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:


Answer:
20 pig callers
Explanation:
Given that:
A pig caller produced intensity level of a sound = 107 dB
To find how many pig callers required to generate an intensity level of 120 dB;
we have:
120 dB - 107 dB = 13 dB
Taking the logarithm function;

where;
= initial intensity


I = 19.95
I ≅ 20 pig callers
It’s the crest, the crest is the top part of the wave and the trough is the bottom so they correspond
Answer:

Explanation:
Given:
- mass of solid disk,

- radius of disk,

- force of push applied to disk,

- distance of application of force from the center,

<em>For the condition of no slip the force of static friction must be greater than the applied force so that there is no skidding between the contact surfaces at the contact point.</em>

where:
= static frictional force



