The force the escaping gas exerts of the rocket is 10.42 N.
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Force escaping gas exerts</h3>
The force the escaping gas exerts of the rocket is calculated as follows;
F = m(v - u)/t
where;
- m is mass of the rocket
- v is the final velocity of the rocket
- u is the initial velocity of the rocket
- t is time of motion
F = (0.25)(40 - 15)/0.6
F = 10.42 N
Thus, the force the escaping gas exerts of the rocket is 10.42 N.
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He thermal velocity or thermal speed is a typical velocity of the thermal motion of particles which make up a gas, liquid, etc. Thus, indirectly, thermal velocity is a measure of temperature. Technically speaking it is a measure of the width of the peak in the Maxwell–Boltzmann particle velocity distribution.
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:


Reducing the amount of loops will cause a loss of strength, as the loops make the magnet stronger.
Answer:
The magnitudes of the second force is 
The magnitudes of the resultant force is 
Explanation:
From the question we are told that
The force is 
The angle made with second force 
The angle between the resultant force and the first force 
For us to solve problem we are going to assume that
The magnitude of the second force is Z N
The magnitude of the resultant force is R N
According to Sine rule

Substituting values

According to cosine rule

Substituting values

