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4 years ago

12

An air conditioner is a refrigerator with the inside of the house acting as the cold reservoir and the outside atmosphere acting

as the hot reservoir. Assume that an air conditioner consumes 1.25 * 103 W of electrical power and that it can be idealized as a reversible Carnot refrigerator. If the coefficient of performance of this device is 1.75, how much heat can be extracted from the house in a day

1 answer:

Monica [59]4 years ago

8 0

Answer:

1.89*10^8 J

Explanation:

The coefficient of performance of this device is $\frac{Q}{W}$ where Q .is the useful heat supplied or removed by the considered system and W is the work required by the considered system.

Step 1

Coeffiecient of perfromance for cooling (COPC) = 1.75 = $\frac{Q}{W}$ .

Q = 1.75W

Step 2

We convert day into seconds:

1 day = 24 hrs = 86400 seconds

Step 3

Heat that can be extracted from the house in a day is:

Q = 1.75 * 1.25 * 10^3 * 86400 = 189000000 = 1.89*10^8 J

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1 year ago

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Answer:

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3 years ago

you should begin viewing a bacteria specimen with what objective lens? view available hint(s)for part g you should begin viewing

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2 years ago

uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station

Alex787 [66]

Answer:

The magnitude of the tangential velocity is $v= 0.868 m/s$

The magnitude of the resultant acceleration at that point is $a = 4.057 m/s^2$

Explanation:

From the question we are told that

The mass of the uniform disk is $m_d = 40.0kg$

The radius of the uniform disk is $R_d = 0.200m$

The force applied on the disk is $F_d = 30.0N$

Generally the angular speed i mathematically represented as

$w = \sqrt{2 \alpha \theta}$

Where $\theta$ is the angular displacement given from the question as

$\theta = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}$

$=1.257\ rad$

$\alpha$ is the angular acceleration which is mathematically represented as

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

The moment of inertial is mathematically represented as

$I = \frac{1}{2} m_dR^2_d$

Substituting values

$I = 0.5 * 40 * 0.200^2$

$= 0.8kg \cdot m^2$

Considering the equation for angular acceleration

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

Substituting values

$\alph$ $\alpha = \frac{(30.0)(0.200)}{0.8}$

$= 7.5 rad/s^2$

Considering the equation for angular velocity

$w = \sqrt{2 \alpha \theta}$

Substituting values

$w =\sqrt{2 * (7.5) * 1.257}$

$= 4.34 \ rad/s$

The tangential velocity of a given point on the rim is mathematically represented as

$v = R_d w$

Substituting values

$= (0.200)(4.34)$

$v= 0.868 m/s$

The radial acceleration at hat point is mathematically represented as

$\alpha_r = \frac{v^2}{R}$

$= \frac{0.868^2}{0.200^2}$

$= 3.7699 \ m/s^2$

The tangential acceleration at that point is mathematically represented as

$\alpha _t = R \alpha$

Substituting values

$\alpha _t = (0.200) (7.5)$

$= 1.5 m/s^2$

The magnitude of resultant acceleration at that point is

$a = \sqrt{\alpha_r ^2+ \alpha_t^2 }$

Substituting values

$a = \sqrt{(3.7699)^2 + (1.5)^2}$

$a = 4.057 m/s^2$

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3 years ago