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torisob [31]
3 years ago
6

Find the magnitude of the torque produced by a 4.8 N force applied to a door at a perpendicular distance of 0.43 m from the hing

e. Answer in units of N · m
Physics
1 answer:
SVEN [57.7K]3 years ago
6 0

Answer:

2.064 N.m

Explanation:

Torque: This is the force that act on a system, which tends to cause rotation or twisting. The S.I unit of Torque is N.m

Torque is represented mathematically as,

T = F×d........................ Equation 1

Where T = Torque, F = Force, d = perpendicular distance.

Given: F = 4.8 N,  d = 0.43 m.

Substitute into equation 1

T = 4.8×0.43

T = 2.064 N.m

Hence the Torque produced = 2.064 N.m

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Answer:

The moment of inertia of the system decreases and the angular speed increases.

Explanation:

This very concept might not seem to be interesting at first, but in combination with the law of the conservation of angular momentum, it can be used to describe many fascinating physical phenomena and predict motion in a wide range of situations.

In other words, the moment of inertia for an object describes its resistance to angular acceleration, accounting for the distribution of mass around its axis of rotation.

Therefore, in the course of this action, it is said that the moment of inertia of the system decreases and the angular speed increases.

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the system shown above is released from rest. if friction is negligible, the acceleration of the 4.0 kg block sliding on the tab
JulsSmile [24]

The acceleration of the first block (4 kg) is -9.8 m/s².

The given parameters:

  • <em>Mass of the first block, m₁ = 4.0 kg</em>
  • <em>Mass of the second block, m₂ = 2.0 kg</em>

The net force on the system of the two blocks is calculated as follows;

m_2 g - T = m_1 a

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m_2 g - m_1 g = m_1 a\\\\a = \frac{m_2 g - m_1g}{m_1} \\\\a = \frac{g(m_2 - m_1)}{m_1} \\\\a = \frac{9.8(2-4)}{2} \\\\a = -9.8 \ m/s^2

Thus, the acceleration of the first block (4 kg) is -9.8 m/s².

Learn more about net force on two connected blocks here: brainly.com/question/13539944

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3 years ago
A crate with a mass of m = 450 kg rests on the horizontal deck of a ship. The coefficient of static friction between the crate a
Zielflug [23.3K]

Answer:F_{v} =\mu_{k} mg

Magnitude of the force is 2601.9 N

Explanation:

m = 450 kg

coefficient of static friction μs = 0.73

coefficient of kinetic friction is μk = 0.59

The force required to  start crate moving is F_{s} =\mu_{s} mg.

but once crate starts moving the force of friction is reduced  F_{v} =\mu_{k} mg.

Hence  to keep crate moving at constant velocity we have to reduce the  force pushing crate ie F_{v} =\mu_{k} mg.

Then the above pushing force will equal the frictional force due to kinetic friction and constant velocity is possible as  forces are balanced.

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