Answer:
(a) the work done by the student is 110.1 J
(b) The gravitational force that acts on the amplifier is 102.9 N
Explanation:
Given;
mass of the amplifier, m = 10.5 kg
initial position of the amplifier, x₀ = 1.82 m
final position of the amplifier, x₁ =0.75 m
The dispalcement of the amplifier Δx = x₁ - x₀ = 1.82 m - 0.75 m = 1.07 m
(b) The gravitational force that acts on the amplifier;
F = mg
F = 10.5 x 9.8
F = 102.9 N
(a) the work done by the student is calculated as;
W = FΔx
W = 102.9 x 1.07
W = 110.1 J
Answer:
The current is 
Explanation:
From the question we are told that
The radius is 
The current density is 
The distance we are considering is 
Generally current density is mathematically represented as

Where A is the cross-sectional area represented as

=> 
=> 
Now the change in current per unit length is mathematically evaluated as

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows


![I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.](https://tex.z-dn.net/?f=I%20%20%3D%202%5Cpi%20%2A%289.0%2A10%5E%7B6%7D%29%20%5B%5Cfrac%7Br%5E4%7D%7B4%7D%20%5D%20%20%7C%20%5Cleft%20%20%20%200.001585%7D%20%5Catop%200%7D%7D%20%5Cright.)
![I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ]](https://tex.z-dn.net/?f=I%20%20%3D%202%5Cpi%20%2A%289.0%2A10%5E%7B6%7D%29%20%5B%20%5Cfrac%7B0.001585%5E4%7D%7B4%7D%20%5D)
substituting values
![I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ]](https://tex.z-dn.net/?f=I%20%20%3D%202%20%2A%20%203.142%20%20%2A%20%209.00%20%2A10%5E6%20%2A%20%20%20%5B%20%5Cfrac%7B0.001585%5E4%7D%7B4%7D%20%5D)

Answer:
Im answering for free points sry
Explanation:
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