A battery-operated car utilizes a 12.0 V system. Find the charge the batteries must be able to move in order to accelerate the 7
1 answer:
Answer:
3894531 coulombs
Explanation:
1 hour = 3600 seconds
Let g = 10m/s2
The distance that the car travel at 25 m/s over an hour is
s = 25 * 3600 = 90000 m
The total mechanical energy of the car is the sum of its kinetic energy to reach 25 m/s, its potential energy to climb up 200m high hill and it work to travel a distance of s = 90000m with F = 500 N force:
This energy is drawn from the battery over an hour (3600 seconds), so its power must be
The system is 12V so its current is
or 1081.8 Coulombs/s
The the total charge it needs for 1 hour (3600 s) is
C = 1081.8 * 3600 = 3894531 coulombs
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Part A)
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The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
from which
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
And so, the energy density at r=0.111 m is
Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor:
. We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
And therefore, the energy density at this distance from the center is
First of all, let's convert the radii of the inner and the outer sphere:
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
from which
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
And so, the energy density at r=0.111 m is
Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor:
And therefore, the energy density at this distance from the center is