Answer:
25.71 kgm/s
Explanation:
Let K₁ and K₂ be the initial and final kinetic energies of object A and v₁ and v₂ its initial and final speeds.
Given that K₂ = 0.7K₁
1/2mv₂² = 0.7(1/2mv₁²)
v₂ = √0.7v₁ = √0.7 × 20 m/s = ±16.73 m/s
Since A rebounds, its velocity = -16.73 m/s and its momentum change, p₂ = mΔv = m(v₂ - v₁) = 0.7 kg (-16.73 - 20) m/s = 0.7( -36.73) = -25.71 kgm/s.
Th magnitude of object A's momentum change is thus 25.71 kgm/s
The value of the second charge is 1.2 nC.
<h3>
Electric potential</h3>
The work done in moving the charge from infinity to the given position is calculated as follows;
W = Eq₂
E = W/q₂
<h3>Magnitude of second charge</h3>
The magnitude of the second charge is determined by applying Coulomb's law.

Thus, the value of the second charge is 1.2 nC.
Learn more about electric potential here: brainly.com/question/14306881
Correct answer is letter B. sandstone
Answer:
The magnitude of angular acceleration is
.
Explanation:
Given that,
Initial angular velocity, 
When it switched off, it comes o rest, 
Number of revolution, 
We need to find the magnitude of angular acceleration. It can be calculated using third equation of rotational kinematics as :
So, the magnitude of angular acceleration is
. Hence, this is the required solution.
Answer:
a

b

c
Explanation:
From the question we are told that
The angle of incidence is 
The refractive index of water is 
Generally Snell's law is mathematically represented as

Here
is the refractive index of air with value 
is the angle of refraction
So
![\theta _2 = sin^{-1}[\frac{n_1 * sin(\theta _1)}{n_2} ]](https://tex.z-dn.net/?f=%5Ctheta%20_2%20%20%3D%20%20sin%5E%7B-1%7D%5B%5Cfrac%7Bn_1%20%2A%20sin%28%5Ctheta%20_1%29%7D%7Bn_2%7D%20%5D)
=> ![\theta _2 = sin^{-1}[\frac{1.3 * sin(10)}{1} ]](https://tex.z-dn.net/?f=%5Ctheta%20_2%20%20%3D%20%20sin%5E%7B-1%7D%5B%5Cfrac%7B1.3%20%2A%20sin%2810%29%7D%7B1%7D%20%5D)
=> 
Given that the angle should not be greater than
then the angle of incidence will be
![\theta _1 = sin^{-1}[\frac{n_2 * sin(\theta _2)}{n_1} ]](https://tex.z-dn.net/?f=%5Ctheta%20_1%20%20%3D%20%20sin%5E%7B-1%7D%5B%5Cfrac%7Bn_2%20%2A%20sin%28%5Ctheta%20_2%29%7D%7Bn_1%7D%20%5D)
=> ![\theta _1 = sin^{-1}[\frac{1 * sin(45)}{1.3} ]](https://tex.z-dn.net/?f=%5Ctheta%20_1%20%20%3D%20%20sin%5E%7B-1%7D%5B%5Cfrac%7B1%20%2A%20sin%2845%29%7D%7B1.3%7D%20%5D)
=> 
Generally for critical angle is mathematically represented as
![\theta_c = sin^{-1}[\frac{n_2}{n_1} ]](https://tex.z-dn.net/?f=%5Ctheta_c%20%20%3D%20%20sin%5E%7B-1%7D%5B%5Cfrac%7Bn_2%7D%7Bn_1%7D%20%5D)
=>
=>