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marishachu [46]
3 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]3 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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An alternating source drives a series RLC circuit with an emf amplitude of 6.34 V, at a phase angle of +33.3°. When the potentia
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Answer:

V_{L}  =- 8.78v

Explanation:

E^{2} _{m}= V^{2} _{R} + (V_{L} -V _{C} )^{2}

OR  V_{L} = V_{c}  + \sqrt{E^{2}{m}  - V^{2} {R}  }

where VR = E{m}cosФ

V_{L} = V_{c}  + \sqrt{E^{2}_{m} - E^{2}_{m} cos^{2}   }Ф

V_{L} = V_{c} + {E_{m} \sqrt{1 - cos^{2} }  }Ф

V_{L} = V_{c} + E_{m}sinФ

substituting the given values

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The bigclaw snapping shrimp shown in (Figure 1) is aptly named--it has one big claw that snaps shut with remarkable speed. The p
leva [86]

1) 1.86\cdot 10^6 rad/s^2

2) 2418 rad/s

3) 27000 m/s^2

4) 36.3 m/s

Explanation:

1)

The angular acceleration of an object in rotation is the rate of change of angular velocity.

It can be calculated using the following suvat equation for angular motion:

\theta=\omega_i t +\frac{1}{2}\alpha t^2

where:

\theta is the angular displacement

\omega_i is the initial angular velocity

t is the time

\alpha is the angular acceleration

In this problem we have:

\theta=90^{\circ} = \frac{\pi}{2}rad is the angular displacement

t = 1.3 ms = 0.0013 s is the time elapsed

\omega_i = 0 is the initial angular velocity

Solving for \alpha, we find:

\alpha = \frac{2(\theta-\omega_i t)}{t^2}=\frac{2(\pi/2)-0}{0.0013}=1.86\cdot 10^6 rad/s^2

2)

For an object in accelerated rotational motion, the final angular speed can be found by using another suvat equation:

\omega_f = \omega_i + \alpha t

where

\omega_i is the initial angular velocity

t is the time

\alpha is the angular acceleration

In this problem we have:

t = 1.3 ms = 0.0013 s is the time elapsed

\omega_i = 0 is the initial angular velocity

\alpha = 1.86\cdot 10^6 rad/s is the angular acceleration

Therefore, the final angular speed is:

\omega_f = 0 + (1.86\cdot 10^6)(0.0013)=2418 rad/s

3)

The tangential acceleration is related to the angular acceleration by the following formula:

a_t = \alpha r

where

a_t is the tangential acceleration

\alpha is the angular acceleration

r is the distance of the point from the centre of rotation

Here we want to find the tangential acceleration of the tip of the claw, so:

\alpha = 1.86\cdot 10^6 rad/s is the angular acceleration

r = 1.5 cm = 0.015 m is the distance of the tip of the claw from the axis of rotation

Substituting,

a_t=(1.86\cdot 10^6)(0.015)=27900 m/s^2

4)

Since the tip of the claw is moving by uniformly accelerated motion, we can find its final speed using the suvat equation:

v=u+at

where

u is the initial linear speed

a is the tangential acceleration

t is the time elapsed

Here we have:

a=27900 m/s^2 (tangential acceleration)

u = 0 m/s (it starts from rest)

t = 1.3 ms = 0.0013 s is the time elapsed

Substituting,

v=0+(27900)(0.0013)=36.3 m/s

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