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bazaltina [42]
3 years ago
15

A windfarm produces 322 MW of power.

Physics
1 answer:
ryzh [129]3 years ago
7 0

The power added by all 75 of them is (539 - 322) = 217 MW .

The AVERAGE power of each improved turbine is 

                   ( 217 / 75 ) =  2.8933 MW  =  2,893.3 KW/unit .

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6X-6=9<br><br> Solve for X<br><br> Round to TWO decimal places
Brums [2.3K]

Answer:

X=2.50

Explanation:

6x-6=9

6x= 9+6

X=15/6

X= 2.50

8 0
3 years ago
One of the largest planes ever to fly, and the largest to fly frequently, is the Ukrainian-built Antonov An-124 Ruslan. The grav
Lerok [7]

Answer:

m = 236212 [kg]

Explanation:

The potential energy can be determined by means of the product of mass by gravity by height. In this way, we have the following equation.

P=m*g*h\\

where:

P = potential energy = 3360000000 [J]

m = mass [kg]

g = gravity acceleration = 9.81 [m/s²]

h = elevation = 1450 [m]

Now, we can clear the mass from the equation above:

3360000000=m*9.81*1450\\m = 236212 [kg]

4 0
3 years ago
A 2,000 kg rocket is launched 12 km straight up at a constant acceleration into the sky at which point the rocket is travelling
hodyreva [135]

Answer:

797700000 J

Explanation:

From the question,

The work done by the rocket, is given as,

W = Ek+Ep............. Equation 1

Where Ek and Ep are the potential and the kinetic energy of the rocket respectively.

Ep = mgh............ Equation 2

Ek = 1/2mv²............. equation 3

Substitute equation 2 and equation 3 into equation 1

W = mgh+1/2mv².............. Equation 4

Where m = mass of the rocket, h = height, v = velocity of the rocket, g = acceleration due to gravity.

Given: m = 2000 kg, h = 12 km = 12000 m, v = 750 m/s, g = 9.8 m/s²

Substitute into equation 4

W = 2000(12000)(9.8)+1/2(2000)(750²)

W = 235200000+562500000

W = 797700000 J

4 0
3 years ago
at certain times the demand for electric energy is low and electric energy is used to pump water to a reservoir 45 m above the g
Readme [11.4K]

The mass of water that must be raised is 5.25\cdot 10^7 kg

Explanation:

Since the process is 70% efficiency, the power in output to the turbine can be written as

P_{out} = 0.70 P_{in}

where P_{in} is the power in input.

The power in input can be written as

P_{in} = \frac{W}{t}

where

W is the work done in lifting the water

t = 3 h = 10,800 s is the time elapsed

The work done in lifting the water is given by

W=mgh

where

m is the mass of water

g=9.8 m/s^2 is the acceleration of gravity

h = 45 m is the height at which the water is lifted

Combining the three equations together, we get:

P_{out} = 0.70 \frac{mgh}{t}

Where

P_{out} = 150 MW = 150\cdot 10^6 W

And solving for m, we find:

m=\frac{Pt}{0.70gh}=\frac{(1.50\cdot 10^6)(10800)}{(0.70)(9.8)(45)}=5.25\cdot 10^7 kg

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

3 0
3 years ago
3. The expression 0.62 x10^3 is equivalent to...
Korolek [52]

\\ \sf\longmapsto 0.62\times 10^3

\\ \sf\longmapsto 62\times 10^{-2}\times 10^3

\\ \sf\longmapsto 62\times 10^{-2+3}

\\ \sf\longmapsto 62\times 10^1

\\ \sf\longmapsto 62\times 10

\\ \sf\longmapsto 620

5 0
2 years ago
Read 2 more answers
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