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goldenfox [79]
2 years ago
7

Which term describes an advantage of wind energy?

Physics
2 answers:
atroni [7]2 years ago
7 0

The answer is "D". This is because wind is very abundant on earth...

Radda [10]2 years ago
7 0

Answer: D. Abundant

Explanation:

Wind is a mass of gases of different origin and form. It can be define as a flow of gases in a particular direction. Wind is a renewable resource which means that it is abundantly available in nature and can be replenished again by the wind cycle. The abundance of wind makes it suitable for the generation of wind energy by the wind mills.

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The graph represents velocity over time.<br> What is the acceleration?
Semmy [17]

Answer:

I think the answer is 0.2 m/s2

Explanation:

7 0
2 years ago
A ball is launched horizontally at 4 m/s
ArbitrLikvidat [17]

Answer:

3.5 seconds of flight time; 13.9 m from the base of the cliff

Explanation:

3 0
3 years ago
A 4.0-kg object has 72 J of kinetic energy.<br> Determine its speed.
Svetlanka [38]

Answer:

6m/s

Explanation:

v = sqrt of 2KE/m

Where:

KE = kinetic energy

m = mass of a body

v = velocity of a body

= sqrt of 2(72)/4

= sqrt of 144/4

= sqrt of 36

= 6m/s

8 0
2 years ago
A body is travelling with a velocity 30 m/s².what will be its velocity after 4s?​
Ipatiy [6.2K]

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70m/s²

Explanation:

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3 0
3 years ago
You hang a heavy ball with a mass of 10 kg from a gold wire 2.6 m long that is 1.6 mm in diameter. You measure the stretch of th
PolarNik [594]

<u>Answer:</u> The Young's modulus for the wire is 6.378\times 10^{10}N/m^2

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

The equation representing Young's Modulus is:

Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}

where,

Y = Young's Modulus

F = force exerted by the weight  = m\times g

m = mass of the ball = 10 kg

g = acceleration due to gravity = 9.81m/s^2

l = length of wire  = 2.6 m

A = area of cross section  = \pi r^2

r = radius of the wire = \frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m      (Conversion factor:  1 m = 1000 mm)

\Delta l = change in length  = 1.99 mm = 1.99\times 10^{-3}m

Putting values in above equation, we get:

Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2

Hence, the Young's modulus for the wire is 6.378\times 10^{10}N/m^2

3 0
3 years ago
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