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Marta_Voda [28]
3 years ago
7

Average amount of time wind machines operate

Physics
1 answer:
Rasek [7]3 years ago
6 0

The wind machine on an average operates for a time of about \boxed{3-4\text{ hours}}.

Explanation:

The wind energy is the method of using the energy carried by the wind to generate the power. The wind energy is the considered as the conventional  and clean source of energy.

The wind machines require the high power winds to rotate the blades of the windmill and produce power using the energy of the moving wind.

The winds blow at such high speed for a very small time during the day, the wind machine cannot be expected to work for a longer time during the whole day.

The wind energy can be obtained form a wind mill for a limited time unlike the other form of energy like thermal energy or even the solar energy. The thermal energy can be obtained at anytime due to burning of fuel whereas the solar energy can at least be obtained during the whole day time.

Thus, the wind machine on an average operates for a time of about \boxed{3-4\text{ hours}}.

Learn More:

1. Acceleration of the box under friction brainly.com/question/7031524

2. The kinetic energy of an object depends on the brainly.com/question/137098

3. The approximate time taken to walk for 2000 miles brainly.com/question/3785992

Answer Details:

Grade: Middle School

Subject: Physics

Chapter: Energy

Keywords:

wind energy, conventional source, energy, thermal, solar, 3-4 hours, high speed, machine, time, operate.

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vekshin1

Answer:

-39.2m/s

Explanation:

Using the equation of motion;

v = u + at

Since the ball is thrown upward, the acceleration due to gravity acting on it will be negative, hence a = -g

v = u - gt

Since g = 9.8m/s²

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With what minimum speed must you toss a 130 gg ball straight up to just touch the 15-mm-high roof of the gymnasium if you releas
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Answer:

The initial velocity is 0.5114 m/s or 511.4 mm/s

Explanation:

Let the initial velocity be 'v'.

Given:

Mass of the ball (m) = 130 g = 0.130 kg   [ 1 g = 0.001 kg]

Initial height of the ball (h₁) = 1.4 mm = 0.0014 m   [ 1 mm = 0.001 m]

Final height of the ball (h₂) = 15 mm = 0.015 m

Now, from conservation of energy principle, energy can neither be created nor be destroyed but converted from one form to another.

Here, the kinetic energy of the ball is converted to gravitational potential energy of the ball after reaching the final height.

Change in kinetic energy is given as:

\Delta KE=\frac{1}{2}m(v_f^2-v_i^2)\\Where\ v_f\to Final\ velocity\\v_i\to Initial\ velocity

As it just touches the 15 mm high roof, the final velocity will be zero. So,

v_f=0\ m/s.

Now, the change in kinetic energy is equal to:

\Delta KE = \frac{1}{2}\times 0.130\times v^2\\\\\Delta KE = 0.065v^2

Change in gravitational potential energy = Final PE - Initial PE

So,

\Delta U=mg(h_f-h_i)\\\\\Delta U=0.130\times 9.8\times (0.015-0.0014)\\\\\Delta U=0.017\ J                    [ g = 9.8 m/s²]

Now, Change in KE = Change in PE

0.065v^2=0.017\\\\v=\sqrt{\frac{0.017}{0.065}}\\\\v=0.5114\ m/s\\\\1\ m=1000\ mm\\\\So,0.5114\ m=511.4\ mm\\\\\therefore v=511.4\ mm/s

Therefore, the initial velocity is 0.5114 m/s or 511.4 mm/s

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