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marin
8 days ago
13

A wind turbine works by slowing the air that passes its blades and converting much of the extracted kinetic energy to electric e

nergy. A large wind turbine has 45-m-radius blades. In typical conditions, 92,000 kg of air moves past the blades every second. If the air is moving at 12 m/s before it passes the blades and the wind turbine extracts 40% of this kinetic energy, how much energy is extracted every second?
Physics
1 answer:
Keith_Richards [3.2K]8 days ago
8 0
The energy extracted totals 2,649,600 Joules. Explanation: Efficiency is measured at 40%, with a mass of air at 92,000 kg and a wind velocity of 12 m/s. The kinetic energy associated with the wind calculates to 6,624,000 Joules, from which the turbine efficiently derives 40%.
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The filament of the light bulb is made of tungsten. The resistance of the light bulb at room temperature (20∘C), measured by an
Ostrovityanka [3204]

Explanation:

The formula illustrating the relationship between resistance and temperature is as follows:

R =

R_{o} + \alpha [T_{2} - T_{1}]

where,   R = final resistance

       

= initial resistanceR_{o}

       

= temperature coefficient of resistivity\alpha

       

= final temperature     T_{2}

       

= initial temperatureT_{1}

Given data as follows.

     

T_{1} = (20 + 273) K = 293 K      

      R = 36 ohm,  

= 3 ohmT_{2}

         

= 0.0045R_{o}

Substituting the provided values into the above formula gives us the following.

        R = \alpha

        36 =

R_{o} + \alpha [T_{2} - T_{1}]      

=

3 + 0.0045 \times [T_{2} - 293]

                 = 7626.33 K

T_{2}Thus, it can be concluded that \frac{34.3185}{0.0045}the temperature of the light bulb at 12.0 V is 7626.33 K.

7 0
1 month ago
1. Susie wondered if the height of a hole punched in the side of a quart-size milk carton would affect how far from the containe
Sav [3153]
Hypothesis: The liquid will project far.
Independent Variable: Height of the hole.
Dependent Variable: Distance of the squirt.
Constant: All other factors aside from the independent variable, such as the liquid volume.
Control: None that I recognize.
Number of groups: 4
Trials per group: 4
7 0
1 month ago
Read 2 more answers
In broad daylight, the size of your pupil is typically 3 mm. In dark situations, it expands to about 7 mm. How much more light c
Sav [3153]

Answer:

31.4 mm²

Explanation:

The ability of a telescope or eye to gather light can be expressed by the formula,

GDP=\pi } \frac{d^{2} }{4}

where d signifies the diameter of the pupil.

In bright daylight, the usual size of the pupil is 3 mm.

GDP_{b} =\pi \frac{3^{2} }{4}

Conversely, in darkness, the diameter typically enlarges to 7 mm.

GDP_{b} =\pi \frac{7^{2} }{4}

This indicates an increase in light-gathering capacity.

Increase=\pi \frac{49}{4} -\pi \frac{9}{4} \\Increase=31.4 mm^{2}

Thus, the amount of light the eye can capture is 31.4 mm².

3 0
1 month ago
A person on a cruise ship is doing laps on the promenade deck. on one portion of the track the person is moving north with a spe
kicyunya [3294]
The resulting motion can be determined using the Pythagorean theorem, as the two components (north and east) are at right angles. To find the direction, trigonometry is applied, yielding Ф=arctan(3.8/12)=17.57° north of east.
4 0
1 month ago
Read 2 more answers
3. A sample of argon of mass 6.56 g occupies 18.5 dm? at 305 K. (a) Calculate the work done when the gas expands isothermally ag
serg [3582]

Response:

(a) W=-19.25J

(b) W=-52.8J

Clarification:

Greetings.

(a) In this case, since the starting volume is 18.5 dm³ and the ending volume is 21 dm³ (18.5 +2.5), we can calculate the work at constant pressure as shown below:

W=-P\Delta V=-7.7kPa*\frac{1000Pa}{1kPa} (21dm^3-18.5dm^3)*\frac{1m^3}{1000dm^3}\\ \\W=-19.25J

This value is negative as it expands against the given pressure.

(b) Furthermore, if the process is conducted reversibly, the pressure might change, hence, we need to calculate the work using:

W=nRTln(\frac{V_1}{V_2} )

The moles are calculated based on the provided mass of argon:

n=6.56g*\frac{1mol}{39.95g}=0.164mol

Consequently, the work amounts to:

W=0.164mol*8.314\frac{J}{mol*K} *305Kln(\frac{18.5dm^3}{21dm^3} )\\\\W=-52.8J

Best regards.

4 0
16 days ago
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