Einstein and Lorentz, being avid tennis players, play a fast-paced game on a court where they stand 20.0 m from each other. Bein
g very skilled players, they play without a net. The tennis ball has mass 0.0580 kg. You can ignore gravity and assume that the ball travels parallel to the ground as it travels between the two players. Unless otherwise specified, all measurements are made by the two men. (a) Lorentz serves the ball at 80.0 m>s. What is the ball’s kinetic energy? (b) Einstein slams a return at 1.80 * 108 m>s. What is the ball’s kinetic energy? (c) During Einstein’s return of the ball in part (a), a white rabbit runs beside the court in the direction from Einstein to Lorentz. The rabbit has a speed of 2.20 * 108 m>s relative to the two men. What is the speed of the rabbit relative to the ball? (d) What does the rabbit measure as the distance from Einstein to Lorentz? (e) How much time does it take for the rabbit to run 20.0 m, according to the players? (f) The white rabbit carries a pocket watch. He uses this watch to measure the time (as he sees it) for the distance from Einstein to Lorentz to pass by under him. What time does he measure?
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Answer:
The magnitude of the horizontal displacement of the rock is 7.39 m/s.
Explanation:
Given that,
Initial speed = 11.5 m/s
Angle = 50.0
Height = 30.0 m
We need to calculate the horizontal displacement of the rock
Using formula of horizontal component
Put the value into the formula
Hence, The magnitude of the horizontal displacement of the rock is 7.39 m/s.
Answer:
(A) 0.279N at angle 38.02°
(B) 0.701N
(C) 14.19°
Explanation:
(A) The net force on q3 is given as:
F = Fxi + Fyj
Fx is the x component of the force
Fy is the y component of the force
Fx = -F(1, 3)cos(90 - x) + F(2, 3)cos0
Fy = -F(2, 3)cosx - F(2, 3)cos90 = -F(2, 3)cosx
First let us find y and angle x from the diagram.
Using Pythagoras theorem,
y² = 0.3² + 0.4²
y² = 0.25
y = 0.5m
Using SOHCAHTOA to find x,
sinx = 0.4/0.5
x = 53.13°
Electrostatic force, F is given as:
F = kqQ/r²
Where k = Coulumbs constant
F(1,3) = (k*q1*q3) / r²
F(1, 3) = (9 * 10^9 * 2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.5²)
F(1, 3) = 0.288N
F(2,3) = (k*q2*q3) / r²
F(2, 3) = (9 * 10^9 * 2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.4²)
F(2, 3) = 0.45N
Therefore,
Fx = -0.288cos36.87 + 0.45
Fx = 0.22N
Fy = 0.288cos53.13
Fy = 0.172N
=> F = 0.22i + 0.172j
The magnitude of the force will be
F(mag) = √(0.22² + 0.172²)
F(mag) = 0.279N
The direction of the force makes will be
tanθ = Fy/Fx
tanθ = 0.172/0.22 = 0.781
θ = 38.02° to the x axis.
(B) q2 = - 2.0 * 10^(-6)
This implies that:
F(2,3) = (k*q2*q3) / r²
F(2, 3) = (9 * 10^9 * -2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.4²)
F(2, 3) = -0.45N
Therefore,
Fx = -0.288cos36.87 - 0.45
Fx = -0.68N
Fy = 0.172N
=> F = - 0.68i + 0.172j
The magnitude of the force will be
F(mag) = √((-0.68)² + 0.172²)
F(mag) = 0.701N
(C) The direction of the force makes will be
tanθ = 0.172/0.68
θ = 14.19° to the x axis
