Answer:
Explanation:
Let the velocity after first collision be v₁ and v₂ of car A and B . car A will bounce back .
velocity of approach = 1.5 - 0 = 1.5
velocity of separation = v₁ + v₂
coefficient of restitution = velocity of separation / velocity of approach
.8 = v₁ + v₂ / 1.5
v₁ + v₂ = 1.2
applying law of conservation of momentum
m x 1.5 + 0 = mv₂ - mv₁
1.5 = v₂ - v₁
adding two equation
2 v ₂= 2.7
v₂ = 1.35 m /s
v₁ = - .15 m / s
During second collision , B will collide with stationary A . Same process will apply in this case also. Let velocity of B and A after collision be v₃ and v₄.
For second collision ,
coefficient of restitution = velocity of separation / velocity of approach
.5 = v₃ + v₄ / 1.35
v₃ + v₄ = .675
applying law of conservation of momentum
m x 1.35 + 0 = mv₄ - mv₃
1.35 = v₄ - v₃
adding two equation
2 v ₄= 2.025
v₄ = 1.0125 m /s
v₃ = - 0 .3375 m / s
Answer:
There are 756.25 electrons present on each sphere.
Explanation:
Given that,
The force of repression between electrons, 
Let the distance between charges, d = 0.2 m
The electric force of repulsion between the electrons is given by :




Let n are the number of excess electrons present on each sphere. It can be calculated using quantization of charges. It is given by :
q = ne


n = 756.25 electrons
So, there are 756.25 electrons present on each sphere. Hence, this is the required solution.
Answer:
0.2 m
Explanation:
PHASE 1
First, we calculate the distance the tongue moved in the first 20 ms (0.02 secs). We use one of Newton's equations of linear motion:

where u = initial velocity = 0 m/s
a = acceleration = 
t = time = 0.02 s
Therefore:

PHASE 2
Then, for the next 30 ms (0.03 secs), we use the formula:

This speed is the same as the final velocity of the tongue after the first 20 ms.
This can be obtained by using the formula:

Therefore:
distance = 5 * 0.03 = 0.15 m
Therefore, the total distance moved by the tongue in the 50 ms interval is:
0.05 + 0.15 = 0.2 m