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)

<em>There are some placeholders in the expression, but they can be safely assumed</em>
Answer:
(a) 
(b) 
(c) 
(d) 
Explanation:
<u>Sinusoidal Waves
</u>
An oscillating wave can be expressed as a sinusoidal function as follows

Where



The voltage of the question is the sinusoid expression

(a) By comparing with the general formula we have


(b) The period is the reciprocal of the frequency:


Converting to milliseconds

(c) The amplitude is

(d) Phase angle:

Answer:
The acceleration of the ball is 4.18 [m/s^2]
Explanation:
By Newton's second law we can find the acceleration of the ball
![F = m*a\\where:\\F = force applied [N] or [kg*m/s^2]\\m = mass of the ball [kg]\\a = acceleration [m/s^s]](https://tex.z-dn.net/?f=F%20%3D%20m%2Aa%5C%5Cwhere%3A%5C%5CF%20%3D%20force%20applied%20%5BN%5D%20or%20%5Bkg%2Am%2Fs%5E2%5D%5C%5Cm%20%3D%20mass%20of%20the%20ball%20%5Bkg%5D%5C%5Ca%20%3D%20acceleration%20%5Bm%2Fs%5Es%5D)
Now we have:
![a = F/m\\a = \frac{1.8 [kg*m/s^s]}{0.43[kg]} \\a = 4.18 [kg]](https://tex.z-dn.net/?f=a%20%3D%20F%2Fm%5C%5Ca%20%3D%20%5Cfrac%7B1.8%20%5Bkg%2Am%2Fs%5Es%5D%7D%7B0.43%5Bkg%5D%7D%20%5C%5Ca%20%3D%204.18%20%5Bkg%5D)
Answer:
A nuclear winter is a climatic phenomenon that would follow the detonation of several atomic bombs in the event that a nuclear war broke out. These bombs would cause firestorms that would raise smoke, dust and particles into the atmosphere that would end up in the stratosphere and eventually spread throughout the globe.
Explanation:
That idea is far fetched, because even though those same particles would absorb sunlight, it would raise the temperature in the stratosphere and cause a decrease in temperature in the Earth's layer. Unable to seep the sun's rays, many plant species would die and this would affect the entire food chain.
In addition, that temperature rise in the stratosphere would destroy part of the ozone layer, causing greater exposure to ultraviolet rays. This would end up affecting health and further damaging plant species.
Answer:
Acceleration is 0.25m/s^2
Explanation:
Given the following :
Speed = 0.5m/s
Radius(r) of circle = 1m
Acceleration round a circular path is given as :
a = v^2 / r
Where
a = acceleration of the body
v = speed / velocity
r = radius
Therefore,
a = v^2 / r
a = (0.5)^2 / 1
a = 0.25m/s^2