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bazaltina [42]
4 years ago
15

A windfarm produces 322 MW of power.

Physics
1 answer:
ryzh [129]4 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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The velocity of a moving body increases from 10m/s to 15 m/s in 5 sec . calculate its acceleration ​
Gnoma [55]

Answer:

v=u+at

15=10+a x 5

15-10= 5a

5= 5a

a=1 m/sec^2

4 0
3 years ago
If a baseball pitch leaves the pitcher's hand horizontally at a velocity of 150 km/h by what percent will the pull of gravity ch
Slav-nsk [51]
<span>0.52% First, let's convert that speed into m/s. 150 km/h * 1000 m/km / 3600 s/h = 41.667 m/s Now let's see how much time gravity has to work on the ball. Divide the distance by the speed. 18 m / 41.667 m/s = 0.431996544 s Now multiply that time by the gravitational acceleration to see what the vertical component to the ball's speed that gravity adds. 0.431996544 s * 9.8 m/s^2 = 4.233566131 m/s Use the pythagorean theorem to get the new velocity of the ball. sqrt(41.667^2 + 4.234^2) = 41.882 m/s Finally, let's see what the difference is (41.882 - 41.667)/41.667 = 0.005159959 = 0.5159959% Rounding to 2 figures, gives 0.52%</span>
8 0
3 years ago
Which type of mixture could this illustration represent?
Oksanka [162]

Answer:

A homogeneous mixture

Explanation:

4 0
3 years ago
Read 2 more answers
Question 4 of 10 A student measures the time it takes for two reactions to be completed. Reaction A is completed in 39 seconds,
Delicious77 [7]
I think the answer is d but I’m not sure
6 0
3 years ago
How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe
Rus_ich [418]

Explanation:

(a)  For an isothermal process, work done is represented as follows.

             W = -nRT ln(\frac{V_{2}}{V_{1}})

Putting the given values into the above formula as follows.

        W = -nRT ln(\frac{V_{2}}{V_{1}})

             = - 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})

             = -12280.82 \times ln (0.09)

             = -12280.82 \times -2.41

             = 29596.78 J

or,         = 29.596 kJ       (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

         W = \frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}

              = \frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}

              = \frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}

              = 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c)   We know that for an isothermal process,

               P_{1}V_{1} = P_{2}V_{2}

or,       P_{2} = \frac{P_{1}V_{1}}{V_{2}}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})

                    = 11 atm

Hence, the required pressure is 11 atm.

(d)   For adiabatic process,  

          P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}

or,       P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}

                    = 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0
4 years ago
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