Explanation :
It is given that,
Mass of the car, m = 1000 kg
Force applied by the motor, 
The static and dynamic friction coefficient is, 
Let a is the acceleration of the car. Since, the car is in motion, the coefficient of sliding friction can be used. At equilibrium,




So, the acceleration of the car is
. Hence, this is the required solution.
Answer:
wen you stick to mangnetits togater
Explanation:
Answer:
i. The radius 'r' of the electron's path is 4.23 ×
m.
ii. The frequency 'f' of the motion is 455.44 KHz.
Explanation:
The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.
r = 
Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.
From the question, B = 1.63 ×
T, v = 121 m/s, Θ =
(since it enters perpendicularly to the field), q = e = 1.6 ×
C and m = 9.11 ×
Kg.
Thus,
r =
÷ sinΘ
But, sinΘ = sin
= 1.
So that;
r = 
= (9.11 ×
× 121) ÷ (1.6 ×
× 1.63 ×
)
= 1.10231 ×
÷ 2.608 × 
= 4.2266 ×
= 4.23 ×
m
The radius 'r' of the electron's path is 4.23 ×
m.
B. The frequency 'f' of the motion is called cyclotron frequency;
f = 
= (1.6 ×
× 1.63 ×
) ÷ (2 ×
× 9.11 ×
)
= 2.608 ×
÷ 5.7263 × 
= 455442.4323
f = 455.44 KHz
The frequency 'f' of the motion is 455.44 KHz.
Explanation:
It is given that,
An electron is released from rest in a weak electric field of, 
Vertical distance covered, 
We need to find the speed of the electron. Let its speed is v. Using third equation of motion as :

.............(1)
Electric force is
and force of gravity is
. As both forces are acting in downward direction. So, total force is:



Acceleration of the electron, 


Put the value of a in equation (1) as :


v = 0.010 m/s
So, the speed of the electron is 0.010 m/s. Hence, this is the required solution.