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3 years ago

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An automotive air conditioner produces a 1-kW cooling effect while consuming 0.75 kW of power. What is the rate at which heat is

rejected from this air conditioner

1 answer:

Ipatiy [6.2K]3 years ago

8 0

Answer:

The rejected by the air conditioning system is 1.75 kilowatts.

Explanation:

A air conditioning system is a refrigeration cycle, whose receives heat from cold reservoir with the help of power input before releasing it to hot reservoir. The First Law of Thermodynamics describes the model:

$\dot Q_{L} + \dot W - \dot Q_{H} = 0$

Where:

$\dot Q_{L}$ - Heat rate from cold reservoir, measured in kilowatts.

$\dot Q_{H}$ - Heat rate liberated to the hot reservoir, measured in kilowatts.

$\dot W$ - Power input, measured in kilowatts.

The heat rejected is now cleared:

$\dot Q_{H} = \dot Q_{L} + \dot W$

If $\dot Q_{L} = 1\,kW$ and $\dot W = 0.75\,kW$ , then:

$\dot Q_{H} = 1\,kW + 0.75\,kW$

$\dot Q_{H} = 1.75\,kW$

The rejected by the air conditioning system is 1.75 kilowatts.

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A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m an

Ber [7]

Complete Question

A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t)−(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.10 m/s. What is the x-coordinate of the object when t = 10.0 s?

Answer:

The position of the object at t = 10s is $X = 38.3 \ m$

Explanation:

From the question we are told that

The acceleration along the x axis is $a_{x}t = -(0.0320\ m/s^3)(15.0 s- t)- (0.0320\ m/s^3)$

The position of the object at t = 0 is x = -14.0 m

The velocity at t = 0 s is $v_{0}x = 7.10 m/s$

Generally from the equation for acceleration along x axis we have that

$a_x = \frac{dV_{x}}{dt} = -0.032 (15- t)$

=> $\int\limits {dV_{x}} \, = \int\limits {-0.032(15- t)} \, dt$

=> $V_{x} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1$

At t =0 s and $v_{0}x = 7.10 m/s$

=> $7.10 = -0.032 [15(0) - \frac{(0)^2 }{2} ]+ K_1$

=> $K_1 = 7.10$

So

$\frac{dX}{dt} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1$

=> $\int\limits dX = \int\limits [-0.032 [15t - \frac{t^2 }{2} ]+ K_1] }{dt}$

=> $X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ K_1t +K_2$

At t =0 s and x = -14.0 m

$-14 = -0.032 [ 15\frac{0^2}{2} - \frac{0^3 }{6} ]+ K_1(0) +K_2$

=> $K_2 = -14$

So

$X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ 7.10 t -14$

At t = 10.0 s

$X = -0.032 [ 15\frac{10^2}{2} - \frac{10^3 }{6} ]+ 7.10 (10) -14$

=> $X = 38.3 \ m$

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3 years ago

What is transferred when a wave hits the beach

Degger [83]

A tsunami or a high tide wave
hoep this helped u!!

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3 years ago

Two long straight wires enter a room through a window. One carries a current of 3.0 ???? into the room while the other carries a

Tju [1.3M]

Answer:

$\begin{equation}\\\oint_LB.dl\\\end{equation}$ = -8πx $10^{-7}$

Explanation:

If you need calculate

$\begin{equation}\\\oint_L B.dl\\ \end{equation}$

You can use the Ampere's Law

$\begin{equation}\\\oint_L B.dl\\ \end{equation}$ = $I_{in}$ μ

Where $I_{in}$ : Current passing through the window

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μ = 4πx $10^{-7} T.m.A^{-1}$

Then

$\begin{equation}\\\oint_L B.dl\\ \end{equation}$ = (3-5)4πx $10^{-7}$

$\begin{equation}\\\oint_L B.dl\\ \end{equation}$ = -8πx $10^{-7}$

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3 years ago

A pendulum oscillates 60 times in 5 seconds. Find its time period.

Afina-wow [57]

Answer:

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so time period of the pendulum is 0.1 secs.

