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igor_vitrenko [27]
3 years ago
6

A solar cell, 3.0 cm square, has an output of 350 mA at 0.80 V when exposed to full sunlight. A solar panel that delivers close

to 1.0 A of current at an emf of 120 V to an external load is needed. How many cells will you need to create the panel? How big a panel will you need, and how should you connect the cells to one another? How can you optimize the output of your solar panel?

Physics
1 answer:
Vesna [10]3 years ago
8 0

Answer:

You will need 450 cells (3 cm each) to meet the voltage/current requirement.

The panel must be 3 cells in one side, by 150 cell in another side. 1350 cm^2 or 0.135 m^2. They must be connected 3 in row in parallel (to add current), then each of the former group must be connected in series to meet the voltage, so it would be 150 rows of connected in series.

The panel can be optimized using a voltage inverter, to convert current to voltage. In this way, less cells can be used achieving the same output specs.

Explanation:

To meet the voltage:

120 [v] required voltage

0.8 [v] voltage of each cell

\frac{120}{0.8} =150[v]\\

So we need 150 cells in series for the voltage.

To meet the current

1.0 [A]      Required current

350[mA]=0.35[A] cell current

1/0.35=3 cell So we need 3 cells in parallel to add the currents and meet the requirement.

See the attached figure

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QC In ideal flow, a liquid of density 850 kg / m³ moves from a horizontal tube of radius 1.00cm into a second horizontal tube of
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The  volume flow rates for ∆P is 6.81m³/s .

<h3>What is pressure?</h3>

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4 0
1 year ago
A 600kg car is moving at 5 m/s to the right and elastically collided with a stationary 900 kg car. What is the velocity of the 9
11111nata11111 [884]

Answer:

\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}

Explanation:

It says “Momentum before the collision is equal to momentum after the collision.” Elastic Collision formula is applied to calculate the mass or velocity of the elastic bodies.

m_{1} v_{1}=m_{2} v_{2}

\mathrm{m}_{1} \text { and } \mathrm{m}_{2} \text { are masses of the object }

\mathrm{v}_{1} \text { velocity before the collision }

\mathrm{v}_{2} \text { velocity after the collision }

\mathrm{m}_{1}=600 \mathrm{kg}

\mathrm{m}_{2}=900 \mathrm{kg}

\text { Velocity before the collision } v_{1}=5 \mathrm{m} / \mathrm{s}

600 \times 5=900 \times v_{2}

3000=900 \times v_{2}

\mathrm{v}_{2}=\frac{3000}{900}

\mathrm{v}_{2}=3.3 \mathrm{m} / \mathrm{s}

\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}

5 0
3 years ago
A 500kg car skids to a stop at a traffic light, leaving behind a 18.25m skid mark as it comes to a rest. Assuming that the car i
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Answer:

Coefficient of friction will be 0.587

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We have given mass of the car m = 500 kg

Distance s = 18.25 m

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From second equation of motion

v^2=u^2+2as

0^2=14.5^2+2\times a\times 18.25

a=-5.76m/sec^2

We know that acceleration is given by

a=\mu g

5.76=\mu\times  9.81

\mu =0.587

So coefficient of friction will be 0.587

6 0
3 years ago
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