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Oksana_A
1 month ago
14

The Palo Verde nuclear power generator of Arizona has three reactors that have a combined generating capacity of 3.937×109 W . H

ow many years would it take the Palo Verde nuclear power generator to produce the same amount of energy as the Sun does in one day? The energy generation rate, or luminosity, of the Sun is 3.839×1026 W
Physics
1 answer:
Softa [2.9K]1 month ago
8 0

Answer:

t = 2.68 x 10¹⁴ years

Explanation:

First, we need to determine the energy the Sun generates in a single day.

Energy = Power * Time

Energy produced by the Sun in one day = (3.839 x 10²⁶ W)(1 day)(24 hr/1 day)(3600 s/1 hr)

Energy produced by the Sun in one day = 3.32 x 10³¹ J

Next, to find out how long the nuclear power generator would take to match this energy output, in years, we have:

Energy of power generator = Energy of the Sun in one day = 3.32 x 10³¹ J

3.32 x 10³¹ J = Power * Time

3.32 x 10³¹ J = (3.937 x 10⁹ W)(t years)(365 days/1 year)(24 hr/1 day)(3600 s/ 1 hr)

t = 3.32 x 10³¹ / 1.24 x 10¹⁷

t = 2.68 x 10¹⁴ years

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5 0
1 month ago
Observe: Up until now, all the problems you have solved have involved converting only one unit. However, some conversion problem
Softa [2943]

Response:

It is important to note that for units like time and angles, the transformations aren't based on ten, as shown below:

        60 s = 1 min

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Hence, for this conversion, special attention is needed

the length conversion is based on ten

Clarification:

In numerous problems, the units utilized undergo transformations via equations into alternate units referred to as derived units, and the transformation of these derived units generally results from the multiplication of the transformations of the basic units.

For instance, in the context of velocity, its derived unit is expressed as m / s; therefore, the transformation operates similarly to that of length and time, meaning if they are multiplied in the equation, it continues to be multiplied, and if divided, it remains divided.

It is important to note that for units like time and angles, the transformations aren't based on ten, as shown below:

        60 s = 1 min

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        24 h = 1 day

Hence, for this conversion, special attention is needed

the length conversion is based on ten

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7 0
6 days ago
For each of the motions described below, determine the algebraic sign (+, -, or 0) of the velocity and acceleration of the objec
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Part A: -,+

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Part B: +, -

The ball ascends, which determines the positive direction of the velocity. Hence, the velocity is positive.

Once the ball is thrown, the only force acting on it is gravity, opposing the ball’s ascent and causing it to descend. This means the acceleration is directed downwards, resulting in a negative sign.

Part C: 0, -

Throughout the ball's trajectory from the moment it's thrown until it drops to the ground, gravity constantly exerts downward acceleration (-).

After the throw, the ball's velocity will decline due to gravity. When it reaches a velocity of 0, it achieves its peak height. At this specific moment, the ball begins to descend again under the influence of gravity. However, at the peak height, the ball's velocity is 0.

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1 month ago
Which of the following statements about ycarrier(x,t) is correct?
serg [3469]

Answer: Option D: indicates rapid travel with slow oscillation.

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8 0
15 days ago
Water drops fall from the edge of a roof at a steady rate. a fifth drop starts to fall just as the first drop hits the ground. a
Keith_Richards [3146]

The height from which the drops fall is 3.57m

Assuming the drops fall at a rate of 1 drop every t seconds. The initial drop takes 5t seconds to hit the ground. The second drop reaches the bottom of the 1.00 m window in 4t seconds, while the third drop reaches the top of the window in 3t seconds.

Calculate the distances covered by the second and third drops s₂ and s₃, which begin from rest at the building's roof.

s_2=\frac{1}{2} g(4t)^2=8gt^2\\ s_3=\frac{1}{2} g(3t)^2=(4.5)gt^2

The height of the window s is described by,

s=s_2-s_3\\ (1.00 m)=8gt^2-4.5gt^2=3.5gt^2\\ t^2=\frac{1.00 m}{(3.5)(9.8m/s^2)} =0.02915s^2

The first drop has reached the base after taking 5t seconds.

The roof height h represents the distance covered by the first drop and is expressed as,

h=\frac{1}{2} g(5t)^2=\frac{25t^2}{2g} =\frac{25(0.02915s^2)}{2(9.8m/s^2)} =3.57 m

the height of the roof remains at 3.57 m



8 0
1 month ago
Read 2 more answers
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