarrow y=0 \text{ or } y=6 \] - From (2): \[ x=8 \Rightarrow (8^2 - 8 \times 8)(2 y - 6) = (64 - 64)(2 y - 6) = 0 \] which is always zero. **Critical points:** - \((8,0)\) - \((8,6)\) --- ### **Summary of critical points:** \[ \boxed{ \begin{aligned} &(4,3) \\ &(0,0) \\ &(0,6) \\ &(8,0) \\ &(8,6) \end{aligned} } \] --- ### **Step 6: Classify each critical point** Compute the second derivatives: \[ z_{xx} = \frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x} \left[ (2x - 8)(y^2 - 6 y) \right] \] \[ z_{xx} = 2 (y^2 - 6 y) \] Similarly, \[ z_{yy} = \frac{\partial}{\partial y} \left[ (x^2 - 8x)(2 y - 6) \right] \] \[ z_{yy} = (x^2 - 8x) \times 2 \] And the mixed partial: \[ z_{xy} = \frac{\partial}{\partial y} \left[ (2x - 8)(y^2 - 6 y) \right] \] \[ z_{xy} = (2x -8) \times (2 y - 6) \] --- ### Step 7: Evaluate second derivatives at each critical point and compute the discriminant: \[ D = z_{xx} z_{yy} - (z_{xy})^2 \] --- ### **At \((4,3)\):** \[ z_{xx} = 2(3^2 - 6 \times 3) = 2(9 - 18) = 2(-9) = -18 \] \[ z_{yy} = (4^2 - 8 \times 4) \times 2 = (16 - 32) \times 2 = (-16) \times 2 = -32 \] \[ z_{xy} = (2 \times 4 -8)(2 \times 3 - 6) = (8 -8)(6 - 6) = 0 \times 0 = 0 \] Discriminant: \[ D = (-18)(-32) - 0^2 = 576 > 0 \] Since \(z_{xx} < 0\), the critical point at \((4,3)\) is a **local maximum**. --- ### **At \((0,0)\):** \[ z_{xx} = 2(0 - 0) = 0 \] \[ z_{yy} = (0 - 0) \times 2 = 0 \] \[ z_{xy} = (2 \times 0 -8)(0 - 6) = (-8)(-6) = 48 \] Discriminant: \[ D = 0 \times 0 - (48)^2 = -2304 < 0 \] **Saddle point** at \((0,0)\). --- ### **At \((0,6)\):** \[ z_{xx} = 2(36 - 36) = 0 \] \[ z_{yy} = (0 - 0) \times 2= 0 \] \[ z_{xy} = (2 \times 0 -8)(12 - 6) = (-8)(6) = -48 \] Discriminant: \[ D = 0 - (-48)^2 = -2304 < 0 \] **Saddle point** at \((0,6)\). --- ### **At \((8,0)\):** \[ z_{xx} = 2(0 - 0) = 0 \] \[ z_{yy} = (64 - 64) \times 2 = 0 \] \[ z_{xy} = (16 -8)(0 -6) = (8)(-6) = -48 \] Discriminant: \[ D = 0 - (-48)^2 = -2304 < 0 \] **Saddle point** at \((8,0)\). --- ### **At \((8,6)\):** \[ z_{xx} = 2(36 - 36) = 0 \] \[ z_{yy} = (64 - 64) \times 2 = 0 \] \[ z_{xy} = (16 -8)(12 - 6) = (8)(6) = 48 \] Discriminant: \[ D = 0 - 48^2 = -2304 < 0 \] **Saddle point** at \((8,6)\). --- ### **Final classification:** | Critical Point | Type | |------------------|---------------| | \((4,3)\) | **Local maximum** | | \((0,0)\) | **Saddle point** | | \((0,6)\) | **Saddle point** | | \((8,0)\) | **Saddle point** | | \((8,6)\) | **Saddle point** | --- **Summary:** - One local maximum at \((4,3)\). - All other critical points are saddle points. --- If you'd like further explanation or details, let me know!

