<h3><u>Answer</u>;</h3>
≈ 5 Kgm²/sec
<h3><u>Explanation</u>;</h3>
Angular momentum is given by the formula
L = Iω, where I is the moment of inertia and ω is the angular speed.
I = mr², where m is the mass and r is the radius
= 0.65 × 0.7²
= 0.3185
Angular speed, ω = v/r
= (2 × 3.142 × r × 2.5) r
= 15.71
Therefore;
Angular momentum = Iω
= 0.3185 × 15.71
= 5.003635
<u>≈ 5 Kgm²/sec</u>
Answer:
Yes it will move and a= 4.19m/s^2
Explanation:
In order for the box to move it needs to overcome the maximum static friction force
Max Static Friction = μFn(normal force)
plug in givens
Max Static friction = 31.9226
Since 36.6>31.9226, the box will move
Mass= Wieght/g which is 45.8/9.8= 4.67kg
Fnet = Fapp-Fk
= 36.6-16.9918
=19.6082
=ma
Solve for a=4.19m/s^2
Answer:

Explanation:
From the question we are told that
The electric filed is
Generally according to Gauss law
=> 
Given that the electric field is pointing downward , the equation become

Here
is the excess charge on the surface of the earth
is the surface area of the of the earth which is mathematically represented as

Where r is the radius of the earth which has a value 
substituting values


So

Here
s the permitivity of free space with value

substituting values


Radio waves are the longest