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Misha Larkins [42]

3 years ago

8

A nuclear power plant operates at 79 percent of its maximum theoretical (Carnot) efficiency between temperatures of 700° and 330

°C. If the plant produces electric energy at the rate of 1.3 GW, how much exhaust heat is discharged per hour?

1 answer:

ololo11 [35]3 years ago

6 0

Answer:

Explanation:

T₁ = 700 + 273 = 973 k

T₂ = 330 + 273 = 603 k

Theoretical efficiency = T₁ - T₂ / T₁

= (973 - 603) / 973

= .38 OR 38%

Operating efficiency = .79 x 38

= 30.02 %

Heat input Q₁ , Heat output to sink Q₂ , conversion into power = Q₁ - Q₂

given Q₁ - Q₂ = 1.3 x 10⁹ W

efficiency = Q₁ - Q₂ / Q₁

Q₁ - Q₂ / Q₁ = 30.02 / 100

100Q₁ - 100Q₂ = 30.02Q₁

69.98 Q₁ = 100Q₂

Q₁ = 1.429 Q₂

Putting this in the relation

Q₁ - Q₂ = 1.3 x 10⁹ W

1.429Q₂ - Q₂ = 1.3 x 10⁹ W

.429Q₂ = 1.3 x 10⁹

Q₂ = 3.03 x 10⁹W

= 3.03 GW.

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50 points !! I need help asap.......Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to t

r-ruslan [8.4K]

1) At the top of the building, the ball has more potential energy

2) When the ball is halfway through the fall, the potential energy and the kinetic energy are equal

3) Before hitting the ground, the ball has more kinetic energy

4) The potential energy at the top of the building is 784 J

5) The potential energy halfway through the fall is 392 J

6) The kinetic energy halfway through the fall is 392 J

7) The kinetic energy just before hitting the ground is 784 J

Explanation:

1)

The potential energy of an object is given by

$PE=mgh$

where

m is the mass

g is the acceleration of gravity

h is the height relative to the ground

While the kinetic energy is given by

$KE=\frac{1}{2}mv^2$

where v is the speed of the object

When the ball is sitting on the top of the building, we have

$h=40 m$ , therefore the potential energy is not zero
$v=0$ , since the ball is at rest, therefore the kinetic energy is zero

This means that the ball has more potential energy than kinetic energy.

2)

When the ball is halfway through the fall, the height is

$h=20 m$

So, half of its initial height. This also means that the potential energy is now half of the potential energy at the top (because potential energy is directly proportional to the height).

The total mechanical energy of the ball, which is conserved, is the sum of potential and kinetic energy:

$E=PE+KE=const.$

At the top of the building,

$E=PE_{top}$

While halfway through the fall,

$PE_{half}=\frac{PE_{top}}{2}=\frac{E}{2}$

And the mechanical energy is

$E=PE_{half} + KE_{half} = \frac{PE_{top}}{2}+KE_{half}=\frac{E}{2}+KE_{half}$

which means

$KE_{half}=\frac{E}{2}$

So, when the ball is halfway through the fall, the potential energy and the kinetic energy are equal, and they are both half of the total energy.

3)

Just before the ball hits the ground, the situation is the following:

The height of the ball relative to the ground is now zero: $h=0$ . This means that the potential energy of the ball is zero: $PE=0$

The kinetic energy, instead, is not zero: in fact, the ball has gained speed during the fall, so $v\neq 0$ , and therefore the kinetic energy is not zero

Therefore, just before the ball hits the ground, it has more kinetic energy than potential energy.

