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3 years ago

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A cylinder 8 in in diameter and 3 ft long is concentric with a pipe of 8.25 in internal diameter. Between the cylinder and the p

ipe there is an oil film. What force is required to move the cylinder along the pipe at a constant velocity of 3 ft/s

1 answer:

Aleks [24]3 years ago

7 0

Answer:

623.5lb

Explanation:

The answer is detailed in the attached photo

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Which of the following is an example of rotation

SOVA2 [1]

I believe it is D. Earth spinning on it's axis.

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3 years ago

A professional cyclist rides a bicycle that is 92 percent efficient. For every 100 joules of energy he exerts as input work on t

emmainna [20.7K]

Efficiency = Work Output / Work Input

92% = Work Output / 100

0.92 = Work Output / 100

Work Output = 0.92 * 100

Work Output = 92 joules.

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2 years ago

Which of the statements best describes density?

Vesnalui [34]

Answer:

The answer to your question is: letter D.

Explanation:

a.The mass that a mole of substance has, measured in grams per mole. Density is not measure in moles, so this is not the correct answer.

b.The amount of substance dissolved in a liquid, measured in moles per liter. The substance dissolved in a liquid must be measure in grams not in moles, so this answer is incorrect.

c.The mass of substance dissolved in a liquid, measured in grams per milliliter. I think that this definition is correct but is incomple, so this answer is wrong.

d.The ratio of a substance's mass to its volume, measured in grams per milliliter and also equivalent to grams per cubic centimeter. This is the right description to density, so this is the correct answer.

8 0

2 years ago

The gravitational force of attraction between two students sitting at their desks in physics class is 2.59 × 10−8 N. If one stud

motikmotik

<h2>The distance between students is 2.46 m</h2>

Explanation:

The force of attraction due to Newton's gravitation law is

F = $\frac{Gm_1m_2}{r^2}$

Here G is the gravitational constant

m₁ is the mass of one student

m₂ is the mass of second student .

and r is the distance between them

Thus r = $\sqrt{\frac{Gm_1m_2}{F} }$

If we substitute the values in the above equation

r = $\sqrt{\frac{6.673x10^-^1^1x31.9x30.0}{2.59x10^-^8} }$

= 2.46 m

3 0

3 years ago

Read 2 more answers

Suppose you wish to fabricate a uniform wire out of 1.10 g of copper. If the wire is to have a resistance R = 0.390 Ω, and if al

Citrus2011 [14]

To solve this problem we will apply the concepts related to volume, as a function of length and area, as of mass and density. Later we will take the same concept of resistance and resistivity, equal to the length per unit area. Once obtained from the known constants it will be possible to obtain the area by matching the two equations:

Mass of copper wire $(m) = 1.10g = 1.10*10^{-3} kg$

Density $(\rho)= 8.92*10^3kg/m^3$

Resistively of copper $(\gamma) = 1.7*10^{-8}\Omega \cdot m$

Resistance (R) = 0.390\Omega

Volume is defined as,

$V= lA \text{ and }\frac{m}{\rho}$

$lA= \frac{1.10*10^{-3}}{8.92*10^3}$

$lA = 1.233*10^{-7} m^3$ (1)

We know that,

$\frac{l}{A} = \frac{R}{\gamma}$

$\frac{l}{A}= \frac{0.390\Omega}{1.7*10^{-8}\Omega m}$

$\frac{l}{A} = 2.2941*10^7 m^{-1}$ (2)

Multiplying equation we have

$l^2 = (1.233*10^{-7})( 2.2941*10^7)$

$l^2 = 2.8286m^2$

$l =\sqrt{2.8286m^2}$

$l = 1.68m$

Therefore the length of the wire is 1.68m

6 0

2 years ago