Answer:
Q = 4 Q₀
Explanation:
This is an exercise on capacitors, where the capacitance is
C =
if we apply the given conditions
C = \epsilon_{o} \ \frac{2A}{0.5d}
C = 4 \epsilon_{o} \ \frac{A}{d}
let's call the capacitance Co with the initial values
C₀ = \epsilon_{o} \ \frac{A}{d}
C = 4 C₀
The charge on each plate of a capacitor is
Q = C ΔV
If the potential difference is maintained, the new charge is
Q = 4 C₀ ΔV
let's call
Q₀ = C₀ ΔV
we substitute
Q = 4 Q₀
A car driving up a hill at a constant speed experiences no change in its kinetic energy while it's potential energy increases with increasing height, thus none of the options are correct.
Understanding the concept
Consider a car moving up the hill at a constant speed as shown in the figure below. The following forces act on the car:
Here as the car is moving up the hill at a constant speed, the net force exerted on the car is zero. Also, the kinetic energy of the car will not change as its velocity is constant and the potential energy will change with increasing height. Thus, none of the given options are correct.
Learn more about motion on an incline here:
<u>brainly.com/question/13513083</u>
#SPJ4
PART a)
As we know that gravitational potential energy is given by the formula
here we can see that gravitational potential energy inversely varies with the distance
so here when distance from the sun is minimum then magnitude of gravitational potential energy is maximum while since it is given with negative sign so its overall value is minimum at that position
So gravitational potential energy is minimum at the nearest point and maximum at the farthest point
PART b)
Since we know that sum of kinetic energy and potential energy is constant here
so the points of minimum potential energy is the point where kinetic energy is maximum which means speed is maximum
So here speed is maximum at the nearest point
Part C)
since gravitational potential energy inversely varies with distance so it's graph will be like hyperbolic graph with distance
