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3 years ago

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For a specific volume of 0.2 m3/kg, find the quality of steam if the absolute pressure is (a) 40 kPa and (b) 630 kPa. What is th

e temperature of each case?

1 answer:

ICE Princess25 [194]3 years ago

4 0

Answer:

$x=0.0498$

$x'=0.659$

Explanation:

Specific Volume $V=0.2m_3/kg$

Absolute Pressure (a) $P_a= 40kpa$

Giving

$T_a=75.87$

$v_f=1.265*10^{-3}m^3/kg$

$v_g=3.993m^3/kg$

(b) $P_a= 630kpa$

Giving

$T_b=160.13C$

$v_f'=1.10282*10^{-3} m^3/kg$

$v_g'=0.30286 m^3/kg$

(a)

Generally the equation for quality of Steam X is mathematically given by

$x=\frac{v-v_f}{v_g-v_f}$

$x=\frac{0.2-1.0265*10^{-3}}{3.993-1.0265*10^{-3}}$

$x=0.0498$

(b)

Generally the equation for quality of Steam X is mathematically given by

$x'=\frac{v-v_f'}{v_g'-v_f'}$

$x'=\frac{0.2-1.10*10^{-3}}{3.30-1.1*10^{-3}}$

$x'=0.659$

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Say I have a series circuit with 20v and four 65 ohm resistors, what is the current in each resistor?

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Data:

$E = 20 V$
$R_{1} = 65\Omega$
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<span>Like this:
</span>
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$I_{T} = \frac{20}{ 260 }$
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3 years ago

Coherent light that contains two wavelengths, 660 nm and 470 nm , passes through two narrow slits with a separation of 0.280 mm

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Answer:

λ1 = 0.0129m = 1.29cm

λ2 = 0.00923m = 0.92 cm

Explanation:

To find the distance between the first order bright fringe and the central peak, can be calculated by using the following formula:

$y_m=\frac{m\lambda D}{d}$ (1)

m: order of the bright fringe = 1

λ: wavelength of the light = 660 nm, 470 nm

D: distance from the screen = 5.50 m

d: distance between slits = 0.280mm = 0.280 *10^⁻3 m

ym: height of the m-th fringe

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For λ = 660 nm = 660*10^-9 m

$y_1=\frac{(1)(660*10^{-9}m)(5.50m)}{0.280*10^{-3}m}=0.0129m=1.29cm$

For λ = 470 nm = 470*10^-9 m

$y_1=\frac{(1)(470*10^{-9}m)(5.50m)}{0.280*10^{-3}m}=0.00923m=0.92cm$

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dm = λ ds

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dm = R λ₀ sin θ dθ

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X = ∫ x dm / ∫ dm

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X = R ∫₀ᵖ sin θ cos θ dθ / ∫₀ᵖ sin θ dθ

X = R ∫₀ᵖ ½ sin 2θ dθ / ∫₀ᵖ sin θ dθ

X = ¼R ∫₀ᵖ 2 sin 2θ dθ / ∫₀ᵖ sin θ dθ

X = ¼R (-cos 2θ)|₀ᵖ / (-cos θ)|₀ᵖ

X = ¼R (-cos 2π − (-cos 0)) / (-cos π − (-cos 0))

X = ¼R (-1 + 1) / (1 + 1)

X = 0

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Y = ∫ y dm / ∫ dm

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Y = ½R ∫₀ᵖ (1 − cos 2θ) dθ / ∫₀ᵖ sin θ dθ

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