4)

The potential energy of the ball as it sits on top of the building is given by

$PE=mgh$

where:

m = 2 kg is the mass of the ball

$g=9.8 m/s^2$ is the acceleration of gravity

h = 40 m is the height of the building, where the ball is located

Substituting the values, we find the potential energy of the ball at the top of the building:

$PE=(2)(9.8)(40)=784 J$

5)

The potential energy of the ball as it is halfway through the fall is given by

$PE=mgh$

where:

m = 2 kg is the mass of the ball

$g=9.8 m/s^2$ is the acceleration of gravity

h = 20 m is the height of the ball relative to the ground

Substituting the values, we find the potential energy of the ball halfway through the fall:

$PE=(2)(9.8)(20)=392 J$

6)

The kinetic energy of the ball halfway through the fall is given by

$KE=\frac{1}{2}mv^2$

where

m = 2 kg is the mass of the ball

v = 19.8 m/s is the speed of the ball when it is halfway through the fall

Substituting the values into the equation, we find the kinetic energy of the ball when it is halfway through the fall:

$KE=\frac{1}{2}(2)(19.8)^2=392 J$

We notice that halfway through the fall, half of the initial potential energy has converted into kinetic energy.

7)

The kinetic energy of the ball just before hitting the ground is given by

$KE=\frac{1}{2}mv^2$

where:

m = 2 kg is the mass of the ball

v = 28 m/s is the speed of the ball just before hitting the ground

Substituting the values into the equation, we find the kinetic energy of the ball just before hitting the ground:

$KE=\frac{1}{2}(2)(28)^2=784 J$

We notice that when the ball is about to hit the ground, all the potential energy has converted into kinetic energy.

Learn more about kinetic and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

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3 years ago

Q7) A box sliding with a velocity of 5 m/s accelerates at 2 m/s^2. How

grigory [225]

Answer:

The box displacement after 6 seconds is 66 meters.

Explanation:

Let suppose that velocity given in statement represents the initial velocity of the box and, likewise, the box accelerates at constant rate. Then, the displacement of the object ( $\Delta s$ ), in meters, can be determined by the following expression:

$\Delta s = v_{o}\cdot t+\frac{1}{2}\cdot a\cdot t^{2}$ (1)

Where:

$v_{o}$ - Initial velocity, in meters per second.

$t$ - Time, in seconds.

$a$ - Acceleration, in meters per square second.

If we know that $v_{o} = 5\,\frac{m}{s}$ , $t = 6\,s$ and $a = 2\,\frac{m}{s^{2}}$ , then the box displacement after 6 seconds is:

$\Delta s = 66\,m$

The box displacement after 6 seconds is 66 meters.

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2 years ago

Explain how the potential energy of two charged particles depends on the distance between the charged particles and on the magni

maks197457 [2]

Answer:

According to Coulomb's Law, the potential energy of two charged particles is directly proportional to the product of the two charges and inversely proportional to the distance between the charges

Explanation:

According to Coulomb's Law, the potential energy of two charged particles is directly proportional to the product of the two charges and inversely proportional to the distance between the charges. Since the potential energy of two charged particles is directly proportional to the product of the two charges, its magnitude increases as the charges of the particles increases. For like charges, the potential energy is positive(the product of the two alike charges must be positive) and since potential energy is inversely proportional to the distance between the charges therefore it decreases as the particles get farther apart . For opposite charges, the potential energy is negative(the product of the two opposite charges must be negative) and since potential energy is inversely proportional to the distance between the two charges, it becomes more negative as the particles get closer together.

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3 years ago

Which example best represents the use of creativity in a scientific inquiry

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3 years ago

Read 2 more answers

Suppose that in a lightning flash the potential difference between a cloud and the ground is 0.96×109 V and the quantity of char

Dvinal [7]

(a) $2.98\cdot 10^{10} J$

The change in energy of the transferred charge is given by:

$\Delta U = q \Delta V$

where

q is the charge transferred

$\Delta V$ is the potential difference between the ground and the clouds

Here we have

$q=31 C$

$\Delta V = 0.96\cdot 10^9 V$

So the change in energy is

$\Delta U = (31 C)(0.96\cdot 10^9 V)=2.98\cdot 10^{10} J$

(b) 7921 m/s

If the energy released is used to accelerate the car from rest, than its final kinetic energy would be

$K=\frac{1}{2}mv^2$

where

m = 950 kg is the mass of the car

v is the final speed of the car

Here the energy given to the car is

$K=2.98\cdot 10^{10} J$

Therefore by re-arranging the equation, we find the final speed of the car:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(2.98\cdot 10^{10})}{950}}=7921 m/s$

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3 years ago