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lozanna [386]
2 years ago
11

This problem follows up on a discussion from lecture. A wind turbine with an efficiency of 45% for converting wind energy into e

lectrical energy is designed to yield 10 kW of electrical power in a 20 mi/h wind. [The density of air is 1.2 kg/m3.]
What length rotor blades does this wind turbine have?
Physics
1 answer:
Volgvan2 years ago
4 0

Answer:

4.1 m

Explanation:

10 kW = 10000 W

20mi/h = 20*1.6 km/mi = 32 km/h = 32 * 1000 (m/km) *(1/3600) hr/s = 8.89 m/s

The power yielded by the wind turbine can be calculated using the following formula

P = \frac{1}{2} \rho v^3 A C_p

where \rho = 1.2 kg/m^3 is the air density, v = 8.89 m/s is the wind speed, A is the swept area and C_p = 0.45 is the efficiency

10000 = 0.5 * 1.2 * 8.89^3 * A * 0.45

10000 = 190A

A = 10000 / 190 = 52.7 m^2

The swept area is a circle with radius r being the blade length

\pi r^2 = A = 52.7

r^2 = 52.7 / \pi = 16.79

r = \sqrt{16.79} = 4.1 m

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Motion map has the points spaced farther apart (because the car would go a further distance in each second), and the velocity vectors (arrows) are longer, because the car is moving faster. So 'with longer vectors' is the correct answer
8 0
2 years ago
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Capillary action in trees can transport water from the roots to the tree's branches. The capillaries (Xyelem) in a certain tree
lesantik [10]

Answer:

h=14.2857\,m

Explanation:

Given:

radius of capillary, r=10^{-6}\,m

angle of contact, \theta=0^{\circ}

density of water, \rho=1000\,kg.m^{-3}

surface tension of water, T=0.07 \,N.m^{-1}

height, h = ?

We have the equation for the height of meniscus as:

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h=\frac{2\times 0.07\times cos\,0^{\circ}}{1000\times 9.8\times 10^{-6}}

h=14.2857\,m

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6 0
2 years ago
A firecracker is thrown downward from a height of 2.75m above the ground, with a speed of 3.15m/s. Ignore air resistance, determ
3241004551 [841]

Here in this question as we can see there is no air friction so we can use the principle of energy conservation

PE_i + KE_i = PE_f + KE_f

mgh_1 + \frac{1}{2}mv_i^2 = mgh_2 + \frac{1}{2}mv_f^2

now here we know that

h_1 = 2.75 m

v_i = 0

v_f = 5.23 m/s

now plug in all values in above equation

mg*2.75 + 0 = mgh + \frac{1}{2}m(5.23)^2

divide whole equation by mass "m"

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4 0
2 years ago
A sample of 4.50 g of methane occupies 12.7 dm3 at 310 K. (a) Calculate the work done when the gas expands isothermally against
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Answer:

(A) Work done will be 87.992 KJ

(B) Work done will be 167.4 KJ            

Explanation:

We have given mass of methane m = 4.5 gram = 0.0045 kg

Volume occupies V_1=12.7dm^3=12.7liters

And volume is increased by 3.3dm^3 so V_2=12.7+3.3=16liters

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(b) At reversible process work done is given by W=nRTln\frac{V_2}{V_1}

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2 years ago
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However, he used his pulse beats as timer during the experiment. This method is unreliable because the pulse beats of a person changes depending on the person's state of mind. A stop clock could have been a more reliable timer than pulse beats.

Learn more: brainly.com/question/7201885

7 0
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