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jeka94
3 years ago
7

In a rear-end collision, occupants are thrown towards the __________ of the vehicle.

Physics
2 answers:
Harrizon [31]3 years ago
7 0

In a rear-end collision, occupants are thrown towards the BACK of the vehicle.

dsp733 years ago
5 0

Answer:

Back

Explanation:

As the vehicle is moving forward the <em>inertia</em> is keeping the occupants move at the same speed and with respect to the vehicle the occupants appear to be at rest. In case of rear collision, the vehicle will suddenly move forward while the occupants will be thrown towards the back of the vehicle.

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Let the masses of blocks A and B be 4.50 kg and 2.00 kg , respectively, the moment of inertia of the wheel about its axis be 0.4
Free_Kalibri [48]

Answer:

Accelerations of both the sides is 0.6125 m/s^{2}, A moves downwards whereas B moves upwards.

\alpha=6.125 rad/s^{2}

Tension on side A = 4.5 × g= 44.1 m/s^{2}

Tension on side B= 2.0 × g=  19.6 m/s^{2}

Explanation:

As both, the blocks A and B are attached due to the constraint they can only possess a single acceleration a.

Observe the figure attached, let the tension with Block A be T_{2} and the tension attached with Block B be T_{1} .

Tensions will be only be due to the weight of the blocks as no other force is present.

T_{2} = 4.5 × g= 44.1 m/s^{2}

T_{1} = 2.0 × g=  19.6 m/s^{2}

Now, lets make a torque equation about the center of the wheel and find the alpha

T_{2}×R- T_{1}×R= MI( Moment of Inertia of Wheel)× Alpha

where, R= Radius of the wheel=0.100m  and

           Alpha(\alpha)= Angular acceleration of the wheel

MI of the wheel= 0.400 kg/m^{2}

(44.1-19.6)R=0.400\alpha

\alpha = \frac{24.5 * 0.100}{0.400}

\alpha=6.125 rad/s^{2}

Acceleration = R ×\alpha

                    = 0.1 * 6.125

                    =0.6125 m/s^{2}

Accelerations of both the sides is 0.6125 m/s^{2}, A moves downwards whereas B moves upwards.

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3 years ago
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