Let's carefully analyze and compute the volume of the solid bounded below by the cone \( z = 2.5 \sqrt{x^2 + y^2} \) and above by the sphere \( x^2 + y^2 + z^2 = 6.5 z \). --- ### Step 1: Rewrite the sphere equation in standard form Given: \[ x^2 + y^2 + z^2 = 6.5 z \] Bring all to one side: \[ x^2 + y^2 + z^2 - 6.5 z = 0 \] Complete the square for \(z\): \[ z^2 - 6.5 z = z^2 - 6.5 z + \left(\frac{6.5}{2}\right)^2 - \left(\frac{6.5}{2}\right)^2 \] Calculations: \[ \frac{6.5}{2} = 3.25 \] \[ (3.25)^2 = 10.5625 \] Rewrite: \[ x^2 + y^2 + \left(z^2 - 6.5 z + 10.5625\right) = 10.5625 \] which is: \[ x^2 + y^2 + (z - 3.25)^2 = 10.5625 \] This is a sphere centered at \((0, 0, 3.25)\) with radius: \[ r = \sqrt{10.5625} \approx 3.245 \] --- ### Step 2: Convert the cone and sphere equations to cylindrical coordinates Set: \[ x = r \cos \theta,\quad y = r \sin \theta,\quad z = z \] - **Cone:** \[ z = 2.5 r \] - **Sphere:** \[ x^2 + y^2 + (z - 3.25)^2 = 10.5625 \] In cylindrical coordinates: \[ r^2 + (z - 3.25)^2 = 10.5625 \] --- ### Step 3: Find the intersection of the cone and sphere At the intersection, both equations hold: \[ z = 2.5 r \] and \[ r^2 + (z - 3.25)^2 = 10.5625 \] Substitute \(z = 2.5 r\): \[ r^2 + (2.5 r - 3.25)^2 = 10.5625 \] Expand: \[ r^2 + \left(2.5 r - 3.25\right)^2 = 10.5625 \] \[ r^2 + (6.25 r^2 - 2 \times 2.5 r \times 3.25 + 3.25^2) = 10.5625 \] Calculate each term: - \(6.25 r^2\) - \(2 \times 2.5 \times 3.25 = 16.25\), so the middle term: \[ -2 \times 2.5 r \times 3.25 = -16.25 r \] - \(3.25^2 = 10.5625\) Putting it together: \[ r^2 + 6.25 r^2 - 16.25 r + 10.5625 = 10.5625 \] Combine like terms: \[ (1 + 6.25) r^2 - 16.25 r + 10.5625 = 10.5625 \] \[ 7.25 r^2 - 16.25 r + 10.5625 = 10.5625 \] Subtract 10.5625 from both sides: \[ 7.25 r^2 - 16.25 r = 0 \] Factor: \[ r (7.25 r - 16.25) = 0 \] Solutions: - \(r=0\) (the axis) - \(7.25 r = 16.25 \Rightarrow r = \frac{16.25}{7.25} \approx 2.24\) Corresponding \(z\): \[ z = 2.5 r \approx 2.5 \times 2.24 \approx 5.60 \] --- ### **Step 4: Set up the volume integral** The volume is bounded between the cone \(z=2.5 r\) (below) and the sphere \(r^2 + (z - 3.25)^2 = 10.5625\) (above). - For each fixed \(r\), \(z\) ranges from the cone \(z = 2.5 r\) up to the sphere: \[ z = 3.25 \pm \sqrt{10.5625 - r^2} \] But the sphere is centered at \(z=3.25\), and the "top" of the sphere corresponds to: \[ z_{sphere} = 3.25 + \sqrt{10.5625 - r^2} \] The "bottom" is at: \[ z_{bottom} = 3.25 - \sqrt{10.5625 - r^2} \] Since the cone is below the sphere (and the intersection occurs at \(z \approx 5.6\)), the **upper boundary** for the volume is the sphere's top, and the **lower boundary** is the cone. --- ### **Step 5: Determine the limits** - \(r\) ranges from 0 to approximately 2.24 (the intersection point). - For each \(r \leq 2.24\), \(z\) goes from \(z = 2.5 r\) (cone) up to \(z = 3.25 + \sqrt{10.5625 - r^2}\) (sphere). --- ### **Step 6: Write the volume integral** In cylindrical coordinates, the differential volume element: \[ dV = r\, dr\, d\theta\, dz \] So, the volume: \[ V = \int_{0}^{2\pi} \int_{0}^{r_{max}} \int_{z_{cone}}^{z_{sphere}} r\, dz\, dr\, d\theta \] where: \[ r_{max} \approx 2.24 \] \[ z_{cone} = 2.5 r \] \[ z_{sphere} = 3.25 + \sqrt{10.5625 - r^2} \] Integrate over \(\theta\): \[ V = 2\pi \int_{0}^{r_{max}} r \left[ z_{sphere} - z_{cone} \right] dr \] \[ V = 2\pi \int_{0}^{2.24} r \left[ 3.25 + \sqrt{10.5625 - r^2} - 2.5 r \right] dr \] --- ### **Step 7: Approximate the integral** While the integral is somewhat complex, numerical approximation yields the volume. Alternatively, you can leave it in integral form. --- ### **Final answer:** **The volume of the solid is approximately:** \[ \boxed{ V \approx 2\pi \int_{0}^{2.24} r \left[ 3.25 + \sqrt{10.5625 - r^2} - 2.5 r \right] dr } \] which can be evaluated numerically for a precise value. --- **Summary:** - The sphere is centered at \((0,0,3.25)\) with radius \(\approx 3.245\). - The intersection occurs at \(r \approx 2.24\), \(z \approx 5.60\). - The volume bounded below by the cone \(z=2.5 r\) and above by the sphere is given by the integral above. Let me know if you'd like me to perform the numerical evaluation!

To identify which cube is made of glass B among the three similar-looking cubes, you can use the property of **refraction** — specifically, how light bends when passing through different materials. **Method:** 1. **Set up a laser pointer** to shine a narrow, coherent beam of light through each cube individually. 2. **Observe the bending (refraction) of the light beam** as it passes through each cube: - When light passes from air into glass, it bends toward the normal (a line perpendicular to the surface) because glass has a higher refractive index than air. - Different types of glass have different refractive indices, which affect **how much** the light bends. 3. **Compare the amount of bending:** - For the two cubes made of glass A, the refraction will be consistent. - The cube made of glass B (with a different refractive index) will cause the light to bend either more or less than the others. 4. **Identify the different cube:** - The cube that causes the **most** or **least** deviation of the laser beam indicates a different refractive index, revealing it is the different type of glass B. **Property of light used:** - **Refraction (bending of light)** at the interface between air and glass, which depends on the **refractive index** of the material. --- **Summary:** - Use a laser beam to pass through each cube. - Observe how much the beam bends due to refraction. - The cube that causes a different bending (more or less) is made of glass B. This method relies on the **refractive index** property, which varies among different types of glass and affects how light propagates through them.

Let's carefully work through each part of the problem involving the Brownian motion of a spherical talcum particle in water. --- ### **Given Data:** - Particle density: \(\rho_{particle} = 300\, \text{kg/m}^3\) - Temperature: \(T = 23^\circ C = 23 + 273.15 = 296.15\, K\) - Boltzmann constant: \(k_B \approx 1.38 \times 10^{-23}\, \text{J/K}\) - The particle's average kinetic energy: \(K = \frac{3}{2} k_B T\) - Diameter of particle: \(d\) (variable) - Assume particle is spherical. --- ## Part (a): Derive an expression for the RMS speed \(v_{rms}\) ### Step 1: Find the particle's mass \(m\) The volume of the sphere: \[ V = \frac{\pi}{6} d^3 \] Mass: \[ m = \rho_{particle} \times V = \rho_{particle} \times \frac{\pi}{6} d^3 \] ### Step 2: Use the kinetic energy relation The average kinetic energy: \[ K = \frac{1}{2} m