Asked by melissa
In an emergency stop to avoid an accident,a shoulderstrap seat belt holds a 60-kg passenger firmly in place. If the car were initally traveling at 90 km/h and came to a stop in 5.5s along a straight, level road, what was the average force applied to the passenger by the seatbelt?
Answers
Answered by
drwls
First convert 90 km/h to ___meters per second.
The conversion factor is
(1000 m/km) * (1 hr/3600s)
Divide that speed in m/s by 5.5 s to get the deceleration rate in m/s^2.
Mass(kg) x acceleration (m/s^2)
= Force (Newtons)
If you want it in pounds, multiply Newtons by 0.225 lb/N
The conversion factor is
(1000 m/km) * (1 hr/3600s)
Divide that speed in m/s by 5.5 s to get the deceleration rate in m/s^2.
Mass(kg) x acceleration (m/s^2)
= Force (Newtons)
If you want it in pounds, multiply Newtons by 0.225 lb/N
Answered by
Anonymous
272.7
Answered by
Juanito alcachofa
2.7×10²
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