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lozanna [386]
3 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:
Volgvan3 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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4 0
3 years ago
For the given current of 300 A and time interval 0.250 s, how many conduction electrons are transferred?
Anna71 [15]

Answer : The number of electrons transferred are, 4.68\times 10^{20}

Explanation :

First we have to calculate the charge on electron.

Formula used :

Q=I\times t

where,

Q = charge on electron = ?

I = current = 300 A

t = time = 0.250 s

Now put all the given values in the above formula, we get:

Q=300A\times 0.250s

Q=75C

Now we have to calculate the number of electrons are transferred.

Formula used :

Q=ne\\\\n=\frac{Q}{e}

where,

n = number of electrons transferred = ?

Q = charge on electrons = 75 C

e = charge on 1 electron = 1.602\times 10^{-19}C

Now put all the given values in the above formula, we get:

n=\frac{75C}{1.602\times 10^{-19}C}

n=4.68\times 10^{20}

Thus, the number of electrons transferred are, 4.68\times 10^{20}

6 0
3 years ago
7.00 kg piece of solid copper metal at an initial temperature T is placed with 2.00 kg of ice that is initially at -20.0°C. The
kipiarov [429]

Answer:

122°C

Explanation:

From the data Final temperature is 0 deg C since there is 0.9kg of ice and 1.10kg of liquid water.

That means that 1.10kg of the ice undergoes Heat of Fusion which is 3.34x10^5 J/kg...

Heat lost by copper = Heat gained by ice + Heat of fusion

-> (7.0kg)(390J/kg*C)(0-T) = (2.00kg)(2100J/kg*C)(0 - (-20) + (1.10kg)(3.34x10^5 J/kg)

-> T(2730) = 334001

-> T = 122°C

5 0
3 years ago
A 430.0 g mass is being pushed by a 1.1 N net force. What is the acceleration?
NeTakaya

Answer:

a = 2.56\ m/s^2

Explanation:

<u>Mechanical Force</u>

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

F = m.a

Where a is the acceleration of the object.

The object has a mass of m=430 g = 0.43 Kg and is pushed by a net force of F=1.1 N. To calculate the acceleration, we solve for a:

\displaystyle a=\frac{F}{m}

\displaystyle a=\frac{1.1}{0.43}

\mathbf{a = 2.56\ m/s^2}

3 0
3 years ago
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