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2 years ago

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What is the maximum frictional force in the knee joint of a person who supports 76. 0 kg of her mass on that knee?

1 answer:

Bad White [126]2 years ago

4 0

The maximum frictional force in the knee joint of a person who supports 76. 0 kg of her mass on that knee 119.28N.

Given,

m=76kg, μ=0.16, g=9.81 m/ $s^{2}$

frictional force = mμg= 76*0.16*9.81= 119.28 N

<h3>Frictional force</h3>

An opposing force to the relative motion of two bodies in contact is known as frictional force. Always acting in the opposite direction from the direction of motion, frictional force is applied to a moving body. Because it resists motion, it aids in lowering the moving object's speed. The force is one of touch. Four broad categories can be made for the force of friction depending on the sort of motion that occurs between the two objects. Static friction is the force of friction between an object and the surface it is put on.

What is the maximum frictional force in the knee joint of a person who supports 76. 0 kg of her mass on that knee?

Learn more about frictional force here:

brainly.com/question/13707283

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1. During the Middle Ages, armies often attacked castles using large siege engines such as the counterweight trebuchet at left.

fredd [130]

Answer: Some challenges that I could see would be the walls, possibly a moat, tar, and the towers.

Explanation: The wall is obliviously a main problem, trying to get over it or through it is a difficult challenge. The moat (if it has one) means that there is more than likely only one way to get in or out. If it does have tar it means that the attackers are going to be put in a "sticky situation" And finally the towers, they have people at the top shooting arrows down at you, or throwing things at you.

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3 years ago

Car A (mass 1100 kg) is stopped at a traffic light when it is rearended by car B(mass 1400 kg). Both cars then slide with locke

viva [34]

Answer:

Part a)

$v_a = 3.94 m/s$

Part b)

$v_b = 3.35 m/s$

Part C)

$v_b = 6.44 m/s$

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

Explanation:

Part a)

As we know that car A moves by distance 6.1 m after collision under the frictional force

so the deceleration due to friction is given as

$a = -\frac{F_f}{m}$

$a = -\frac{\mu mg}{m}$

$a = - \mu g$

now we will have

$v_f^2 - v_i^2 = 2ad$

$0 - v_i^2 = 2(-\mu g)(6.1)$

$v_a = \sqrt{(2(0.13)(9.81)(6.1)}$

$v_a = 3.94 m/s$

Part b)

Similarly for car B the distance of stop is given as 4.4 m

so we will have

$v_b = \sqrt{2(0.13)(9.81)(4.4)}$

$v_b = 3.35 m/s$

Part C)

By momentum conservation we will have

$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

$1400 v_b = 1100(3.94) + 1400(3.35)$

$v_b = 6.44 m/s$

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

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3 years ago

Convert 4 kilograms into grams with process

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3 years ago

when you place your feet near the fireplace and they become warm what type of energy conversion occurs

JulsSmile [24]

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3 years ago

Read 2 more answers

1) A train accelerates from 36 km/hr to 54 km/hr in 10<br> s. Find acceleration?

Crank

Answer: Given:

Initial velocity= 36km/h=36x5/18=10m/s

Final velocity =54km/h=54x5/18=15m/s

Time =10sec

Acceleration = v-u/ t

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Distance =s=?

From second equation of motion:

S=ut +1/2 at^2

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=100+25

=125m

So distance travelled 125m

Hope it helps you

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3 years ago