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frez [133]
3 years ago
11

A small car has a head-on collision with a large truck. Which of the following statements concerning the magnitude of the averag

e force due to the collision is correct? A small car has a head-on collision with a large truck. Which of the following statements concerning the magnitude of the average force due to the collision is correct? The small car and the truck experience the same average force. The small car experiences the greater average force. The truck experiences the greater average force. It is impossible to tell since the masses are not given. It is impossible to tell since the velocities are not given.
Physics
1 answer:
Veronika [31]3 years ago
4 0

Answer:

The small car and the truck experience the same average force.

Explanation:

The average net force will be the resultant force of the average forces of both vehicles. On collision, they'll both experience the same impact force, but the deceleration and deformation felt by the individual vehicle will be proportional to the mass of the vehicle. This is why it will seem like the car will have more force but is not actually so.

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7 0
3 years ago
A 1300 kg steel beam is supported by two ropes. (Figure
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Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

  • the net horizontal force acting on the beam is

R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0

where R_1,R_2 are the magnitudes of the tensions in ropes 1 and 2, respectively;

  • the net vertical force acting on the beam is

R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0

where m=1300\,\rm kg and g=9.8\frac{\rm m}{\mathrm s^2}.

Eliminating R_2, we have

\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)

R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2

R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2

-R_1 \sin(50^\circ) = -\dfrac{mg}2

R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}

Solve for R_2.

\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0

\dfrac{R_2}2 = -mg\cot(110^\circ)

R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}

8 0
2 years ago
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