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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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A simple experiment to measure the speed of sound doesn't involve a stopwatch. You can fill up along tube with water and put a t
Serhud [2]

Answer:

Explanation:

In order to answer this problem you have to know the depth of the column, we say R, this information is important because allows you to compute some harmonic of the tube. With this information you can compute the depth of the colum of air, by taking tino account that the new depth is R-L.

To find the fundamental mode you use:

f_n=\frac{nv_s}{4L}

n: mode of the sound

vs: sound speed

L: length of the column of air in the tube.

A) The fundamental mode id obtained for n=1:

f_1=\frac{v_s}{4L}

B) For the 3rd harmonic you have:

f_3=\frac{3v_s}{4L}

C) For the 2nd harmonic:

f_2=\frac{2v_s}{4L}

7 0
3 years ago
I’m sorry this is a quiz and we’re correcting it and I still don’t get it anyone know ?
kotegsom [21]

Answer:

3.0M

Explanation:

Thats two wavelengths,not one.

Pretty honest mistake I would've made the same if I was rushing

5 0
3 years ago
A ray diagram without the produced image is shown.
Goryan [66]

Answer:

B) inverted and real

Explanation:

6 0
3 years ago
A virtual image produced by a lens is always
Sever21 [200]
C. located in front of the lens
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Read 2 more answers
A projectile is launched from ground level with an initial speed of 47 m/s at an angle of 0.6 radians above the horizontal. It s
Zarrin [17]

Answer:

30.67m

Explanation:

Using one of the equations of motion as follows, we can describe the path of the projectile in its horizontal or vertical displacement;

s = ut ± \frac{1}{2} at^2               ------------(i)

Where;

s = horizontal/vertical displacement

u = initial horizontal/vertical component of the velocity

a = acceleration of the projectile

t = time taken for the projectile to reach a certain horizontal or vertical position.

Since the question requires that we find the vertical distance from where the projectile was launched to where it hit the target, equation (i) can be made more specific as follows;

h = vt ± \frac{1}{2} at^2               ------------(ii)

Where;

h = vertical displacement

v = initial vertical component of the velocity = usinθ

a = acceleration due to gravity (since vertical motion is considered)

t = time taken for the projectile to hit the target

<em>From the question;</em>

u = 47m/s, θ = 0.6rads

=> usinθ = 47 sin 0.6

=> usinθ = 47 x 0.5646 = 26.54m/s

t = 1.7s

Take a = -g = -10.0m/s   (since motion is upwards against gravity)

Substitute these values into equation (ii) as follows;

h = vt - \frac{1}{2} at^2

h = 26.54(1.7) - \frac{1}{2} (10)(1.7)^2

h = 45.118 - 14.45

h = 30.67m

Therefore, the vertical distance is 30.67m        

7 0
3 years ago
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