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 ratio is 
Explanation:
Given
The RMS velocity of molecules in a gas is given by
where T=temperature
For T = 387K
For T = 774
dividing eqn 1 and eqn 2
Thus,the ratio is
Answer:
5308.34 N/C
Explanation:
Given:
Surface density of each plate (σ) = 47.0 nC/m² = 
Separation between the plates (d) = 2.20 cm
We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:
Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,
Now, plug in for 'σ' and 
for 
and solve for the electric field. This gives,
Therefore, the electric field between the plates has a magnitude of 5308.34 N/C
Answer:
the direction of acceleration of the vehicle is the same direction of its velocity of car
s acceleration has the opposite direction to the car speed.
Explanation:
The initial acceleration of the car can be calculated with
v = v₀ + a t
a = (v-v₀) t
indicate that the initial velocity is zero (v₀ = 0 m / s)
a = v / t
a = 300 / t
the direction of acceleration of the vehicle is the same direction of its acceleration movement.
When the car collides with the wall, it exerts a force in the opposite direction that stops the vehicle, therefore this acceleration has the opposite direction to the car speed. But your module must be much larger since the distance traveled to stop is small
If the mass of the cylinder increases, the temperature of the water increases, because a greater mass means the cylinder has more potential energy that can be converted to thermal energy, increasing the temperature of the water.
