The correct answer is 40.5 Newtons just finished the quiz and 36.5 was incorrect.
Answer:

Explanation:
Since
, we calculate the resistance rate by deriving this formula with respect to time:

Deriving what is left (remember that
):

So we have:

Which for our values is (the rate of <em>I(t)</em> is decreasing so we put a negative sign):

Answer:
Electric potential = 0.00054 V
Explanation:
We are given;
Charge; q = 3 pC = 3 × 10^(-12) C
Radius; r = 2 cm = 0.02 m
Formula for the electric potential of this surface will be;
V = kqr
Where;
K is a constant = 9 × 10^(9) N⋅m²/C².
Thus;
V = 9 × 10^(9) × 3 × 10^(-12) × 0.02
V = 0.00054 V
Given:
m = 4 kg, the mass of the object
h = 5 m, distance fallen
Neglect air resistance.
The PE (potential energy) is
PE = mgh = (4 kg)*(9.8 m/s²)*(5 m) = 196 J
The PE is converted into KE (kinetic energy) after the fall.
Therefore the PE decreased by 196 J ≈ 200 J
Answer: d. It has decreased by 200 J