The minimum potential difference must be supplied by the ignition circuit to start a car is -1800 V
<u>Explanation:</u>
Given data,
E= 3 ×10 ⁶ Δx=0.06/100
We have to find the minimum potential difference
E= -ΔV/Δx
ΔV=- E × Δx
ΔV =-3 ×10 ⁶ . 0.06/100
ΔV=-1800 V
The minimum potential difference must be supplied by the ignition circuit to start a car is -1800 V
Answer:
a charge Q is transferred from an initially uncharged
Explanation:
Hope this helps!
(a) Let's convert the final speed of the car in m/s:

The kinetic energy of the car at t=19 s is

(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:

But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into

(c) The instantaneous power is given by

where F is the force exerted by the engine, equal to F=ma.
So we need to find the acceleration first:

And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
The strength of the gravitational forces between two masses depends on
-- the product of the masses,
-- the distance between their centers of mass.
C) Acceleration is the rate of change of velocity of an object. Velocity is the speed and direction of an object so acceleration is used to describe the rate of change. I hope this helps!!