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Ksju [112]
1 year ago
12

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

Physics
1 answer:
Bad White [126]1 year 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

#SPJ4

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The given parameters;

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The acceleration of the car is calculated as;

s = ut + \frac{1}{2} at^2\\\\1.2 = 0 + 0.5\times a\times (0.67)^2\\\\1.2 = 0.225a\\\\a = \frac{1.2}{0.225} \\\\a = 5.33 \ m/s^2

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v_f^2 = u^2 + 2as\\\\v_f^2 = 0 + 2\times 5.33 \times 1.2\\\\v_f^2 = 12.792\\\\v_f = \sqrt{12.792} \\\\v_f = 3.58 \ m/s

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Water flows through a 4.50-cm inside diameter pipe with a speed of 12.5 m/s. At a later position, the pipe has a 6.25-cm inside
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Given,

The initial inside diameter of the pipe, d₁=4.50 cm=0.045 m

The initial speed of the water, v₁=12.5 m/s

The diameter of the pipe at a later position, d₂=6.25 cm=0.065 m

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\begin{gathered} A_1v_1=A_2v_2 \\ \pi(\frac{d_1}{2})^2v_1=\pi(\frac{d_2}{2})^2v_2 \\ \Rightarrow d^2_1v_1=d^2_2v_2 \end{gathered}

Where A₁ is the area of the cross-section at the initial position, A₂ is the area of the cross-section of the pipe at a later position, and v₂ is the flow rate of the water at the later position.

On substituting the known values,

\begin{gathered} 0.045^2\times12.5=0.065^2\times v_2 \\ \Rightarrow v_2=\frac{0.045^2\times12.5}{0.065^2} \\ =5.99\text{ m/s} \end{gathered}

Thus, the flow rate of the water at the later position is 5.99 m/s

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