Below are the choices that can be found from other source:
a. 0.3 AU
<span>b. 3.8 AU </span>
<span>c. 30 AU </span>
<span>d. 3 million AU
</span>
Answer is c. 30 AU.
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Answer:
a) Energy stored in the capacitor, 
b) Q = 45 µC
c) C' = 1.5 μF
d) 
Explanation:
Capacitance, C = 1 µF
Charge on the plates, Q = 45 µC
a) Energy stored in the capacitor is given by the formula:
b) The charge on the plates of the capacitor will not change
It will still remains, Q = 45 µC
c) Electric field is non zero over (1-1/3) = 2/3 of d
From the relation V = Ed,
The voltage has changed by a factor of 2/3
Since the capacitance is given as C = Q/V
The new capacitance with the conductor in place, C' = (3/2) C
C' = (3/2) * 1μF
C' = 1.5 μF
d) Energy stored in the capacitor with the conductor in place
When resistance is constant, current is proportional to voltage. When 1/3 the voltage is applied, 1/3 the current will result.
(1/3)*(1.2 A) = 0.4 A
The resulting current will be 0.4 A.
Answer:
250,000 Joules
Explanation:
The work required is the amount of energy added, so the answer is the difference in kinetic energy before and after the acceleration of the car. (The problem doesn't say it's flat where the car is, but well assume it since there is no information about an elevation change that would require us to concern ourselves with potential energy!)
W = E_final - E_initial
= 1/2 m v_final^2 - 1/2 m v_initial^2
= 1/2 m (v_final^2 - v_initial^2)
= 1/2 (1000 kg) ( (30 m/s)^2 - (20 m/s^2) )
= 1/2 (1000 kg) (500 m^2/s^2)
= 250,000 kg m^2/s^2
= 250,000 Joules
