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3 years ago

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Refrigerant 134a enters a compressor with a mass flow rate of 5 kg/s. The working fluid enters the compressor as a saturated vap

or at 15ºC and negligible velocity and leaves the compressor at 1400 kPa with an enthalpy of 281.39 kJ/kg and a velocity of 50 m/s. The power of the compressor is measured to be 132.4 kW. Determine the rate of net heat associated with this process in kW (with three significant figures).

1 answer:

wariber [46]3 years ago

7 0

Answer:see the attachment below

Explanation:

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A flat coil of wire consisting of 20 turns, each with an area of 50 cm2, is positioned perpendicularly to a uniform magnetic fie

Scilla [17]

Answer:

Induced current, I = 0.5 A

Explanation:

It is given that,

number of turns, N = 20

Area of wire, $A=50\ cm^2=0.005\ m^2$

Initial magnetic field, $B_i=2\ T$

Final magnetic field, $B_f=6\ T$

Time taken, t = 2 s

Resistance of the coil, R = 0.4 ohms

We know that due to change in magnetic field and emf will be induced in the coil. Its formula is given by :

$\epsilon=\dfrac{-d\phi}{dt}$

Where

$\phi=BA$

$\epsilon=\dfrac{-d(NBA)}{dt}$

$\epsilon=NA\dfrac{B_f-B_i}{t}$

$\epsilon=20\times 0.005\times \dfrac{6-2}{2}$

$\epsilon=0.2\ V$

Let I is the induced current in the wire. It can be calculated using Ohm's law as :

$\epsilon=I\times R$

$I=\dfrac{\epsilon}{R}$

$I=\dfrac{0.2}{0.4}$

I = 0.5 A

So, the magnitude of the induced current in the coil is 0.5 A. Hence, this is the required solution.

5 0

3 years ago

Which describes how the same force affects a small mass and a large mass

n200080 [17]

The same force accelerates a small mass faster than
it accelerates a large mass.

It's easier to get a little red wagon going by pushing it
than it is to get a school bus going by pushing it.

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3 years ago

A 1.80-m -long uniform bar that weighs 531 N is suspended in a horizontal position by two vertical wires that are attached to th

Readme [11.4K]

Answer:

(a) 498.4 Hz

(b) 442 Hz

Solution:

As per the question:

Length of the wire, L = 1.80 m

Weight of the bar, W = 531 N

The position of the copper wire from the left to the right hand end, x = 0.40 m

Length of each wire, l = 0.600 m

Radius of the circular cross-section, R = 0.250 mm = $.250\times 10^{- 3}\ m$

Now,

Applying the equilibrium condition at the left end for torque:

$T_{Al}.0 + T_{C}(L - x) = W\frac{L}{2}$

$T_{C}(1.80 - 0.40) = 531\times \frac{1.80}{2}$

$T_{C} = 341.357\ Nm$

The weight of the wire balances the tension in both the wires collectively:

$W = T_{Al} + T_{C}$

$531 = T_{Al} + 341.357$

$T_{Al} = 189.643\ Nm$

Now,

The fundamental frequency is given by:

$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

where

$\mu = A\rho = \pi R^{2}\rho$

(a) For the fundamental frequency of Aluminium:

$f = \frac{1}{2L}\sqrt{\frac{T_{Al}}{\mu}}$

$f = \frac{1}{2L}\sqrt{\frac{T_{Al}}{\pi R^{2}\rho_{Al}}}$

where

$\rho_{l} = 2.70\times 10^{3}\ kg/m^{3}$

$f = \frac{1}{2\times 0.600}\sqrt{\frac{189.643}{\pi 0.250\times 10^{- 3}^{2}\times 2.70\times 10^{3}}} = 498.4\ Hz$

(b) For the fundamental frequency of Copper:

$f = \frac{1}{2L}\sqrt{\frac{T_{C}}{\mu}}$

$f = \frac{1}{2L}\sqrt{\frac{T_{C}}{\pi R^{2}\rho_{C}}}$

where

$\rho_{C} = 8.90\times 10^{3}\ kg/m^{3}$

$f = \frac{1}{2\times 0.600}\sqrt{\frac{341.357}{\pi 0.250\times 10^{- 3}^{2}\times 2.70\times 10^{3}}} = 442\ Hz$

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4 years ago

If the 15.0 cm string is shorter than the 30.0 cm string.

Margaret [11]

Answer:

option c) 2 is the right answer

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3 years ago

When was the first airplane made?

expeople1 [14]

The first ariplanr was made December 17, 1903

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4 years ago

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