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Stells [14]
3 years ago
14

A team of engineers is working with a wind turbine. The team is working with a model and testing different angles for the blade

of the wind turbine. What would the research question be for this team?
Which angle converts the largest percentage of potential energy from the wind into kinetic energy of the turbine?

Which angle converts the largest percentage of kinetic energy from the wind into kinetic energy of the turbine?

Which angle converts the largest percentage of potential energy from the wind into potential energy of the turbine?

Which angle converts the largest percentage of kinetic energy from the wind into potential energy of the turbine?
Physics
2 answers:
Zinaida [17]3 years ago
7 0

Answer:

i think it would be which angle converts the most potential energy to into kinetic energy of the turbine

Explanation: because the windmill makes kinetic energy and converts it into mechanical power. then a generator takes the mechanical power and makes it into electricity

Vladimir79 [104]3 years ago
3 0

Answer:it would be which angle converts the most potential energy to into kinetic energy of the turbine

Explanation:

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Yes, all waves can be distorted, deflected, or changed

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- As you ride in a car, both you and the car are moving
KIM [24]

Answer:

c. Your body is at rest but its inertia puts it in motion

Explanation:

3 0
3 years ago
A ball is thrown directly downward with an initial speed of 7.95 m/s, from a height of 29.0 m. After what time interval does it
Y_Kistochka [10]

Answer: after 1.75 seconds

Explanation:

The only force acting on the ball is the gravitational force, so the acceleration will be:

a = -9.8 m/s^2

the velocity can be obtained by integrating over time:

v = -9.8m/s^2*t + v0

where v0 is the initial velocity; v0 = -7.95 m/s.

v = -9.8m/s^2*t - 7.95 m/s.

For the position we integrate again:

p = -4.9m/s^2*t^2 - 7.95 m/s*t + p0

where p0 is the initial position: p0 = 29m

p =  -4.9m/s^2*t^2 - 7.95 m/s*t + 29m

Now we want to find the time such that the position is equal to zero:

0 = -4.9m/s^2*t^2 - 7.95 m/s*t + 29m

Then we solve the Bhaskara's equation:

t = \frac{7.95 +- \sqrt{7.95^2 +4*4.9*29} }{-2*4.9} = \frac{7.95 +- 25.1}{9.8}

Then the solutions are:

t = (7.95 + 25.1)/(-9.8) = -3.37s

t = (7.95 - 25.1)/(-9.8) = 1.75s

We need the positive time, then the correct answer is 1.75s

4 0
4 years ago
Find the magnitude of the gravitational force a 63.5 kg person would experience while standing on the surface of Earth with a ma
Anastaziya [24]

Answer: 3976N

Explanation:

Using the formula for calculating gravitational force between two masses, we have

F = GMm/r^2

Where G is the gravitational constant

M and m are the masses

r is the distance between the masses

F= 6.673 × 10-¹¹ × 5.98 × 10²⁴ × 63.5/ (6.37 × 10^6)^2

F= 2.533×10^16/6.37×10^12

F= 0.3976×10⁴N

F= 3976N

3 0
3 years ago
A 70.0-kg person throws a 0.0420-kg snowball forward with a ground speed of 35.0 m/s. A second person, with a mass of 57.0 kg, c
ollegr [7]

Answer:

so throwers new velocity = 2.18032m/s

so catchers new velocity = 0.02577m/s

Explanation:

Directly by conservation of momentum we can write

m_1u_1+m_2u_2= m_1v_1+m_2v_2

let x be the thrower's new velocity

(70+0.042)×2.2 + 57×0 = 70× x +0.042×35 +57×0

x = 2.18032m/s

so the velocity of 70 kg man = 2.18032m/s

so throwers new velocity = 2.18032m/s

now again by conservation of momentum

0.042×35 = (57+0.042) ×y

y = 0.02577m/s

so catchers new velocity = 0.02577m/s

3 0
3 years ago
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