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s344n2d4d5 [400]
3 years ago
8

The coefficient of kinetic friction between an object and the surface upon which it is sliding is 0.10. The mass of the object i

s 8.0 kg. What is the force of friction? What is the acceleration of the object?
Physics
1 answer:
denis-greek [22]3 years ago
6 0

Answer:

Explanation:

1) Force Friction = Normal Force * Coefficient of Friction

Force Friction = Mass * Gravity * Coefficient of Friction

2) F = ma

Force = mass * acceleration

Force Friction (from #1) = mass * acceleration

acceleration = Force Friction / Mass

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It is neither false nor true. When they collide some of one of the objects goes to the other object.

Explanation:

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As a stands, her entire weight is momentarily placed on the heels of her high-heeled shoes. Calculate the pressure exerted on th
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Answer:

Pressure will be 6072449.952Pa

Explanation:

We have given mass of the women m = 65 kg

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3 years ago
The pressure drop needed to force water through a horizontal 1-in diameter pipe if 0.60 psi for every 12-ft length of pipe. Dete
oksian1 [2.3K]

Answer:

The shear stress at a distance 0.3-in away from the pipe wall is 0.06012lb/ft²

The shear stress at a distance 0.5-in away from the pipe wall is 0

Explanation:

Given;

pressure drop per unit length of pipe = 0.6 psi/ft

length of the pipe = 12 feet

diameter of the pipe = 1 -in

Pressure drop per unit length in a circular pipe is given as;

\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\

make shear stress (τ) the subject of the formula

\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\\tau = \frac{\delta P *r}{2L}

Where;

τ is the shear stress on the pipe wall.

ΔP is the pressure drop

L is the length of the pipe

r is the distance from the pipe wall

Part (a) shear stress at a distance of  0.3-in away from the pipe wall

Radius of the pipe = 0.5 -in

r = 0.5 - 0.3 = 0.2-in = 0.0167 ft

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Part (b) shear stress at a distance of  0.5-in away from the pipe wall

r = 0.5 - 0.5 = 0

\tau = \frac{\delta P *r}{2L}  = \frac{86.4 *0}{2*12} =0

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