Answer: It hits the brakes and the truck now has only 20000J of kinetic energy
Explanation:
Hope it help you
A. something that happens when a force is applied to an object and its moved
(however all 3 answers are arguebly correct)
Answer:

Explanation:
As we know that the energy stored in the capacitor is given as

here we know that

also we know that

now we have


now we know the formula of capacitance



Answer:
The points 2 and 4 should be connected.
Explanation:
To complete the circuit, we need to connect the two points which when connected, encompass the battery and the bulb in the circuit. The points 2 and 4 do the job, since they connect the terminal of the battery and the terminal of the bulb, and thus complete the circuit.
Therefore, the choice C is correct.
Answer:
The magnitude of the electric field between the plates is half its initial value.
Explanation:
We know the electric field E = V/d where V = voltage applied and d = separation between plates.
Since V is constant and V = Ed,
So, E₁d₁ = E₂d₂ where E₁ = initial electric field at separation d₁, d₁ = initial separation of plates, E₂ = final electric field at separation d₂ and d₂ = final separation of plates.
So, E₂ = E₁d₁/d₂
Now, the distance between the plates is twice their original separation. Thus, d₂ = 2d₁
So, E₂ = E₁d₁/2d₁ = E₁/2
So, E₂ = E₁/2
Thus, the magnitude of the electric field between the plates is half its initial value.