When it crashes into a bridge support that does not move, a car goes from 105 km/hkm/h to 0 in 1.23 mm.Part APart completeWhat is the impulse delivered to the 70-kgkg driver by the seat belt, assuming the belt makes the driver's motion identical to the car's motion? Assume that the initial direction of motion of the car is the positive direction.Express your answer to two significant figures and include the appropriate units.J𝐽 =-2000 N⋅sN⋅sSubmitPrevious Answers CorrectCorrect answer is shown. Your answer −2041.9kg⋅ms−2041.9kg⋅ms(=-2040 N⋅sN⋅s) was either rounded differently or used a different number of significant figures than required for this part.Part BWhat is the average force exerted by the belt on the driver?Express your answer to two significant figures and include the appropriate units.

The question seems to be a bit jumbled, but I'll try to answer it as best as I can.

Part A:

The impulse delivered to the driver by the seat belt can be calculated using the formula:

Impulse = Change in momentum

First, we need to convert the speed of the car from km/h to m/s.

105 km/h = 29.17 m/s

The initial momentum of the driver is:

Momentum = mass x velocity = 70 kg x 29.17 m/s = 2041.9 kg.m/s

Since the car crashes and comes to a stop, the final momentum of the driver is 0.

Therefore, the change in momentum (or impulse) is:

Impulse = Final momentum - Initial momentum = 0 - 2041.9 kg.m/s = -2041.9 kg.m/s

The negative sign indicates that the impulse is in the opposite direction to the initial motion of the car.

Part B:

The average force exerted by the belt on the driver can be calculated using the formula:

Force = Impulse / time

We need to convert the time from mm (which I assume is a typo and should be seconds) to seconds.

1.23 mm = 0.00123 s

Therefore, the average force is:

Force = -2041.9 kg.m/s / 0.00123 s = -1.66 x 10^6 N

The negative sign indicates that the force is in the opposite direction to the initial motion of the car.