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Anvisha [2.4K]
3 years ago
12

A project is located in an area with a demand-response program and on a site that has enough room for a wind-turbine to allow fo

r on-site renewable energy. If the project implements both of these strategies, which of the following will occur?
A. The project's energy costs will decrease B. The project's renewable energy production will decrease C. The project's energy demands will decrease D. the project minimum energy performance will decrease
Physics
1 answer:
Ksju [112]3 years ago
5 0

Answer:

A. The project's energy costs will decrease

Explanation:

Since the project is located in an area with a demand-response program and on a site that has enough room for a wind-turbine to allow for on-site renewable energy.

Hence, the project's energy costs will decrease very well because it's implementing both of these strategies;

- Area with demand-response program.

- On-site renewable energy.

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A car is moving at a speed of 50 km/hour.the driver takes her foot off the accelerator and the car coasts to stop.why?
vodka [1.7K]
Because of the opposing forces on the car like friction because there is no more acceleration 
3 0
4 years ago
What is the adaptation to the arctic fox
arsen [322]
Animal Adaptation 1 Adaptation 2 Arctic Fox It's thick fur and fluffy tail help it survive in it's harsh habitat. Their small, pointy ears can hear their prey moving around in underground tunnels. An Arctic fox's fur changes colors with the seasons of the year. The Arctic Fox has many unique adaptations.
7 0
3 years ago
"a horn emits a frequency of 1000 HZ. A 14 m/s wind is blowing toward a listener. What is the frequency of the sound heard by th
Semenov [28]

Answer:

 f_L = 1000 Hz

Explanation:

given,

speed of wind, = 14 m/s

frequency of horn,f_s = 1000 Hz

speed of sound,V = 344 m/s

frequency heard by the listener

using Doppler effect

f_L = \dfrac{v+v_L}{v+v_s}f_s

f_L is the frequency of the sound heard by the listener

f_s is the frequency of sound emitted by the listener

V is the speed of sound

v_L is the speed of listener

v_s is the speed of source

now,

considering the frame of reference in which wind is at rest now, both listener and the source will be moving at 14 m/s

 f_L = \dfrac{v+14}{v+14}\times 1000

now on solving we will get

 f_L = 1000 Hz

hence, the frequency heard by the listener is equal to  1000 Hz

 

7 0
3 years ago
The earth has a vertical electric field at the surface, pointing down, that averages 100 N/C. This field is maintained by variou
zlopas [31]

Answer:

q = 3.6 10⁵  C

Explanation:

To solve this exercise, let's use one of the consequences of Gauss's law, that all the charge on a body can be considered at its center, therefore we calculate the electric field on the surface of a sphere with the radius of the Earth

          r = 6 , 37 106 m

          E = k q / r²

          q = E r² / k

          q = \frac{100 \ (6.37 \ 10^6)^2}{9 \ 10^9}

          q = 4.5 10⁵ C

Now let's calculate the charge on the planet with E = 222 N / c and radius

           r = 0.6 r_ Earth

           r = 0.6 6.37 10⁶ = 3.822 10⁶ m

           E = k q / r²

            q = E r² / k

            q = \frac{222 (3.822 \ 10^6)^2}{ 9 \ 10^9}

            q = 3.6 10⁵  C

4 0
3 years ago
Starting from rest, a small block of mass m slides frictionlessly down a circular wedge of mass M and radius R which is placed o
guapka [62]

Answer:

Part a)

V = \sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}

v = \frac{M}{m}\sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}

Part b)

Since on the block wedge system there is no external force in horizontal direction so the Center of mass will not move in horizontal direction but in vertical direction it will move

so displacement in Y direction is given as

y_{cm} = \frac{mR}{m + M}

Explanation:

PART A)

As we know that there is no external force on the system of two masses in horizontal direction

So here the two masses will have its momentum conserved in horizontal direction

So we have

mv + MV = 0

Also we know that here no friction force on the system so total energy will always remains conserved

So we have

\frac{1}{2}mv^2 + \frac{1}{2}MV^2 = mgR

now we have

\frac{1}{2}m(\frac{MV}{m})^2 + \frac{1}{2}MV^2 = mgR

\frac{1}{2}MV^2(\frac{M}{m} + 1) = mgR

so we have

V = \sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}

and another block has speed

v = \frac{M}{m}\sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}

Part b)

Since on the block wedge system there is no external force in horizontal direction so the Center of mass will not move in horizontal direction but in vertical direction it will move

so displacement in Y direction is given as

y_{cm} = \frac{mR}{m + M}

7 0
3 years ago
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