i need solutions for all this problems...help please....
1.) an electron moving at a speed of 4x10 power of 6 m/s, shot through a sheet of paper which is 3x10 to power of -5 m thick. the electron energyfrom the sheet with with a speed of 2x10 to the power of 6 m/s. find the
a.) acceleration experienced y the electron in passing through the sheet and
b.) time take by the electron to pass through the assuming that the acceleration of the electron constant...
2.) the velocity of a train changes from 0 to 60 km/hr in 20 seconds assuming that is uniformly accelerated ,what is the acceleration..?
3.) a body is moving with a velocity of 4 m/s . it is then accelerated for 5 seconds at the rate of 2 m/s squared.
a.) find the speed at the end of the fifth second.
b.)find the distance covered during the first and fifth seconds.
4.)a speeding motorist traveling 120km/h passes a stationary police officer. the officer immediately begins pursuit at a constant acceleration of 10.0 km/h/s (note the mixed units). how much time will it take for the police officer to reach the speeder,assuming that the speeder maintains a constant speed? how fast will the police officer be traveling at this time..?
5.) a light plane must reach a speed of 30 m/s for takeoff.how long a runway is needed if the (constant) acceleration is 3.0m/s squared?
Sure, let me help you with each of these problems one by one:
1.) a.) To find the acceleration experienced by the electron, we can use the equation of motion: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance. Here, the electron's initial speed is 4x10^6 m/s, the final speed is 2x10^6 m/s, and the distance traveled is 3x10^-5 m (thickness of the sheet). Substitute these values into the equation and solve for a:
(2x10^6)^2 = (4x10^6)^2 + 2a(3x10^-5)
Solve for a.
b.) To find the time taken by the electron to pass through the sheet, we can use the equation of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Rearrange the equation to solve for t:
t = (v - u) / a
Substitute the values and solve for t.
2.) To find the acceleration of the train, we can use the equation of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Rearrange the equation to solve for a:
a = (v - u) / t
Substitute the values and solve for a.
3.) a.) To find the speed at the end of the fifth second, we can use the equation of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Substitute the values and solve for v.
b.) To find the distance covered during the first and fifth seconds, we can use the equation of motion: s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time. Substitute the values and solve for s.
4.) To find the time it will take for the police officer to reach the speeder, we can use the equation of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Since the speeder maintains a constant speed, v is constant. Rearrange the equation to solve for t:
t = (v - u) / a
Substitute the values and solve for t.
To find the speed of the police officer at this time, substitute the value of t into the equation of motion and solve for v.
5.) To find the length of the runway needed, we can use the equation of motion: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance. In this case, the initial velocity is 0, the final velocity is 30 m/s, and the acceleration is 3.0 m/s^2. Rearrange the equation to solve for s.
These are the steps to solve each problem. Let me know if you need any further explanation or assistance with any specific step.