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nevsk [136]
3 years ago
7

What is limiting friction​

Physics
1 answer:
Fantom [35]3 years ago
5 0

Answer:

Explanation:

As you start increasing the force on an object from 0 N it the object will stay at rest and at a certain magnitude of force the object will start moving. The friction acting on that object is known as limiting frictional force. The magnitude of this frictional force is equal to the force that we apply on the object as it just start to move.

It's important to note that the limiting frictional force is the largest frictional force act on that object in the above explained process. The dynamic frictional force ( the frictional force acting on an moving object.) is always less than the limiting frictional force.

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Exercising in hot weather can cause what side effects ?
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dehydration, lots of sweating, heat stroke, heat exhaustion: nausea, dizziness, vomitting, diahreaa, headache

Explanation:

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A 15.7 kg block is dragged over a rough, horizontal surface by a constant force of 83.1 N
Kay [80]

Answer:

The work done by the 83.1 N force is 4687.5 J.

The magnitude of the work done by the  force of friction is 1187.5 J.

Explanation:

Given:

Mass of the block, m=15.7 kg

Force acting on it, F=83.1 N

Angle of application of force, \theta = 25.7°

Displacement of the block, d=62.6 m

Coefficient of friction, \mu =0.161

Acceleration due to gravity, g=9.8 m/s²

Work done by a force is given as:

Work,W=F\times d\times \cos\theta

So, work done by the constant force is given as:

W_{force}=83.1\times 62.6\times \cos(25.7)\\W_{force}=4687.5\textrm{ J}

Now, in order to find work done by friction, we need to evaluate friction.

For the vertical direction, the net force is zero as there is no vertical motion.

Therefore,

N + F\sin\theta = mg\\ N = mg-F\sin\theta

Frictional force is given as:

f=\mu N=\mu (mg-F \sin \theta)=0.161\times ((15.7\times 9.8)-(83.1\times \sin(25.7))=18.97\textrm{ N}

Now, friction acts in the direction opposite to the displacement. Thus, angle between frictional force and displacement is 180°.

Therefore, work done by friction is:

W_{fric}=f\times d\times \cos\theta_f\\W_{fric}=12.72\times 62.6\times \cos(-180)\\W_{fric}=18.97\times 62.6\times -1\\W_{fric}=-1187.5\textrm{ J}

Negative sign indicates that frictional force is acting opposite to motion.

So, the magnitude of the work done by the  force of friction is 1187.5 J.

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4 years ago
What is the effect of putting a ferromagnetic material inside the coil of a solenoid?
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For small amplitudes of oscillation the motion of a pendulum is simple harmonic. Consider a pendulum with a period of 0.550 s Fi
Ivenika [448]

To solve this problem we will use the concepts related to the expression of energy for harmonic oscillator. From our given values we have that the period is equivalent to

T = 0.55s

Therefore the frequency will be the inverse of the period and would be given as

f= \frac{1}{T}

f = \frac{1}{0.55}

f = 1.82s^{-1}

The ground state energy of the pendulum is,

E = \frac{1}{2} hv

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\Delta E = 12.1*10^{-34}J

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3 years ago
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