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4 years ago

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The elevator accelerates upward (in the positive direction) from rest at a rate of 1.95 m/s2 for 2.15 s. Calculate the tension i

n the cable supporting the elevator in newtons.

1 answer:

sammy [17]4 years ago

3 0

The mass is missing. The mass of the elevator is 1650 kg.

Answer:

The tension in the cable is 19387.5 N.

Explanation:

Given:

Initial velocity of the elevator (u) = 0 m/s

Acceleration in the upward direction (a) = 1.95 m/s²

Time taken by the elevator (t) = 2.15 s

Mass of the elevator and persons (m) = 1650 kg

Let the tension in the cable wire be 'T' Newtons.

Now, there are 2 forces acting in the vertical direction. One is tension in the upward direction and the other the weight of the elevator in the downward direction.

As the elevator is accelerating upward, the net force acts in the upward direction.

So, net force on the elevator is given as:

$F_{net}=T-mg$

Now, from Newton's second law, net force equals mass times acceleration.

$F_{net}=ma\\\\T-mg=ma\\\\T=m(g+a)$

Plug in the given values and solve for 'T'. This gives,

$T=1650\ kg(9.8+1.95)\ m/s^2\\\\T=1650\times11.75\ N\\\\T=19387.5\ N$

Therefore, the tension in the cable is 19387.5 N.

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Answer:

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$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

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$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

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$\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm$

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