Complete Question
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.
I = 1.2 A at time 5 secs.
Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.
Answer:
The charge is 
Explanation:
From the question we are told that
The diameter of the wire is 
The radius of the wire is 
The resistivity of aluminum is 
The electric field change is mathematically defied as

Generally the charge is mathematically represented as

Where A is the area which is mathematically represented as

So

Therefore

substituting values
![Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5Cint%5Climits%5E%7Bt%7D_%7B0%7D%20%7B%20%5B%200.0004t%5E2%20-%200.0001t%20%2B0.0004%5D%20%7D%20%5C%2C%20dt)
![Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004t%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20t%5E2%7D%7B2%7D%20%2B0.0004t%5D%20%7D%20%20%5Cleft%20%7C%20t%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
From the question we are told that t = 5 sec
![Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004t%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20t%5E2%7D%7B2%7D%20%2B0.0004t%5D%20%7D%20%20%5Cleft%20%7C%205%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
![Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004%285%29%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20%285%29%5E2%7D%7B2%7D%20%2B0.0004%285%29%5D%20%7D)

Answer:

Explanation:
During the exchange of applied force, thermal energy is generated by the friction that exists between the ground and the tire.
Said force according to the statement is the reaction of half the force on the rear tire. In this way the normal force acted is,

The work done is given by the friction force and the distance traveled,

Where ![\mu_k [/ tex] is the coefficient of kinetic frictionN is the normal force previously found d is the distance traveled,Replacing,[tex]W_f = (0.80)(441)(0.42)](https://tex.z-dn.net/?f=%20%5Cmu_k%20%5B%2F%20tex%5D%20is%20the%20coefficient%20of%20kinetic%20friction%3C%2Fp%3E%3Cp%3EN%20is%20the%20normal%20force%20previously%20found%20d%20is%20the%20distance%20traveled%2C%3C%2Fp%3E%3Cp%3EReplacing%2C%3C%2Fp%3E%3Cp%3E%5Btex%5DW_f%20%3D%20%280.80%29%28441%29%280.42%29)
The thermal energy released through the work done is,

Answer:
Acceleration generate by punk = 3 m/s²
Explanation:
Given:
Weight of punk = 100 Kg
Force applied on punk = 300 N
Find:
Acceleration generate by punk = ?
Computation:
Acceleration = Force / Mass
Acceleration generate by punk = Force applied on punk / Weight of punk
Acceleration generate by punk = 300 N / 100 Kg
Acceleration generate by punk = 3 m/s²
Answer: -2.5
Explanation:
1/2(-5)= -2.5
-2.5(1)= -2.5
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Answer:

Explanation:
The computation of the weight of the paper in newtons is shown below:
On the paper, the induced charge is of the same magnitude as on the initial charges and in sign opposite.
Therefore the paper charge is

Now the distance from the charge is

Now, to raise the paper, the weight of the paper acting downwards needs to be managed by the electrostatic force of attraction between both the paper and the charge, i.e.



