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1 year ago

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A machine is used to convert heat energy into work energy, and is connected to a cooling system which keeps the machine at a con

stant temperature. If 2000 kJ of heat energy enter the machine, and the machine performs 700 kJ of work,
how much heat energy must still be removed by the cooling system?

Remember What the law of Conservation of energy says about the total energy before and after a transformation.

1 answer:

vampirchik [111]1 year ago

4 0

The amount of heat energy that must still be removed by the cooling system is 1300 kJ.

<h3>Conservation energy</h3>

The total energy before and after a transformation must be equal.

Based on this principle, the heat energy that must still be removed by the cooling system is calculated as follows;

Input heat energy = out put heat energy

E = h + w

2000 kJ = h + 700 kJ

1300 kJ = h

Thus, the amount of heat energy that must still be removed by the cooling system is 1300 kJ.

Learn more about heat energy here: brainly.com/question/7541718

#SPJ1

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Answer:

H = 3.9 m

Explanation:

mass (m) = 48 kg

initial velocity (initial speed) (U) = 8.9 m/s

final velocity (V) = 1.6 m/s

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find the height she raised her self to as she crosses the bar (H)

from energy conservation, the change in kinetic energy = change in potential energy

0.5m(V^{2} - [test]U^{2}[/tex]) = mg(H-h)

where h = initial height = 0 since she was on the ground

the equation becomes

0.5m(V^{2} - [test]U^{2}[/tex]) = mgH

0.5 x 48 x (1.6^{2} - [test]8.9^{2}[/tex]) = 48 x 9.8 x H

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3 years ago

The potential energy stored in the compressed spring of a dart gun, with a spring constant of 32.50 N/m, is 0.640 J. Find by how

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Answer:

A

$x = 0.198456 \ m$

B

$h = 1.3061 \ m$

C

$v = 5.06 \ m/s$

D

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Explanation:

Considering the first question

From the question we are told that

The spring constant is $k = 32.50 N/m$

The potential energy is $PE = 0.640 \ J$

Generally the potential energy stored in spring is mathematically represented as $PE = \frac{1}{2} * k * x^2$

=> $0.640= \frac{1}{2} * 32.50 * x^2$

=> $x = \sqrt{0.03938}$

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Considering the second question

From the question we are told that

The mass of the dart is m = 0.050 kg

Generally from the law of energy conservation

$PE = mgh$

=> $0.640 = 0.050 * 9.8 * h$

=> $h = 1.3061 \ m$

Considering the third question

The height at which the dart was fired horizontally is $H = 3.90\ m$

Generally from the law of energy conservation

$PE = KE$

Here KE is kinetic energy of the dart which is mathematical represented as

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

=> $0.640 = \frac{1}{2} * 0.050 * v^2$

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=> $v = 5.06 \ m/s$

Considering the fourth question

Generally the total time of flight of the dart is mathematically represented as

$t = \frac{ 2 * H }{g}$

=> $t = \frac{ 2 * 3.90 }{9.8 }$

=> $t = 0.7959 \ s$

Generally the horizontal distance from the equilibrium position to the ground is mathematically represented as

$d = v * t$

=> $d = 5.06 * 0.7959$

=> $d = 4.0273 \ m$

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