# physics

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?

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1. sqrt (130^2 + 4500^2)

tan angle from straight ahead = 4500/130

seat belt equal and opposite

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2. 4500/130= 34.6N

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3. sqrt (130^2 + 4500^2) = 4501.8 N

tan(angle) (4500.0/130.0)

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4. you subtract 130 from 4500, and you get 4470 :)

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5. sqrt (130^2 + 4500^2) = 4501.9N
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

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6. 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?​

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7. Fx=130 N
Fy=4500 N
Fnet=fx+fy
4500+130=4.630
Therefor force of sensor report = 4.630

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