Asked by i love polka

the sensor in the torso of a crash-test dummy records the magnitude and direction of the net force acting on the dummy. if the dummy is thrown forward with a force of 130.0 N while simultaneously being hit from side with a force of 4500.0 N, what force will the sensor report?

what force will the seat belt have to exert on the dummy to hold the dummy in the seat?

Answers

Answered by Damon
sqrt (130^2 + 4500^2)

tan angle from straight ahead = 4500/130

seat belt equal and opposite
Answered by Emanuela
4500/130= 34.6N
Answered by Krys
sqrt (130^2 + 4500^2) = 4501.8 N

tan(angle) (4500.0/130.0)
Answered by Kay Crum
you subtract 130 from 4500, and you get 4470 :)
Answered by Kyser
sqrt (130^2 + 4500^2) = 4501.9N
tan(angle) = (opp/adj)
Angle = (tan^-1)(4500/130)
Angle = 88.3 degrees to the side of forward

Answer: -4501.9N @-88.3 degrees to the side of forward

(answer due to opposite of what sensor reads)
Answered by Anonymous
5. The sensor in the torso of a crash-test dummy records the magnitude and direction of the net force acting on the dummy. If the dummy is thrown forward with a force of 130.0 N while simultaneously being hit from the side with a force of 4500.0 N, what force will the sensor report?​
Answered by Fahad
Fx=130 N
Fy=4500 N
Fnet=fx+fy
4500+130=4.630
Therefor force of sensor report = 4.630
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