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2 years ago

How does the wind turbine generator work?

1 answer:

5 0

Answer: wind pushes directly against the blades of the turbine, which converts the linear motion of the wind into the rotary motion, then it spins the generator's rotor and the harder the wind pushes, the more electrical energy can be generated.

Explanation:

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PLEASE HELP WITH THIS QUESTION.

Sophie [7]

Answer:

#2 and #3 respectively

Explanation:

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2 years ago

Explain the difference between the three types of friction

USPshnik [31]

Static Friction

It is the friction that exists between a stationary object and the surface on which it's resting.

Sliding friction

It is the resistance created by two objects sliding against each other.

Rolling friction:-

It is the force resisting the motion when a body rolls on a surface.

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2 years ago

What happens to a sheet of copper as kinetic energy of copper molecules decrease

liberstina [14]

First is melts then it expands next it gets cooler Finally it gains ener. Hope this helps you out.

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3 years ago

Example 3 :

kherson [118]

The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.

<h3>How to determine the friction factor</h3>

Using the formula

μ = viscosity = 0. 06 Pas

d = diameter = 120mm = 0. 12m

V = velocity = 1m/s and 3m/s

ρ = density = 0.9

a. Velocity = 1m/s

friction factor = 0. 52 × $\frac{0. 06}{0. 12* 1* 0. 9}$

friction factor = 0. 52 × $\frac{0. 06}{0. 108}$

friction factor = 0. 52 × 0. 55

friction factor $= 0. 289$

b. When V = 3mls

Friction factor = 0. 52 × $\frac{0. 06}{0. 12 * 3* 0. 9}$

Friction factor = 0. 52 × $\frac{0. 06}{0. 324}$

Friction factor = 0. 52 × 0. 185

Friction factor $= 0.096$

Loss When V = 1m/s

Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter

Head loss = 0. 289 × $\frac{1}{2*6. 6743 * 10^-11}$ × $\frac{1^2}{0. 120}$ × $\frac{1}{100}$

Head loss = 1. 80 × 10^8

Head loss When V = 3m/s

Head loss = $0. 096$ × $\frac{1}{1. 334 *10^-10}$ × $\frac{3^2}{0. 120}$ × $\frac{1}{100}$

Head loss = 5. 3× 10^8

Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.

Learn more about friction here:

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1 year ago

A spring whose spring constant is 260 lbf/in has an initial force of 100 lbf acting on it. Determine the work, in Btu, required

Readme [11.4K]

Answer:

A spring whose spring constant is 200 lbf/in has an initial force of 100 lbf acting on it. Determine the work, in Btu, required to compress it another 1 inch.

Step 1 of 4

The force at any point during the deflection of the spring is given by,

where is the initial force

and x is the deflection as measured from the point where the initial force occurred.

The work required to compress the spring is

Therefore work required to compress the spring is

The work required to compress the spring in Btu is calculated by

Where 1Btu =778

The work required to compress the spring,

eman Asked on February 19, 2018 in thermal fluid Sciences 4th solutions.

Explanation:

7 0

2 years ago