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

Answer:(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 energyUsing formula of kinetic energyPut the value into the formula

We need to calculate the output powerUsing formula of powerWe need to calculate the currentUsing formula of electric powerPut the value into the formula

The current is 323.4 A.(b). We need to calculate the distancePut the value into the formula

We need to calculate the forceUsing formula of forcePut the value into the formula

We need to calculate the powerUsing formula of powerPut the value into the formula

We need to calculate the currentUsing formula of currentPut 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:a)

b)

Explanation:Given:

a)

initial velocity,

final velocity,

time taken for the acceleration,

Now we know by the Newton’s second law of motion:

Now the power will be :

According to the question:0.95 times of the electrical power should yield this mechanical power.

b)

height climbed by the car,

velocity of climb,

time taken to climb the height,

force exerted to overcome air and frictional resistances,

Now the Power required to climb the hill:

Now according to the electrical efficiency: