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:
The distance of stars and the earth can be averagely measured by using the knowledge of geometry to estimate the stellar parallax angle(p).
From the equation below, the stars distances can be calculated.
D = 1/p
Distance = 1/(parallax angle)
Stellar parallax can be used to determine the distance of stars from an observer, on the surface of the earth due to the motion of the observer. It is the relative or apparent angular displacement of the star, due to the displacement of the observer.
Explanation:
Parallax is the observed apparent change in the position of an object resulting from a change in the position of the observer. Specifically, in the case of astronomy it refers to the apparent displacement of a nearby star as seen from an observer on Earth.
The parallax of an object can be used to approximate the distance to an object using the formula:
D = 1/p
Where p is the parallax angle observed using geometry and D is the actual distance measured in parsecs. A parsec is defined as the distance at which an object has a parallax of 1 arcsecond. This distance is approximately 3.26 light years
//////Correct answer is C.///////
Answer:
Time taken by A and B is 1.2 hr.
Explanation:
Given that
Time taken by tank when all(A+B+C) are open = 1 hr
Time taken by tank when A+C are open = 1.5 hr
Time taken by tank when B+C are open = 2 hr
If we treat as filling of tank is a work then
Work = time x rate
Lets take work is 1 unit
1 = 1(1/a+1/b+1/c) ---------1
1 = 1.5(1/a+1/c) ----------2
1 = 2(1/b+1/c) --------3
From equation 1 and 3
1=1(1/a+1/2)
a=2
Form equation 2
1 = 1.5(1/2+1/c)
c=6
From equation 3
1 = 2(1/b+1/6)
b=3
So time taken by
A is alone to fill tank is 2 hr
B is alone to fill tank is 3 hr
C is alone to fill tank is 6 hr
So 
Time taken by A and B is 1.2 hr.
C is correct. Only a closed circuit can carry current, opening a switch disconnects the circuit and stops current from flowing. Remember, all current flows in loops, if you can't make a loop out of it, current can't flow, and opening a switch, assuming no other paths for current exist, will completely stop the current.
