Answer:

Negative sign shows that velocity of the car is decreases at a constant rate
Explanation:
We have given velocity of the car is decreases from 32 m /sec to 24 m/sec in 4 second
So initial velocity of the car u = 32 m /sec
And finally car reaches to a velocity of 24 m/sec
Time taken to change in velocity = 4 sec
So final velocity v = 24 m/sec
From first equation of motion v = u+at
So 

Negative sign shows that velocity of the car is decreases at a constant rate
Answer:
Explanation:
Maximum force of friction possible = μmg
= .65 x 3.8 x 9.8
= 24.2 N
u = 72 x 1000 / 60 x 60
= 20 m /s
v² = u² - 2as
a = 20 x 20 / (2 x 30)
= 6.67 m / s²
force acting on it
= 3.8 x 6.67
= 25.346 N
Friction force possible is less .
So friction will not be able to prevent its slippage
It will slip off .
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:
