Answer:
h' = 603.08 m
Explanation:
First, we will calculate the initial velocity of the pellet on the surface of Earth by using third equation of motion:
2gh = Vf² - Vi²
where,
g = acceleration due to gravity on the surface of earth = - 9.8 m/s² (negative sign due to upward motion)
h = height of pellet = 100 m
Vf = final velocity of pellet = 0 m/s (since, pellet will momentarily stop at highest point)
Vi = Initial Velocity of Pellet = ?
Therefore,
(2)(-9.8 m/s²)(100 m) = (0 m/s)² - Vi²
Vi = √(1960 m²/s²)
Vi = 44.27 m/s
Now, we use this equation at the surface of moon with same initial velocity:
2g'h' = Vf² - Vi²
where,
g' = acceleration due to gravity on the surface of moon = 1.625 m/s²
h' = maximum height gained by pellet on moon = ?
Therefore,
2(1.625 m/s²)h' = (44.27 m/s)² - (0 m/s)²
h' = (1960 m²/s²)/(3.25 m/s²)
<u>h' = 603.08 m</u>
Answer:
m is expressed in kilograms and r in metres, with I (moment of inertia) having the dimension kilogram-metre square.
No friction present means: Ek = Ep
So Ek = mgh = 10 * 9.8 * 2 = 196 J
Protons and neutrons in an atom are held together by a nuclear energy also called the strong force.
Answer:
Probability of tunneling is 
Solution:
As per the question:
Velocity of the tennis ball, v = 120 mph = 54 m/s
Mass of the tennis ball, m = 100 g = 0.1 kg
Thickness of the tennis ball, t = 2.0 mm = 
Max velocity of the tennis ball,
= 89 m/s
Now,
The maximum kinetic energy of the tennis ball is given by:

Kinetic energy of the tennis ball, KE' = 
Now, the distance the ball can penetrate to is given by:


Thus



Now,
We can calculate the tunneling probability as:



Taking log on both the sides:

