A toy car has a mass of 120g, including two 1.5 V batteries connected in series. Suppose the motor that drives the car is 80\% efficient (i.e., 80\% of the electric energy goes to the drive wheels, 20\% is dissipated as heat) and that friction and air resistance are negligible. How much charge passes through the batteries during the time it takes the car to accelerate from rest to 1.5 m/s?

Change in kinetic energy of car = (1/2) m v^2

divide that by 0.8
The result is the energy taken from the battery, q V.
V is 1.5*2 = 3 volts
solve for q

0.036 Coulombs

real answer=0.05625 coulombs

To find out how much charge passes through the batteries during the time it takes the car to accelerate, we can use the formula:

Q = I * t

Where:
Q is the charge in coulombs (C)
I is the electric current in amps (A)
t is the time in seconds (s)

First, let's calculate the current flowing through the circuit. We know that the motor is 80% efficient, so only 80% of the electric energy goes to the drive wheels. This means that 20% of the energy is dissipated as heat.

We can calculate the electric energy used by the motor using the following formula:

E = 0.8 * V * Q

Where:
E is the electric energy in joules (J)
V is the voltage in volts (V)
Q is the charge in coulombs (C)

We are given the voltage of the batteries, which is 1.5 V. And we know that the electric energy is equal to the kinetic energy of the car:

E = 0.5 * m * v^2

Where:
m is the mass of the car in kilograms (kg)
v is the final velocity of the car in meters per second (m/s)

We are given the mass of the car, which is 120g. Let's convert it to kilograms:

m = 120g / 1000 = 0.12 kg

We are also given the final velocity of the car, which is 1.5 m/s.

Now we can substitute the values into the equation to find the electric energy:

E = 0.5 * 0.12 * (1.5^2) = 0.135 J

Now we can rearrange the equation to solve for the charge passing through the batteries:

Q = E / (0.8 * V)

Substituting the values:

Q = 0.135 / (0.8 * 1.5) = 0.1125 C

Therefore, during the time it takes the car to accelerate, approximately 0.1125 coulombs of charge pass through the batteries.