v_{rms}^2 = \frac{3}{2} k_B T \] Solve for \(v_{rms}\): \[ v_{rms} = \sqrt{\frac{3 k_B T}{m}} \] Substitute \(m\): \[ v_{rms} = \sqrt{\frac{3 k_B T}{\frac{\pi}{6} \rho_{particle} d^3}} = \sqrt{\frac{3 k_B T \times 6}{\pi \rho_{particle} d^3}} \] Simplify: \[ v_{rms} = \sqrt{\frac{18 k_B T}{\pi \rho_{particle} d^3}} \] --- ### **Express as a multiple of \(d^{-3/2}\):** Observe that: \[ v_{rms} \propto \left( \frac{1}{d^3} \right)^{1/2} = d^{-3/2} \] Thus, \[ \boxed{ v_{rms} = \sqrt{\frac{18 k_B T}{\pi \rho_{particle}}} \times d^{-3/2} } \] **Answer (boxed):** \[ \boxed{ v_{rms} \propto d^{-3/2} } \] or explicitly: \[ v_{rms} = \left( \sqrt{\frac{18 k_B T}{\pi \rho_{particle}}} \right) \times d^{-3/2} \] --- ## Part (b): Time to travel a distance equal to its diameter at RMS speed ### Step 1: Assume \[ t = \frac{\text{distance}}{\text{speed}} = \frac{d}{v_{rms}} \] From part (a), \(v_{rms} \propto d^{-3/2}\), so: \[ t \propto \frac{d}{d^{-3/2}} = d^{1 + 3/2} = d^{5/2} \] ### **Answer:** \[ \boxed{ t = \text{constant} \times d^{5/2} } \] or explicitly: \[ t = C \times d^{5/2} \] where the constant \(C\) includes the factors from part (a): \[ C = \frac{1}{\sqrt{\frac{18 k_B T}{\pi \rho_{particle}}}} = \text{a constant} \] --- ### **In terms of notation:** \[ t \propto d^{5/2} \] or \[ t = \text{[some constant]} \times d^{5/2} \] --- ## Part (c): Numerical calculation for \(d=4.00\, \mu m\) ### Step 1: Convert diameter to meters: \[ d = 4.00\, \mu m = 4.00 \times 10^{-6}\, \text{m} \] --- ### Step 2: Calculate the constant factor \[ v_{rms} = \sqrt{\frac{18 \times 1.38 \times 10^{-23} \times 296.15}{\pi \times 300}} \times d^{-3/2} \] Calculate numerator: \[ 18 \times 1.38 \times 10^{-23} \times 296.15 \approx 18 \times 1.38 \times 296.15 \times 10^{-23} \] Stepwise: \[ 1.38 \times 296.15 \approx 408.9 \] \[ 18 \times 408.9 \approx 7360 \] \[ \text{Numerator} \approx 7360 \times 10^{-23} = 7.36 \times 10^{-20} \] Denominator: \[ \pi \times 300 \approx 942.4778 \] Compute: \[ \sqrt{\frac{7.36 \times 10^{-20}}{942.4778}} \approx \sqrt{7.81 \times 10^{-23}} \approx 8.84 \times 10^{-12} \] --- ### Step 3: Calculate \(d^{-3/2}\) \[ d = 4.00 \times 10^{-6}\, \text{m} \] \[ d^{-3/2} = (4.00 \times 10^{-6})^{-3/2} \] Calculate the exponent: \[ (4.00)^{-3/2} \times (10^{-6})^{-3/2} \] - \(4^{-3/2} = (4^{1/2})^{-3} = (2)^{-3} = \frac{1}{8}\) - \(10^{-6}\) to the \(-3/2\): \[ (10^{-6})^{-3/2} = 10^{6 \times 3/2} = 10^{9} \] Thus, \[ d^{-3/2} = \frac{1}{8} \times 10^{9} = 1.25 \times 10^{8} \] ### **Calculate \(v_{rms}\):** \[ v_{rms} \approx 8.84 \times 10^{-12} \times 1.25 \times 10^{8} = (8.84 \times 1.25) \times 10^{-12 + 8} \approx 11.05 \times 10^{-4} = 1.105 \times 10^{-3}\, \text{m/s} \] Convert to mm/s: \[ 1.105 \times 10^{-3}\, \text{m/s} = 1.105\, \text{mm/s} \] --- ### **Calculate the time \(t\):** \[ t = \frac{d}{v_{rms}} = \frac{4.00 \times 10^{-6}\, \text{m}}{1.105 \times 10^{-3}\, \text{m/s}} \approx 3.62 \times 10^{-3}\, \text{s} = 3.62\, \text{ms} \] --- ### **Final answers:** \[ \boxed{ v_{rms} \approx \mathbf{1.105\, \text{mm/s}} \\ t \approx \mathbf{3.62\, \text{ms}} } \] --- ## **Summary:** - **RMS speed:** approximately **1.11 mm/s** - **Time to travel one diameter:** approximately **3.62 ms** --- If you'd like further details or clarifications, let me know!