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3 years ago

The shape of the earth's orbit around the sun is essentially a circle. Assuming the earth maintains a constant speed as it orbit

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This question and the list of answer-choices were written by somebody who is unclear on the concepts, and has no business being involved with teaching Physics. This is not harmless. It's misleading to any student who is trying to make sense of it, and perhaps also to his Physics teacher.

The writer of this question thinks that 'acceleration' means speeding up.

IT DOESN'T !

'Acceleration' means ANY change of speed OR DIRECTION. Speeding up, slowing down, and turning or curving are all 'acceleration'.

An object moving with constant speed is still accelerating if it isn't moving in a STRAIGHT LINE, because its direction keeps changing.

An object moving with constant speed in a circle is experiencing an acceleration that is directed toward the center of the circle. Its magnitude is

(the object's speed)² divided by (the radius of the circle) .

It's called "centripetal acceleration".

Going through the choices . . .

<em>a). Its acceleration is zero since it is moving at constant speed.</em> NO. Speed is not the only thing that determines acceleration.

b). Its acceleration is not zero but is very small. What does "very small" mean ? Could it be small without being very small ? What are the numbers for acceleration that would be called 'small', 'very small', and 'not small' ? This choice could not legitimately be the answer to anything.

<em>c). Its acceleration is 9.8 m/s²</em>. NO. That number is the acceleration of things that fall down on the Earth. It's the acceleration that results from Earth's gravity on the surface. It points toward the center of the Earth. It's not related to the Earth's centripetal acceleration in orbit.

We could stop here. The question only wants us to say which choice is "most correct", and we already know that a)., b)., and c). are all definitely NOT correct. So if there IS a correct one, it'll have to be d). But this is a sleazy way out. We're honorable people, so we have to continue, and see what the deal is with choice-d).

<em>d). Its acceleration is 1.0 m/s²</em>. Maybe it is, and maybe it isn't. I suspect it isn't, because it would be a huge coincidence if it happened to be exactly this number. But there's no way to know without calculating it.

Centripetal acceleration = (speed)² / radius

Radius of the orbit = 93 million miles

Speed = one circumference per year

This is gonna be a #%$&@ to calculate ... we have to convert everything to consistent units. I'll tell you what: YOU just relax, and I'LL do the work.

Radius = (93 x 10⁶ miles) · (1609 meters/mile) = 150 gigameters (not exact, but close enough for right now; I only have to find out whether the acceleration is anywhere close to 1.0 m/s², to find out whether choice-d). is any good at all.)

Circumference = (2π) · (radius) = 9.4 x 10¹¹ meters

Speed = (1 circumference/year) = (9.4 x 10¹¹ meters) / (365 days)

Speed = (9.4 x 10¹¹ meters / 365 days) · (1 day / 86,400 seconds)

Speed = (9.4 x 10¹¹ / 365 · 86,400) (meters-days / days-seconds)

Speed = 2.98 x 10⁴ meters/sec

OK. I'm seeing light at the end of the tunnel.

[Centripetal] acceleration = (speed)² / (radius)

Acceleration = (2.98 x 10⁴ m/s)² / (150 x 10⁹ meters)

Acceleration = (8.89 x 10⁸ m²/s²) / (150 x 10⁹ meters)

Acceleration = (8.89 x 10⁸ / 150 x 10⁹) (meters² / second-meters)

Centripetal Acceleration = 0.0059 m/s²

And there you have it. I'll admit that I did a lot of rounding and approximating on the way. But unless my answer is at least 100 times too small, choice-d). is as worthless as the other three choices listed.

So I'm going to say that even though choice-d) is technically the 'most correct', that's only because the other three choices are so terrible. The only nice thing you can say about choice-d) is that maybe it's the least bad of all the choices. But it's still poor, and seriously misleading.

<em>None of the choices describes the Earth's motion in its orbit</em>. <em>The correct answer to the question is not included </em>among the choices. The question itself is almost certainly based on a fundamental misunderstanding of the underlying concept of acceleration.

So there !

This is not your fault, Jenny ! I'm grateful to you for the generous bounty of 5 points. And the green crusts and tepid cloudy water were also delicious.

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3 years ago