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WITCHER [35]
3 years ago
14

An air-conditioning system is to be filled from a rigid container that initially contains 5 kg of liquid R-134a at 24°C. The val

ve connecting this container to the air-conditioning system is now opened until the mass in the container is 0.25 kg, at which time the valve is closed. During this time, only liquid R-134a flows from the container. Presuming that the process is isothermal while the valve is open, determine the final quality of the R-134a in the container and the total heat transfer. Use data from the tables.

Physics
1 answer:
OverLord2011 [107]3 years ago
3 0

Answer:

Answer

The Final Quality of teh R-134a in the container  is  0.5056

The Total Heat transfer is Q_{in} = 22.62 KJ

Explanation:

Explanation is  in the following  attachments

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If 478 watts of power are used in 14 seconds,how much work was done
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Answer:

6692J

Explanation:

Power is defined as the rate at which work is being done.

So,

    Power  = \frac{workdone}{time }  

  Work done  = Power x time

Given parameters:

Power  = 478watts

Time  = 14s

So;

 Work done  = 478 x 14  = 6692J

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3 years ago
Determine the vector perpendicular to the plane of A= 31+ 6j - 2k and B=4i-j +3k
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The vector perpendicular to the plane of A = 3i+ 6j - 2k and B = 4i-j +3k is 16 i - 17 j - 27 k

Let r be the vector perpendicular to A and B,

r = A * B

A = 3i + 6j - 2k

B = 4i - j + 3k

a1 = 3

a2 = 6

a3 = - 2

b1 = 4

b2 = - 1

b3 = 3

a * b = ( a2 b3 - b2 a3 ) i + ( a3 b1 - b3 a1 ) j + ( a1 b2 - b1 a2 ) k

a * b = [ ( 6 * 3 ) - ( - 1 * - 2 ) ] i + [ ( - 2 * 4 ) - ( 3 * 3 ) ] j + [ ( 3 * - 1 ) - ( 4 * 6 ) ] k

a * b = 16 i - 17 j - 27 k

The perpendicular vector, r = 16 i - 17 j - 27 k

Therefore, the vector perpendicular to the plane of A = 3i + 6j - 2k and B = 4i - j + 3k is 16 i - 17 j - 27 k

To know more about perpendicular vectors

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1 year ago
If an object is placed at the center of carvature of a convance mirror the image formed is called
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Answer:

When the object is placed between centre of curvature and principal focus of a concave mirror the image formed is beyond C as shown in the figure and it is real, inverted and magnified.

3 0
2 years ago
A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo
-BARSIC- [3]

Answer:

It would take \tau(\ln 9 - \ln 8) time for the capacitor to discharge from q_0 to \displaystyle \frac{8}{9} \, q_0.

It would take \tau(\ln 9 - \ln 7) time for the capacitor to discharge from q_0 to \displaystyle \frac{7}{9}\, q_0.

Note that \ln 9 = 2\,\ln 3, and that\ln 8 = 3\, \ln 2.

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is \tau, and the initial charge of the capacitor be q_0. Then at time t, the charge stored in the capacitor would be:

\displaystyle q(t) = q_0 \, e^{-t / \tau}.

<h3>a)</h3>

\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{8}{9}\, q_0.

Apply the equation \displaystyle q(t) = q_0 \, e^{-t / \tau}:

\displaystyle \frac{8}{9}\, q_0 = q_0 \, e^{-t/\tau}.

The goal is to solve for t in terms of \tau. Rearrange the equation:

\displaystyle e^{-t/\tau} = \frac{8}{9}.

Take the natural logarithm of both sides:

\displaystyle \ln\, e^{-t/\tau} = \ln \frac{8}{9}.

\displaystyle -\frac{t}{\tau} = \ln 8 - \ln 9.

t = - \tau \, \left(\ln 8 - \ln 9\right) = \tau(\ln 9 - \ln 8).

<h3>b)</h3>

\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{7}{9}\, q_0.

Apply the equation \displaystyle q(t) = q_0 \, e^{-t / \tau}:

\displaystyle \frac{7}{9}\, q_0 = q_0 \, e^{-t/\tau}.

The goal is to solve for t in terms of \tau. Rearrange the equation:

\displaystyle e^{-t/\tau} = \frac{7}{9}.

Take the natural logarithm of both sides:

\displaystyle \ln\, e^{-t/\tau} = \ln \frac{7}{9}.

\displaystyle -\frac{t}{\tau} = \ln 7 - \ln 9.

t = - \tau \, \left(\ln 7 - \ln 9\right) = \tau(\ln 9 - \ln 7).

7 0
3 years ago
Calculate the wavelength (in nm) of the blue light emitted by a mercury lamp with a frequency of 6.88 × 1014 Hz. The speed of li
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Answer: 430 nm.

Explanation:

The relation of wavelength and frequency is:

Formula used : \nu=\frac{c}{\lambda}

where,

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c = speed of light = 3.00\times 10^{8}m/s

Now put all the given values in this formula, we get

6.88\times 10^{14}=\frac{3.00\times 10^{8}m/s}{\lambda}

\lambda=\frac{3.00\times 10^{8}m/s}{6.88\times 10^{14}}=0.43\times 10^{-6}m=430m        (1nm=10^{-9}m)

Thus the wavelength (in nm) of the blue light emitted by a mercury lamp is 430 nm.

6 0
3 years ago
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