1.In a cathode ray tube, electrons are accelerated from rest by a
constant electric force of magnitude 6.56 x 10*-17 N during the
first 1.87 cm of tube's length; then they move at essentially
constant velocity another 45.0 cm before hitting the screen.
Find the speed of the electrons when they hit the screen.
How long does it take them to travel the length of the tube?
2.It takes 2.60 s for a small ball with a mass of 0.060 kg released
from rest from a tall building to reach the ground. Calculate
the height from which the ball is released.
If that ball had been released from the same height, but this
time above the surface of the moon, how long would it have
taken for the ball to hit the ground?
1; you know acceleration (F/mass) and distance vf^2=2*acceleration*distance
2; You know g, time, and mass.
h=1/2 g t^2
1. To find the speed of the electrons when they hit the screen, we need to use the equations of motion.
First, we can use the equation of motion for constant acceleration, which is:
v^2 = u^2 + 2as
Where:
v = final velocity
u = initial velocity (in this case, 0 since the electrons are accelerated from rest)
a = acceleration
s = displacement
In this case, the displacement (s) is 1.87 cm. The acceleration (a) can be found using Newton's second law, F = ma, where F is the force acting on the electrons.
Given that the force (F) is 6.56 x 10^-17 N, we can rearrange the equation to solve for acceleration (a):
a = F/m = (6.56 x 10^-17 N) / (mass of the electron)
Using the mass of an electron (9.10938356 x 10^-31 kg), we can calculate the acceleration:
a = (6.56 x 10^-17 N) / (9.10938356 x 10^-31 kg) ≈ 7.19 x 10^13 m/s^2
Now we can substitute the values into the equation of motion:
v^2 = 0 + 2 * (7.19 x 10^13 m/s^2) * 1.87 * 10^-2 m
v^2 ≈ 2 * (7.19 x 10^13 m/s^2) * 1.87 * 10^-2 m
v ≈ √((2 * (7.19 x 10^13 m/s^2) * 1.87 * 10^-2 m))
v ≈ 5.60 x 10^6 m/s
Therefore, the speed of the electrons when they hit the screen is approximately 5.60 x 10^6 m/s.
2. To find the height from which the ball is released, we can use the equation of motion for freefall:
s = ut + 0.5 * a * t^2
Where:
s = height (displacement)
u = initial velocity (0 since the ball is released from rest)
a = acceleration due to gravity (approximately 9.8 m/s^2 on the surface of the Earth)
t = time taken to reach the ground (2.60 s)
Since the ball is released from rest, u = 0, the equation simplifies to:
s = 0 + 0.5 * a * t^2
s = 0.5 * 9.8 m/s^2 * (2.60 s)^2
s ≈ 33.92 m
Therefore, the height from which the ball is released is approximately 33.92 meters.
If the ball were released from the same height above the surface of the moon, we can assume that the acceleration due to gravity on the moon is 1/6th of the acceleration on Earth, approximately 1.63 m/s^2.
Using the same equation of motion:
s = 0.5 * a * t^2
Substituting the values:
33.92 m = 0.5 * 1.63 m/s^2 * t^2
Simplifying the equation:
t^2 ≈ (33.92 m) / (0.5 * 1.63 m/s^2)
t^2 ≈ 41.32 s^2
Taking the square root of both sides of the equation:
t ≈ √(41.32 s^2)
t ≈ 6.43 s
Therefore, if the ball were released from the same height above the surface of the moon, it would take approximately 6.43 seconds for the ball to hit the ground.
To solve these problems, we need to use the equations of motion to analyze the motion of the objects involved. Specifically, we will use the equations for uniformly accelerated motion:
1. s = ut + (1/2)at^2
2. v = u + at
3. v^2 = u^2 + 2as
Problem 1:
We know the initial velocity (u) is 0 m/s because the electrons are initially at rest. The acceleration (a) is given by dividing the force (F) by the mass (m) of an electron: a = F/m. You can calculate the value of acceleration a.
To find the speed (v) of the electrons when they hit the screen, we need to find the time it takes for them to travel the length of the tube. We can use the equation of motion s = ut + (1/2)at^2 to solve for s by substituting the given values for s as 45.0 cm and a as the calculated value for acceleration. Rearranging the equation, we get t^2 + (2u/a)t - (2s/a) = 0. We can solve this quadratic equation to find the value of t. Once we have the value of t, we can find the speed (v) using the equation v = u + at, where u is 0 m/s and a is the calculated acceleration.
Problem 2:
To find the height from which the ball is released, we can use the equation for free fall motion: s = ut + (1/2)gt^2, where s is the height, u is the initial velocity (0 m/s because it is released from rest), g is the acceleration due to gravity (approximately 9.8 m/s^2). Rearranging the equation, we get t^2 + (2u/g)t - (2s/g) = 0. We can solve this quadratic equation using the given time (2.60 s) and g to find the value of s.
If the ball is released from the same height above the surface of the moon, the acceleration due to gravity (g) will be different because it is approximately one-sixth of that on Earth. We can use the same equation of motion s = ut + (1/2)gt^2 with the new value of g to find the time it takes for the ball to hit the ground on the moon. Rearranging the equation, we get t^2 + (2u/g_moon)t - (2s/g_moon) = 0. We can solve this quadratic equation to find the value of t.