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Natasha2012 [34]
3 years ago
8

A power plant burns coal at 830 K, and it exhaust to air at 288 K. If it runs at the Carnot efficiency, how much heat must it ex

haust into the air to produce 230,000 J of work?
Physics
2 answers:
statuscvo [17]3 years ago
8 0

Correct Answer:

122000 J

ElenaW [278]3 years ago
7 0

Answer:

122241.02 J

Explanation:

Work Done = 230000 J

T_H=\text {High Temperature Reservoir}=830\ K\\T_L=\text {Low Temperature Reservoir}=288\ K\\\text{In case of Carnot Cycle}\\\text{Efficiency}=\eta\\\eta=1-\frac{T_L}{T_H}\\\Rightarrow \eta =1-\frac{288}{830}\\\Rightarrow \eta=0.65\\

\eta=\frac{\text{Work}}{\text{Heat}}\\\Rightarrow \eta=\frac{W}{Q_H}\\\\\Rightarrow 0.65=\frac{230000}{Q_H}\\\Rightarrow Q_H=\frac{230000}{0.65}\\\Rightarrow Q_H=352214.02

Q_L=Q_H-W\\\Rightarrow Q_L=352214.02-230000\\\Rightarrow Q_L=122241.02\ J

∴Heat exhausted into the air to produce 230,000 J of work is 122241.02 Joule

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The auditory nerve carries an electrical signal to the brain, which turns it into a sound that we recognize and understand.
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3 years ago
a trampoline launches a 50kg person 2m into the air. if the springs push with 1960N of force, how much displacement was there in
Lina20 [59]

Answer: 0.5 m

Explanation:

Given

Mass of the person is m=50\ kg

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Force experience by springs is F=1960\ N

Here, the work done on displacing the springs is equivalent to the Potential energy acquired by the person i.e.

\Rightarrow F\cdot x=mgh\quad [\text{x=displacement of the trampoline}]\\\\\text{Insert the values}\\\\\Rightarrow x=\dfrac{50\times 9.8\times 2}{1960}\\\\\Rightarrow x=\dfrac{980}{1960}\\\\\Rightarrow x=0.5\ m

6 0
3 years ago
Read 2 more answers
A baseball is hit that just goes over a wall that is 45.4m high. If the baseball is traveling at 46.2 m/s at an angle of 32.7° b
mario62 [17]

Answer:

54.9 m/s at 44.9 degrees

Explanation:

If the ball has a total velocity of 46.2 m/s, at an angle of -32.7 degrees, we can decompose its speed into its horizontal and vertical components.

Vx = V * cos(a) = 46.2 * cos(-32.7) = 38.9 m/s

Vy = V * sin(a) = 46.2 * sin(-32.7) = -25 m/s

SInce there is no force on the horizontal direction (omitting air drag), we can assume constant horizontal speed.

Since a ball thrown is at free fall, only affected by gravity (omitting air drag), we can say it is affected by constant acceleration, therefore we can use

Y(t) = Y0 + Vy0 *t + 1/2 * a * t^2

We consider t=0 as the moment when the ball was hit, so in this case Y0 = 1 m

If we take the first derivative of the equation of position, we get the equation for speed

V(t) = Vy0 + a * t

We know that being t2 the moment the ball goes over the wall

V(t2) = -25 m/s

Y(t2) = 45.4 m

So:

45.4 = 1 + Vy0 * t2 + 1/2 * a * t2^2

-25 = Vy0 + a * t2

Then:

Vy0 = -25 - a * t2

So:

45.4 = 1 + (-25 - a * t2) * t2 + 1/2 * a * t2^2

0 = -44.4 - 25 * t2 - 1/2 * a * t2^2

a = -9.81 m/s^2

0 = -44.4 - 25 * t2 + 4.9 * t2^2

Solving this quadratic equation we get:

t1 = -1.39 s

t2 = 6.5 s

Since we are looking for a positive value we disregard t1.

Now we can obtain Vy0:

Vy0 = -25 + 9.81 * 6.5 = 38.76 m/s

Since horizontal speed is constant Vx0 = 38.9 m/s

By Pythagoras theorem we obtain the value of the initial speed:

V0 = \sqrt{Vx0^2 + Vy0^2} = \sqrt{38.9^2 + 38.76^2} = 54.9 m/s

The angle is in the the first quadrant because both comonents ate positive, so: 0 < a < 90

a = atan(Vy0/Vx0) = 44.9 degrees

5 0
3 years ago
For questions 1 - 4 complete the representations for the four patterns below. Provide the mathematical
larisa [96]

Answer:

Explanation:

1

2

3

6 0
3 years ago
Please help and I don't mean to sound rude but, ONLY ANSWER IF YOUR GOING TO DO ALL 4 QUESTIONS
anzhelika [568]

Answer:

for first question is 2

for second question 1

for third question 2

for forth question 1

Explanation:

i hope i helped

6 0
2 years ago
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