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3 years ago

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A 500-m^3 rigid tank is filled with saturated liquid-vapor mixture of water at 200 kPa. If 20% of the mass is liquid and the 80%

of the mass is vapor, the total mass in the tank is: (a) 635 kg (b) 2809 kg (c) 258 kg (d) 500 kg (e) 705 kg

1 answer:

guapka [62]3 years ago

3 0

Answer:

(a) 705 kg

Explanation:

it is given that volume =500 $m^3$

pressure = 200 kPa

it is given that 80% mass is vapor so dryness fraction =x=0.8

from the standard table at 200 kPa

$v_f=0.001061 \frac{m^3}{kg}$

$v_g=0.88578\frac{m^3}{kg}$

specific volume $v=v_f+x\left ( v_g-v_f \right )$

$v=0.00106+0.8\left ( 0.88578-0.00106 \right )=0.7088\frac{m^3}{kg}$

we know that $specific\ volume =\frac{volume}{mass}$

$0.7088=\frac{500}{mass}$

$mass=\frac{500}{0.7088}=705.456 \ kg$

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Nimfa-mama [501]

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a)

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$q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C$

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$V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V$

b)

Here we have to calculate the net potential at point B, located at

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The distance of charge 1 from point B is

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$\Delta V=-26,800 V$ is the potential difference

Therefore, the work required on the electron is

$W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J$

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