Answer:
energy is converted into mass
Explanation:
Answer:
The work done by the applied force is 259.22 J.
Explanation:
The work done by the applied force is given by:

Where:
F: is the applied horizontal force = 108.915 N
d: is the distance = 2.38 m
Hence, the work is:

Therefore, the work done by the applied force is 259.22 J.
I hope it helps you!
metamorphic, sedimentary, igneous
Car X traveled 3d distance in t time. Car Y traveled 2d distance in t time. Therefore, the speed of car X, is 3d/t, the speed of car Y, is 2d/t. Since speed is the distance taken in a given time.
In figure-2, they are at the same place, we are asked to find car Y's position when car X is at line-A. We can calculate the time car X needs to travel to there. Let's say that car X reaches line-A in t' time.

Okay, it takes t time for car X to reach line-A. Let's see how far does car Y goes.

We found that car Y travels 2d distance. So, when car X reaches line-A, car Y is just a d distance behind car X.
A) d. 10T
When a charged particle moves at right angle to a uniform magnetic field, it experiences a force whose magnitude os given by

where q is the charge of the particle, v is the velocity, B is the strength of the magnetic field.
This force acts as a centripetal force, keeping the particle in a circular motion - so we can write

which can be rewritten as

The velocity can be rewritten as the ratio between the lenght of the circumference and the period of revolution (T):

So, we get:

We see that this the period of revolution is directly proportional to the mass of the particle: therefore, if the second particle is 10 times as massive, then its period will be 10 times longer.
B) 
The frequency of revolution of a particle in uniform circular motion is

where
f is the frequency
T is the period
We see that the frequency is inversely proportional to the period. Therefore, if the period of the more massive particle is 10 times that of the smaller particle:
T' = 10 T
Then its frequency of revolution will be:
