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3 years ago

7

A Carnot air conditioner operates between an indoor temperature of 20°C and an outdoor temperature of 39°C. How much energy does

it need to remove 2000 J of heat from the interior of the house?A. 340 J
B. 780 J
C. 105 J
D. 130 J
E. 520 J

1 answer:

Gelneren [198K]3 years ago

8 0

Answer:

D. 130 J

Explanation:

The coefficient of performance for a machine that is being used to cool, is given by:

$COP=\frac{Q_C}{W}=\frac{T_C}{T_H-T_C}$

Here $Q_C$ is the heat removed from the cold reservoir, W is the work required, that is, the energy required to remove the heat from the interior of the house, $T_C$ is the cold temperature and $T_H$ is the hot temperature. Recall use absolutes temperatures( $273.15+^\circ C$ ). Replacing and solving for W:

$W=Q_c\frac{T_H-T_C}{T_C}\\W=2000J\frac{312.15K-293.15K}{293.15K}\\W=129.63J$

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Answer:

After 1 sec = 4.9 m

After 2 sec = 19.6 m

After 3 sec = 44.1 m

After 4 sec = 78.4 m

After 5 sec = 122.5 m

Explanation:

After 1 sec:

<em>u=0m/s t=1 s a=9.8m/s²</em>

s = ut + (1/2)at²

=0(1) + (1/2)(9.8)(1²) = 4.9m

After 2 sec:

<em>u=0m/s t=2 s a=9.8m/s²</em>

s = ut + (1/2)at²

=0(2) + (1/2)(9.8)(2²) = 19.6m

After 3 sec:

<em>u=0m/s t=3 s a=9.8m/s²</em>

s = ut + (1/2)at²

=0(3) + (1/2)(9.8)(3²) = 44.1m

After 4 sec:

<em>u=0m/s t=4 s a=9.8m/s²</em>

s = ut + (1/2)at²

=0(4) + (1/2)(9.8)(4²) = 78.4m

After 5 sec:

<em>u=0m/s t=5 s a=9.8m/s²</em>

s = ut + (1/2)at²

=0(5) + (1/2)(9.8)(5²) = 122.5m

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2 years ago

Energy from the Sun travels to Earth as ______. a. mechanical energy a. mechanical energy b. chemical energy c. radiant energy d

shepuryov [24]

Answer:

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Explanation:

Radiant energy is energy that travels by waves or particles, particularly electromagnetic radiation such as heat or x-rays.

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Answer:

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3 years ago

When a golf club hits a 0.0459 kg ball at rest, it exerts a 2380 N force for 0.00100 s. What is the speed of the ball afterwards

aksik [14]

Answer:

51.85m/s

Explanation:

Given parameters:

Mass of ball = 0.0459kg

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Time taken = 0.001s

Unknown:

Speed of the ball afterwards = ?

Solution:

To solve this problem, we use Newton's second law of motion:

F = m x $\frac{v - u}{t}$

F is the force

m is the mass

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8 0

3 years ago

Read 2 more answers

A 70.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 33.5 m/s. A second person, with a mass of 55.0 kg, c

saw5 [17]

Answer:

The final velocity of the thrower is $\bf{3.88~m/s}$ and the final velocity of the catcher is $\bf{0.029~m/s}$ .

Explanation:

Given:

The mass of the thrower, $m_{t} = 70~Kg$ .

The mass of the catcher, $m_{c} = 55~Kg$ .

The mass of the ball, $m_{b} = 0.0480~Kg$ .

Initial velocity of the thrower, $v_{it} = 3.90~m/s$

Final velocity of the ball, $v_{fb} = 33.5~m/s$

Initial velocity of the catcher, $v_{ic} = 0~m/s$

Consider that the final velocity of the thrower is $v_{ft}$ . From the conservation of momentum,

$&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s$

Consider that the final velocity of the catcher is $v_{fc}$ . From the conservation of momentum,

$&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s$

Thus, the final velocity of thrower is $3.88~m/s$ and that for the catcher is $0.029~m/s$ .

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2 years ago