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3 years ago

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What are some of the challenges that face wave energy production?

2 answers:

valentina_108 [34]3 years ago

4 0

Answer:

One of the major obstacles with any wave resource study is lack of long-term ocean wave measurements inside the 100-meter-depth contour, where refraction effects result in spatially inhomogeneous wave parameters. Lack of data makes it difficult or impossible to mark the optimum locations for WECs.

Explanation:

dlinn [17]3 years ago

3 0

Answer:

WEC Challenges

One of the major obstacles with any wave resource study is lack of long-term ocean wave measurements inside the 100-meter-depth contour, where refraction effects result in spatially inhomogeneous wave parameters. Lack of data makes it difficult or impossible to mark the optimum locations for WECs

Explanation:

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How do I Change the following to decimal numbers.<br> a. 5.93 x 10-5

PSYCHO15rus [73]

Explanation:

To convert from scientific notation to decimal, move the decimal point 5 places to the left.

5.93×10⁻⁵ = 0.0000593

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3 years ago

3. A certain wire, 3 m long, stretches by 1.2 mm when under tension of 200 N. By how much does

nikitadnepr [17]

Answer:

The extension of the second wire is $e_2 = 0.0024 \ m = 2.4 mm$

Explanation:

From the question we are told that

The length of the wire is $L = 3 \ m$

The elongation of the wire is $e = 1.2mm = \frac{1.2}{1000} = 0.0012 m$

The tension is $F = 200 \ N$

The length of the second wire is $L_2 = 6 \ m$

Generally the Young's modulus(Y) of this material is

$Y = \frac{stress}{strain }$

Where $stress = \frac{F}{A}$

Where A is the area which is evaluated as

$A = \pi r^2$

and $strain = \frac{extention}{length} = \frac{e}{L}$

So

$Y = \frac{\frac{F}{\pi r^2 } }{ \frac{e}{L} }$

Since the wire are of the same material Young's modulus(Y) is constant

So we have

$\frac{F * L }{r^2 e} = \pi * Y = constant$

$F * L = constant * r^2 e$

Now the ration between the first and the second wire is

$\frac{F_1}{F_2} * \frac{L_1}{L_2} = \frac{r*2_1}{r^2} * \frac{e_1}{e_2}$

Since tension , radius are constant

We have

$\frac{L_1}{L_2} = \frac{e_1}{e_2}$

substituting values

$\frac{3}{6} = \frac{0.0012}{e_2}$

$0.5 e_2 = 0.0012$

$e_2 = \frac{ 0.0012 }{0.5}$

$e_2 = 0.0024 \ m = 2.4 mm$

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3 years ago

How is your guys day going?

faltersainse [42]

Good i’m tired how about you

4 0

3 years ago

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A 55-kg woman is wearing high heels. If each heel has a circular cross-section 6.0 mm in diameter and she puts all her weight on

bogdanovich [222]

Answer:

The pressure exerted by the woman on the floor is 1.9061 x 10⁷ N/m²

Explanation:

Given;

mass of the woman, m = 55 kg

diameter of the circular heel, d = 6.0 mm

radius of the heel, r = 3.0 mm = 0.003 m

Cross-sectional area of the heel is given by;

A = πr²

A = π(0.003)²

A = 2.8278 x 10⁻⁵ m²

The weight of the woman is given by;

W = mg

W = 55 x 9.8

W = 539 N

The pressure exerted by the woman on the floor is given by;

P = F / A

P = W / A

P = 539 / (2.8278 x 10⁻⁵ )

P = 1.9061 x 10⁷ N/m²

Therefore, the pressure exerted by the woman on the floor is 1.9061 x 10⁷ N/m²

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2 years ago

A car jack with a mechanical advantage of 6 needs to produce an output force of 600 N to raise a car. What input force is requir

Mkey [24]

B.) 100 N is the CORRECT answer

Please mark as brainliest! :)

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3 years ago

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