Assuming 95.0% efficiency for the conversion of electrical power by the motor, what current must the 12.0 V batteries of a 740 kg electric car be able to supply:

(a) To accelerate from rest to 25.0 m/s in 1.00 min?

(b) To climb a 200 m high hill in 2.00 min at a constant 25.0 m/s speed while exerting 475 N of force to overcome air resistance and friction?

(c) To travel at a constant 25.0 m/s speed, exerting a 475 N force to overcome air resistance and friction?

To find the current required by the electric car's batteries, we need to use the power equation P = IV, where P is power, I is current, and V is voltage.

First, we need to calculate the power required for each scenario:

(a) To accelerate from rest to 25.0 m/s in 1.00 min:
We need to find the work done to accelerate the car using the formula W = (1/2)mv^2, where W is work, m is mass, and v is velocity.
Given:
m = 740 kg,
v = 25.0 m/s

W = (1/2)(740 kg)(25.0 m/s)^2
= 231,250 J (Joules)

Since we are given the time taken in minutes, we need to convert it to seconds. 1 min = 60 s.

Time, t = 1.00 min = 1.00 x 60 s
t = 60 s

To find power, we divide the work by the time:
Power, P = W / t
P = 231,250 J / 60 s
P ≈ 3854.17 W

Since the motor has an efficiency of 95.0%, the actual power needed is higher, given by:
Actual Power = Power / Efficiency
Actual Power = 3854.17 W / 0.95
Actual Power ≈ 4051.34 W

Now we can use the power equation to find the current:
Current, I = Actual Power / Voltage
Voltage, V = 12.0 V (given)

I = 4051.34 W / 12.0 V
I ≈ 337.61 A

Therefore, the current the 12.0 V batteries of the electric car must be able to supply to accelerate from rest to 25.0 m/s in 1.00 min is approximately 337.61 A.

(b) To climb a 200 m high hill in 2.00 min at a constant 25.0 m/s speed while exerting 475 N of force to overcome air resistance and friction:
To calculate power, we use the equation P = F * v, where F is force and v is velocity.
Given:
F = 475 N
v = 25.0 m/s

Power, P = (475 N) * (25.0 m/s)
P = 11,875 W

Considering the motor's efficiency of 95.0%:
Actual Power = Power / Efficiency
Actual Power = 11,875 W / 0.95
Actual Power ≈ 12,500 W

Current, I = Actual Power / Voltage
I = 12,500 W / 12.0 V
I ≈ 1,041.67 A

Therefore, the current the 12.0 V batteries of the electric car must be able to supply to climb a 200 m high hill in 2.00 min at a constant 25.0 m/s speed while exerting 475 N of force is approximately 1,041.67 A.

(c) To travel at a constant 25.0 m/s speed, exerting a 475 N force to overcome air resistance and friction:
Power, P = (475 N) * (25.0 m/s)
P = 11,875 W

Considering the motor's efficiency of 95.0%:
Actual Power = Power / Efficiency
Actual Power = 11,875 W / 0.95
Actual Power ≈ 12,500 W

Current, I = Actual Power / Voltage
I = 12,500 W / 12.0 V
I ≈ 1,041.67 A

Therefore, the current the 12.0 V batteries of the electric car must be able to supply to travel at a constant 25.0 m/s speed, exerting a 475 N force to overcome air resistance and friction is approximately 1,041.67 A.