Question

Assuming 84.0% efficiency for the conversion of electrical power by the motor, what current must the 13.0-V batteries of a 716 kg electric car be able to supply to climb a 3.00 x 102 m high hill in 2.00 min at a constant 22.0 m/s speed while exerting 7.00 x 102 N of force to overcome air resistance and friction

Answer 1

Answer:

$\mathbf{ current(I) =1766.67 \ A}$

Explanation:

Given that:

The air resistance and friction = 700 N

The gravity caused force = 716 × 9.8 = 7016.8

Total force = (7016.8 + 700) N

Total force = 7716.8 N

∴

$13 \times current(I) \times 0.84 = \dfrac{7716.8 \times 300}{2 \times 60}$

$current(I) \times 10.92= 19292$

$current(I) = \dfrac{19292}{10.92}$

$\mathbf{ current(I) =1766.67 \ A}$