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2 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

1 answer:

Ganezh [65]2 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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Question 1 (1 point)

Juli2301 [7.4K]

Hello!

$\large\boxed{800Ns}$

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2 years ago

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Explanation

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2 years ago

Read 2 more answers

A 1870 kg car traveling at 13.5 m/s collides with a 2970 kg car that is initally at rest at a stoplight. The cars stick together

Inga [223]

Answer:

The value is $\mu = 0.72$

Explanation:

From the question we are told that

The mass of the first car is $m_1 = 1870\ kg$

the initial speed of the car is $u = 13.5 \ m/s$

The mass of the second car is $m_2 = 2970\ kg$

The distance move by both cars is s = 1.93 m

Generally from the law of momentum conservation

$m_1 * u_1 + m_2 * u_2 = (m_1 + m_2 ) * v_f$

Here $u_2 = 0$ because the second car is at rest

and $v_f$ is the final velocity of the the two car

So

$1870* 13.5+ 0= ( 1870 + 2970 ) * v_f$

=> $v_f = 5.22\ m/s$

Generally from kinematic equation

$v_f^2 = u_2^2 + 2as$

here a is the deceleration

So

$5.22^2 = 0 + 2 *a * 1.93$

=> $a = 7.06 \ m/s^2$

Generally the frictional force is equal to the force propelling the car , this can be mathematically represented as

$F_f = F$

Here F is mathematically represented as

$F = (m_1 + m_2) * a$

$F = (1870 + 2970) * 7.06$

$F =34170.4 \ N$

and

$F_f = \mu * (m_1 + m_2 ) * g$

$F_f = 47432 * \mu$

So

$47432 * \mu = 34170.4$

=> $47432 * \mu = 34170.4$

=> $\mu = 0.72$

6 0

2 years ago