The concept required to perform this exercise is given by the coulomb law.
The force expressed according to this law is given by

Where,

q = charges of the objects
r= distance/radius
Our values are previously given, so

Replacing,



The force acting on the block are given by,




Therefore the box is accelerated upward.
Answer:
The vertical distance is ![d = \frac{2}{k} *[mg + f]](https://tex.z-dn.net/?f=d%20%3D%20%5Cfrac%7B2%7D%7Bk%7D%20%2A%5Bmg%20%2B%20f%5D)
Explanation:
From the question we are told that
The mass of the cylinder is m
The kinetic frictional force is f
Generally from the work energy theorem

Here E the the energy of the spring which is increasing and this is mathematically represented as

Here k is the spring constant
P is the potential energy of the cylinder which is mathematically represented as

And
is the workdone by friction which is mathematically represented as

So

=> ![\frac{1}{2} * k * d^2 = d[mg + f ]](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%20%2A%20k%20%20%2A%20%20d%5E2%20%3D%20%20d%5Bmg%20%2B%20%20f%20%20%20%20%5D)
=> ![\frac{1}{2} * k * d = [mg + f ]](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%20%2A%20k%20%20%2A%20%20d%20%3D%20%20%5Bmg%20%2B%20%20f%20%20%20%20%5D)
=> ![d = \frac{2}{k} *[mg + f]](https://tex.z-dn.net/?f=d%20%3D%20%5Cfrac%7B2%7D%7Bk%7D%20%2A%5Bmg%20%2B%20f%5D)
Answer:
80% (Eighty percent)
Explanation:
The material has a refractive index (n) of 1.25
Speed of light in a vacuum (c) is 2.99792458 x 10⁸ m/s
We can find the speed of light in the material (v) using the relationship
n = c/v, similarly
v = c/n
therefore v = 2.99792458 x 10⁸ m/s ÷ (1.25) = 239 833 966 m/s
v = 239 833 966 m/s
Therefore the percentage of the speed of light in a vacuum that is the speed of light in the material can be calculated as
(v/c) × 100 = (1/n) × 100 = (1/1.25) × 100 = 0.8 × 100 = 80%
Therefore speed of light in the material (v) is eighty percent of the speed of light in the vacuum (c)
Momentum question. This is an inelastic collision, so
m1v1+m2v2=Vf(m1+m2)
Vf=(m1v1+m2v2)/(m1+m2)=[(120kg)(0m/s)+(60kg)(2m/s)] / (120kg+60kg)
Vf=120kg m/s / 180kg
Vf=0.67m/s
0.67m/s
<span>Answer:
For a disc, the moment of inertia about the perpendicular axis through the center is given by 0.5MR^2.
where M is the mass of the disc and R is the radius of the disc.
For the axis through the edge, use parallel axis theorem.
I = I(axis through center of mass) + M(distance between the axes)^2
= 0.5MR^2 + MR^2 (since the axis through center of mass is the axis through the center)
= 1.5 MR^2</span>