Here is your answer
b) 
REASON :
We know that
Velocity= Frequency× Wavelength
So,
Frequency= Velocity/wavelength
Here,
V= 3× 10^8 m/s
Wavelength= 2×10^-3 m
Hence,
Frequency= 3×10^8/2×10^-3
= 3/2 × 10^11
= 1.5× 10^11 Hz
HOPE IT IS USEFUL
Explanation:
The given data is as follows.
Mass, m = 75 g
Velocity, v = 600 m/s
As no external force is acting on the system in the horizontal line of motion. So, the equation will be as follows.
where,
= mass of the projectile
= mass of block
v = velocity after the impact
Now, putting the given values into the above formula as follows.
![75(10^{-3}) \times 600 = [(75 \times 10^{-3}) + 50] \times v](https://tex.z-dn.net/?f=75%2810%5E%7B-3%7D%29%20%5Ctimes%20600%20%3D%20%5B%2875%20%5Ctimes%2010%5E%7B-3%7D%29%20%2B%2050%5D%20%5Ctimes%20v)
= 
v = 0.898 m/s
Now, equation for energy is as follows.
E = 
= 
= 13500 J
Now, energy after the impact will be as follows.
E' = ^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5B75%20%5Ctimes%2010%5E%7B-3%7D%20%2B%2050%5D%280.9%29%5E%7B2%7D)
= 20.19 J
Therefore, energy lost will be calculated as follows.
= E E'
= (13500 - 20) J
= 13480 J
And, n = 
= 
= 99.85
= 99.9%
Thus, we can conclude that percentage n of the original system energy E is 99.9%.
Answer:
(iv), (v), (vi) would be incorrect.
Explanation:
(iv) Force isn't transferred from one colliding object to another, but momentum can be.
(v) An object doesn't stop immediately a force stops acting on it. Think of a thrown ball.
(vi) For an object not to move, it means that the net force on the object is zero, and not necessarily that there are no forces acting on the object. For example, an object could be pushed on one side, and be pushed on the other side with an equal force in the opposite direction. The forces would cancel each other and the net force would be zero.
The rest should be correct.
Answer:
0.625 c
Explanation:
Relative speed of a body may be defined as the speed of one body with respect to some other or the speed of one body in comparison to the speed of second body.
In the context,
The relative speed of body 2 with respect to body 1 can be expressed as :

Speed of rocket 1 with respect to rocket 2 :



Therefore, the speed of rocket 1 according to an observer on rocket 2 is 0.625 c