1)
The capacitance of a parallel-plate capacitor is given by:
where
is the vacuum permittivity
A is the area of each plate
d is the distance between the plates
Here, the radius of each plate is
so the area is
While the separation between the plates is
So the capacitance is
And now we can find the energy stored,which is given by:
2) 0.71 J/m^3
The magnitude of the electric field is given by
and the energy density of the electric field is given by
and using
, we find
Panel surface area =34m×46m=1,564m^2
total power =1564m^2×1390w/m^2
=2173960watts
now you must calculate total energy.
Energy = power×Time
However time must be in seconds so we multiply 2hrs×60min×60s=7200seconds
7200s×2173960w =15,652,512,000 joules of energy
Answer:
11.8 m/s
Explanation:
At the top of the hill, there are two forces on the car: weight force pulling down (towards the center of the circle), and normal force pushing up (away from the center of the circle).
Sum of forces in the centripetal direction:
∑F = ma
mg − N = m v²/r
At the maximum speed, the normal force is 0.
mg = m v²/r
g = v²/r
v = √(gr)
v = √(9.8 m/s² × 14.2 m)
v = 11.8 m/s
Answer: a) io=233.28 A ( initial current); b) τ=R*C= 22.31 ms; c) 81.7 ms
Explanation: In order to explain this problem we have to use, the formule for the variation of the current in a RC circuit:
I(t)=io*Exp(-t/τ)
and also we consider that io=V/R=(1.5/6.43*10^3)
=233.28 A
then the time constant for the RC circuit is τ=R*C=6.43*10^3*3.47*10^-6
=22.31 ms
Finally the time to reduce the current to 2.57% of its initial value is obtained from:
I(t)=io*Exp(-t/τ) for I(t)/io=0.0257=Exp(-t/τ) then
ln(0.0257)*τ =-t
t=-ln(0.0257)*τ=81.68 ms
The answer would be (A) Protons