An electric car accelerates for 8.0s by drawing energy from its 320-V battery pack. During this time, 1300C of charge pass through the battery pack. Find the minimum horsepower rating of the car.

I need some ideas to do it!!PLEASE HELP!!!THANK YOU SO MUCH!!!!

Vq is energy, Vq/time is power. Convert that to HP

Volts = J/C

and 735.5 J = 1 hp

then divide your answer by time to get the solution

To find the minimum horsepower rating of the car, we need to use the equation:

Power = (Work) / (time)

Where power is measured in watts, work is measured in joules, and time is measured in seconds.

To find the work done by the car, we can use the equation:

Work = Charge × Voltage

Where work is measured in joules, charge is measured in coulombs, and voltage is measured in volts.

First, convert the charge from coulombs to ampere-hours by dividing it by 3600 (since 1 ampere-hour is equal to 3600 coulombs).

1300 C ÷ 3600 = 0.3611 Ah

Next, convert the voltage from volts to kilovolts by dividing it by 1000 (since 1 kilovolt is equal to 1000 volts).

320 V ÷ 1000 = 0.3200 kV

Now, we can calculate the work done:

Work = 0.3611 Ah × 0.3200 kV = 0.1156 kWh

In order to use the power equation, we need to convert the work to joules. Since 1 kilowatt-hour is equal to 3.6 × 10^6 joules, we can multiply the work by this conversion factor:

0.1156 kWh × 3.6 × 10^6 J/kWh = 415,776 J

Next, we need to find the time, which is given as 8.0 seconds. Now we can substitute the values into the power equation:

Power = 415,776 J / 8.0 s = 51,972 W

To convert the power from watts to horsepower, divide it by 746 (since 1 horsepower is equal to 746 watts):

51,972 W ÷ 746 = 69.7 horsepower

Therefore, the minimum horsepower rating of the car is approximately 69.7 horsepower.