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Bond [772]
3 years ago
7

A force 6.3 N is applied at a distance 2.1 m from the point of rotation of a wrench. If the force is applied in a direction para

llel to the position vector, how much torque is exerted by the force
Physics
1 answer:
Ganezh [65]3 years ago
8 0

Answer: 0 (zero)

Explanation:

The formula for torque = force × perpendicular distance between force and point of rotation.

Where force = 6.3 N, distance between force and point of rotation = 2.1m

It has been stated from the question that the force is applied in such a way that it is parallel to the position vector.

If we take the magnitude of this position vector, it will give the distance between the force and the point of rotation which still shows that the force is parallel to the rotation point which does not satisfy the condition for torque to occur hence torque is zero.

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3 years ago
A spring hangs from the ceiling with an unstretched length of x 0 = 0.35 m . A m 1 = 6.3 kg block is hung from the spring, causi
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Answer:

x₂=0.44m

Explanation:

First, we calculate the length the spring is stretch when the first block is hung from it:

\Delta x_1=0.50m-0.35m=0.15m

Now, since the stretched spring is in equilibrium, we have that the spring restoring force must be equal to the weight of the block:

k\Delta x_1=m_1g

Solving for the spring constant k, we get:

k=\frac{m_1g}{\Delta x_1}\\\\k=\frac{(6.3kg)(9.8m/s^{2})}{0.15m}=410\frac{N}{m}

Next, we use the same relationship, but for the second block, to find the value of the stretched length:

k\Delta x_2=m_2g\\\\\Delta x_2=\frac{m_2g}{k}\\\\\implies \Delta x_2=\frac{(3.7kg)(9.8m/s^{2})}{410N/m}=0.088m

Finally, we sum this to the unstretched length to obtain the length of the spring:

x_2=0.35m+0.088m=0.44m

In words, the length of the spring when the second block is hung from it, is 0.44m.

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