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marishachu [46]
2 years ago
13

A solar cell has an open circuit voltage value of 0.60 V with a reverse saturation current density of Jo = 3.9 × 10−9 A/m2 . The

temperature of the cell is 27◦C, the cell voltage is 0.52 V, and the cell area is 28 m2 . If the solar irradiation is 485 W/m2 , determine the power output and the efficiency of the solar cell.

Physics
1 answer:
xenn [34]2 years ago
4 0

Answer:

Explanation:

given data

ocv=0.6 V

Vmax= 0.52 v

J₀= 3.9*10⁻⁹

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A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's in
german

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

\Sigma F = -f = m\cdot a (Eq. 1)

Where:

f - Kinetic friction force, measured in newtons.

m - Mass of the box, measured in kilograms.

a - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight (W = m\cdot g) and uniform accelerated motion (v = v_{o}+a\cdot t), we expand the previous expression:

-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)

And the magnitude of the friction force exerted on the box is calculated by this formula:

f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right) (Eq. 1b)

Where:

W - Weight, measured in newtons.

g - Gravitational acceleration, measured in meters per square second.

v_{o} - Initial speed, measured in meters per second.

v - Final speed, measured in meters per second.

t - Time, measured in seconds.

If we know that W = 52.4\,N, g = 9.807\,\frac{m}{s^{2}}, v_{o} = 1.37\,\frac{m}{s}, v = 0\,\frac{m}{s} and t = 2.8\,s, the magnitud of the kinetic friction force exerted on the box is:

f = -\left(\frac{52.4\,N}{9.807\,\frac{m}{s^{2}} } \right)\cdot \left(\frac{0\,\frac{m}{s}-1.37\,\frac{m}{s}  }{2.8\,s} \right)

f = 2.614\,N

The magnitude of the friction force exerted on the box is 2.614 newtons.

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3 years ago
You have a meteorite sample and you decide to use the uranium-235/lead-207 system to date it. After analysis, you find that it h
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Originally there must been

1,4775E6 + 2.25E4 = 147.75E4 + 2.25E4 = 150E4 present at start

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2 years ago
How do you find the force of gravity?
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1<span>Define the equation for the force of gravity that attracts an object, <span>Fgrav = (Gm1m2)/d2</span>
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7 0
3 years ago
A simple pendulum of length of 1.37 m and mass of 6.66 kg is given an initial speed of 2.85 m/s at its equilibrium position. Det
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Answer:

2.35 s

Explanation:

The period of a simple pendulum is expressed as;

                                T = 2π\sqrt{\frac{L}{g} }

Where

T is the period in seconds

L is the length in metres

g is acceleration due to gravity

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                                 T = 2.349 s

                                 T = 2.35 s

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