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Taya2010 [7]
3 years ago
8

What is/are the energy transformation(s) that take place when using a wind turbine to generate usable energy?

Physics
1 answer:
Cerrena [4.2K]3 years ago
4 0

Answer:

the answer is C

Explanation:

C) friction - mechanical - electrical

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Can someone help me with 45a-g and 46?
LUCKY_DIMON [66]

Idk but it must be hard tho

7 0
3 years ago
A 250 g air-track glider is attached to a spring with springconstant 4.0 N/m. Th damping constant due to air resistance is0.015
vaieri [72.5K]

Answer:

33.33 seconds

Explanation:

N=\dfrac{1}{e}N_0

N_0 = Initial length pulled = 20 cm

b = Damping constant = 0.015 kg/s

k = Spring constant = 4 N/m

m = Mass of glider = 250 g

Time period is given by

T=2\pi\sqrt{\dfrac{m}{k}}\\\Rightarrow T=2\pi\sqrt{\dfrac{0.25}{4}}\\\Rightarrow T=1.57079632679\ s

Using exponential decay formula

N=N_0e^{\dfrac{-bt}{m}}

Final amplitude = Initial times decay

\dfrac{1}{e}0.2=0.2e^{\dfrac{-0.015t}{2\times 0.25}}\\\Rightarrow 0.2=0.2e^{\frac{-0.015t}{2\cdot \:0.25}+1}\\\Rightarrow e^{\frac{-0.015t}{2\cdot \:0.25}+1}=1\\\Rightarrow \ln \left(e^{\frac{-0.015t}{2\cdot \:0.25}+1}\right)=\ln \left(1\right)\\\Rightarrow \left(\frac{-0.015t}{2\cdot \:0.25}+1\right)\ln \left(e\right)=\ln \left(1\right)\\\Rightarrow \frac{-0.015t}{2\cdot \:0.25}+1=\ln \left(1\right)\\\Rightarrow -\frac{0.015t}{0.5}=-1\\\Rightarrow -0.000225t=-0.0075\\\Rightarrow t=33.33\ s

The time taken is 33.33 seconds

7 0
4 years ago
Scientists are expected to share their results by
shepuryov [24]

Answer:

Publishing results of research projects in peer-reviewed journals enables the scientific and medical community to evaluate the findings themselves. It also provides instructions so that other researchers can repeat the experiment or build on it to verify and confirm the results.

7 0
4 years ago
An object of mass 100 kg is accelerated uniformly from a velocity of 5 ms–1
solniwko [45]

Answer:

Initial velocity of the object, u = 5 m/s

Final velocity of the object, v = 8 m/s

Mass of the object, m = 100 kg

Time take by the object to accelerate, t = 6 s

Initial momentum = mu = 100 — 5 = 500 kg m sˆ’1

Final momentum = mv = 100 — 8 = 800 kg m sˆ’1

Force exerted on the object, F = mv – mu / t

= m (v-u) / t

= 800 – 500

= 300 / 6

= 50 N

Initial momentum of the object is 500 kg m sˆ’1.

Final momentum of the object is 800 kg m sˆ’1.

Force exerted on the object is 50 N.

mark this as brainliest!

6 0
3 years ago
A 1 900-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 4.00 m before coming into contact
Leno4ka [110]

Answer:

471392.4 N

Explanation:

From the question,

Just before contact with the beam,

mgh = Fd.................... Equation 1

Where m = mass of the beam, g = acceleration due to gravity, h = height. F =  average Force on the beam, d = distance.

make f the subject of the equation

F = mgh/d................ Equation 2

Given: m = 1900 kg, h = 4 m, d = 15.8 = 0.158 m

Constant: g = 9.8 m/s²

Substitute into equation 2

F = 1900(4)(9.8)/0.158

F = 471392.4 N

6 0
3 years ago
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