So the average force that will need to be applied is um the wait times the average acceleration or the mass times the average acceleration. And that winds up being minus Newton's. And um I actually should technically have written put this in. Um Well yeah, actually I should this all should be um should have drawn this before this breaking for us again, this is positive and this is positive.
So we have um the acceleration is negative, so in our case it's in this way and the forces negatives of that case it's in this way. So if this was positive, so we can see we have a force that's applied in the negative direction, or in this case the velocity is also in, you know, in this direction. So it's opposite to the to the velocity. II What average force is required to stop a kg car in 8. If it … Problem II What average force is needed to accelerate a….
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The car with the smaller mass will require less force to stop. What force is required to stop a kg car in a distance of 0. In an emergency stop to avoid an accident,a shoulderstrap seat belt holds a kg passenger firmly in place.
A car traveling along the highway needs a certain amount of force exerted on it to stop. More stopping force may be required when the car has:. Find a the braking force and b the time required to stop. Assuming the same braking force, find c the distance and. The car's engine delivers 52 kW of power. Calculate the average force of friction that is resisting the motion of the car. My thinking: Convert 52 kW to W. You can view more similar questions or ask a new question.
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