Aperson who is properly constrained by an over-the-shoulder seat belt has a good chance of surviving a car collision if the deceleration does not exceed about 30 "gs" (1.0 g = 9.8 m/s^2). assuming uniform deceleration of this value, what is the distance over which the front end of the car must be designed to collapse if a crash brings the car to rest from 75 km/h?

Distance = 0.738 m

Explanation:

Solution:  

First convert Km/h into m/s.  

75 km/h * 1000 m/km * 1 hr/3600 sec = 20.8333 m/s  

According to third equation of motion:

– = 2 * acceleration * distance

Vf= final velocity

Vi= initial velocity

putting values in third equation of motion....

– = 2 * (-30 * 9.8) * distance

note:

negative sign is due to deceleration

Distance = 0.738 m

when insulating materials rub against each other, they may become electrically charged . electrons , which are negatively charged, may be 'rubbed off' one material and on to the other. the material that gains electrons becomes negatively charged. the material that loses electrons is left with a positive charge.

b mirrors submit a iisner? e. for which optical devices does d sometimes have a positive value? o a a and b o b. a and c o c. b and d 0 d. b and cb