Assuming 95.0% efficiency for the conversion of electrical power by the motor, what current must the 12.0-V batteries of a 708-kg electric car be able to supply to do the following?

(a) accelerate from rest to 25.0 m/s in 1.00 min
1 (answer in Amps)

(b) climb a 200-m high hill in 2.00 min at a constant 25.0 m/s speed while exerting 423 N of force to overcome air resistance and friction

(a). The current is 323.4 A.

(b). The current is 1942 A.

Explanation:

Given that,

Efficiency = 95.0 %

Voltage = 12.0 V

Mass of electric car= 708 Kg

Height = 200 m

We need to calculate the change in kinetic energy

Using formula of kinetic energy

Put the value into the formula

We need to calculate the output power

Using formula of power

We need to calculate the current

Using formula of electric power

Put the value into the formula

The current is 323.4 A.

(b). We need to calculate the distance

Put the value into the formula

We need to calculate the force

Using formula of force

Put the value into the formula

We need to calculate the power

Using formula of power

Put the value into the formula

We need to calculate the current

Using formula of current

Put the value into the formula

The current is 1942 A.

Hence, (a). The current is 323.4 A.

(b). The current is 1942 A.

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg