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?
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
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)
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)
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
Fy=4500 N
Fnet=fx+fy
4500+130=4.630
Therefor force of sensor report = 4.630
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.