Mechanical waves transfer energy by inducing vibrations in the propagation medium.
I think jogging/running/walking because you don't need any equipment and you can structure around it on your own time.
For this problem, we would be using the formula: Vf^2 = Vi^2 + 2ad
where:
Vf = 400m/s
Vi = 300m/s
a = ?
d = 4.0km
= 4000m
400^2 = 300^2 + 2a4000
a = [ 160000 - 90000 ] / 8000
a = 8.75m/s^2
rounding it off to 2 significant figures, will give us 8.8 m/s^2.
The charge of the object must be 
Answer: Option C
<u>Explanation:</u>
Suppose an electric charge can be represented by the symbol Q. This electric charge generates an electric field; Because Q is the source of the electric field, we call this as source charge. The electric field strength of the source charge can be measured with any other charge anywhere in the area. The test charges used to test the field strength.
Its quantity indicated by the symbol q. In the electric field, q exerts an electric, either attractive or repulsive force. As usual, this force is indicated by the symbol F. The electric field’s magnitude is simply defined as the force per charge (q) on Q.

Here, given E = 4500 N/C and F = 0.05 N.
We need to find charge of the object (q)
By substituting the given values, we get

Explanation:
It is given that,
Magnitude of charge, 
It moves in northeast direction with a speed of 5 m/s, 25 degrees East of a magnetic field.
Magnetic field, 
Velocity, 
![v=[(4.53)i+(2.11)j]\ m/s](https://tex.z-dn.net/?f=v%3D%5B%284.53%29i%2B%282.11%29j%5D%5C%20m%2Fs)
We need to find the magnitude of force on the charge. Magnetic force is given by :

![F=15\times 10^{-6}[(4.53i+2.11j)\times 0.08\ j]](https://tex.z-dn.net/?f=F%3D15%5Ctimes%2010%5E%7B-6%7D%5B%284.53i%2B2.11j%29%5Ctimes%200.08%5C%20j%5D)
<em>Since</em>, 
![F=15\times 10^{-6}[(4.53i)\times (0.08)\ j]](https://tex.z-dn.net/?f=F%3D15%5Ctimes%2010%5E%7B-6%7D%5B%284.53i%29%5Ctimes%20%280.08%29%5C%20j%5D)


So, the force acting on the charge is
and is moving in positive z axis. Hence, this is the required solution.