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3 years ago

I've got a test tomorrow morning help!

Physics

1 answer:

topjm [15]3 years ago

3 0

Answer:

The efficiency is just 0.016

Explanation:

The efficiency is given by the useful energy ÷ by the total energy.

1 W is the same that 1 joule per second

If you do the math:

1.6/100=0.016

You might be interested in

What does Newton's third law describe?

Dennis_Churaev [7]

A. Equal and opposite forces

Explanation:
History law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exert an equal and opposite force on object A

7 0

3 years ago

A single-story retail store wishes to supply all its lighting requirement with batteries charged by photovoltaic cells. The PV c

Umnica [9.8K]

Answer:

83.33% of the roof area will be occupied by the PV cells

Explanation:

Given the data in the question;

time-averaged lighting requirement $P_{lighting$ = 10 W/m²

the annual average solar irradiance $q_{solar$ = 150 W/m²

the PV efficiency η $_{pv$ = 10% = 0.1

battery charging/discharging efficiency η $_{battery$ = 80% = 0.8

we know that; Annual average power to the light = $P_{lighting$ × A $_{roof$

Now, the electrical power delivered by the solar cell battery system will be;

⇒ $q_{solar$ × A $_{pv$ × η $_{pv$ × η $_{battery$

$P_{lighting$ A $_{roof$ = $q_{solar$ × A $_{pv$ × η $_{pv$ × η $_{battery$

Such that;

A $_{pv$ = $P_{lighting$ A $_{roof$ / $q_{solar$ × A $_{pv$ × η $_{pv$ × η $_{battery$

A $_{pv$ / A $_{roof$ = $P_{lighting$ / $q_{solar$ × η $_{pv$ × η $_{battery$

so we substitute

A $_{pv$ / A $_{roof$ = 10 W/m² / [ 150 W/m² × 0.1 × 0.8 ]

A $_{pv$ / A $_{roof$ = 10 W/m² / 12 W/m²

A $_{pv$ / A $_{roof$ = 0.8333

A $_{pv$ / A $_{roof$ = (0.8333 × 100)%

A $_{pv$ / A $_{roof$ = 83.33%

Therefore, 83.33% of the roof area will be occupied by the PV cells.

7 0

3 years ago

PLZZ HELP!!! When a star is undergoing nuclear fusion what type of radiation will be found?

zimovet [89]

Answer:

I think the correct answer would be the third option. As the elements around the star begins to emit more and more electromagnetic radiation, the rocky materials are pulled in by the electromagnetic radiation. They are being drawn closer to the star and there would be a very high chance of a nuclear

Explanation:

6 0

3 years ago

A stone is thrown horizontally from 2.4m above the ground at 35m/s. The wall is 14m away and 1m high.At what height the stone wi

KIM [24]

The stone reaches the wall at a height of 1.62 m.

The stone lands at a point 24.5 m from the point of projection.

The stone is projected horizontally with a velocity u at a height h from the ground. The wall is located at a distance x from the point of projection. The stone takes a time t to reach the wall and in the same time the stone falls a vertical distance y.

The horizontal distance x is traveled with a constant velocity u.

$x=ut$

Calculate the time taken t.

$t=\frac{x}{u} \\ =\frac{14m}{35 m/s} \\ =0.40s$

The stone's initial vertical velocity is zero. It falls through a distance y in the time t under the action of acceleration due to gravity g.

$y=\frac{1}{2} gt^2\\ \frac{1}{2} (9.81m/s^2)(0.40s)^2\\ =0.784m$

The height h₁ of the stone above the ground when it reaches the wall is given by,

$h_1=h-y\\ =(2.4m)-(0.784m)\\ =1.616m=1.62m$

When the stone reaches the wall, its height from the ground is 1.62m.

The stone thus crosses over the wall, since the height of the wall is 1 m. It reaches the ground at a distance R from the point of projection. If the time taken by the stone to reach the ground is t₁, then,

$h=\frac{1}{2} gt_1^2$