There is more wire to travel through,farther distance, and a higher possibility of other disruptions. Please Mark Brainliest!!!
Answer:
two people who are not going to be able to make it to class today because of the day and then I will be there at the house and then we can go
Its very dense. Hey, are you homeschooled?
(a) 328.6 kg m/s
The linear impulse experienced by the passenger in the car is equal to the change in momentum of the passenger:

where
m = 62.0 kg is the mass of the passenger
is the change in velocity of the car (and the passenger), which is

So, the linear impulse experienced by the passenger is

(b) 404.7 N
The linear impulse experienced by the passenger is also equal to the product between the average force and the time interval:

where in this case
is the linear impulse
is the time during which the force is applied
Solving the equation for F, we find the magnitude of the average force experienced by the passenger:
