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aliina [53]
4 years ago
7

Water is contained in a closed, rigid 0.2 m 3 tank at an initial pressure of 5 bar and a quality of 50%. Heat transfer occurs un

til the tank contains only saturated vapor. Determine the final mass of vapor in the tank, in kg, and the final pressure, in bar.
Physics
1 answer:
elena55 [62]4 years ago
7 0

Answer:

Final mass=0.89kg

Final pressure=5.6bar

Explanation:

To find mass,m=v/v1

But v1=vf + x(vg-vf)

Vf= 0.001093m^3/kg

Vg= 0.3748m^3/kg

V1= 0.001093+0.5(0.3748-0.001093)

V1= 0.225m^3/kg

M= 0.20/0.225 =0.89kg

Final pressure will be:

V/V1= P/P1

Cross multiply

VP1=V1P

P1= 0.225×5/0.2

P1=:5.6 bar

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A man whose mass is 69 kg and a woman whose mass is 52 kg sit at opposite ends of a canoe 5 m long, whose mass is 20 kg. Suppose
dusya [7]

Answer:

the canoe moved 1.2234 m in the water

Explanation:

Given that;

A man whose mass = 69 kg

A woman whose mass = 52 kg

at opposite ends of a canoe 5 m long, whose mass is 20 kg

now let;

x1 = position of the man

x2 = position of canoe

x3 = position of the woman

Now,

Centre of mass = [m1x1 + m2x2 + m3x3] / m1 + m2 + m3

= ( 69×0 ) + ( 52×5) + ( 20× 5/2) / 69 + 52 + 20

= (0 + 260 + 50 ) / ( 141 )

= 310 / 141

= 2.19858 m

Centre of mass is 2.19858 m

Now, New center of mass will be;

52 × 2.5 / ( 69 + 52 + 20 )

= 130 / 141

= 0.9219858 m  { away from the man }

To get how far, the canoe moved;

⇒ 2.5 + 0.9219858 - 2.19858

= 1.2234 m

Therefore, the canoe moved 1.2234 m in the water

5 0
3 years ago
Which options correctly describe the velocity of the object represented in the graph?
kirza4 [7]
There is no graph ...
7 0
2 years ago
How high does the water rise in the bell after enough time has passed for the air to reach thermal equilibrium
Minchanka [31]

The height risen by water in the bell after enough time has passed for the air to reach thermal equilibrium is 3.8 m.

<h3>Pressure and temperature at equilibrium </h3>

The relationship between pressure and temperature can be used to determine the height risen by the water.

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

where;

  • V₁ = AL
  • V₂ = A(L - y)
  • P₁ = Pa
  • P₂ = Pa + ρgh
  • T₁ = 20⁰C = 293 K
  • T₂ = 10⁰ C = 283 k

\frac{PaAL}{T_1} = \frac{(P_a + \rho gh)A(L-y)}{T_2} \\\\\frac{PaL}{T_1} = \frac{(P_a + \rho gh)(L-y)}{T_2} \\\\L-y = \frac{PaLT_2}{T_1(P_a + \rho gh)} \\\\y = L (1 - \frac{PaT_2}{T_1(P_a + \rho gh)})\\\\y = 4.2(1 - \frac{101325 \times 283}{293(101325\  +\  1000 \times  9.8 \times  100)} )\\\\y = 3.8 \ m

Thus, the height risen by water in the bell after enough time has passed for the air to reach thermal equilibrium is 3.8 m.

The complete question is below:

A diving bell is a 4.2 m -tall cylinder closed at the upper end but open at the lower end. The temperature of the air in the bell is 20 °C. The bell is lowered into the ocean until its lower end is 100 m deep. The temperature at that depth is 10°C. How high does the water rise in the bell after enough time has passed for the air to reach thermal equilibrium?

Learn more about thermal equilibrium here: brainly.com/question/9459470

#SPJ4

3 0
2 years ago
How might the changes in tide affect a fishing community?
artcher [175]

Answer:

THEY WOULD FIND DED FISH:)))))

Explanation:

6 0
3 years ago
An airplane is being pushed West by its engines with a force of 350 N as it starts to take off the runway at the Atlanta airport
Marizza181 [45]

Answer:

option C

Explanation:

given,

Force by the engine on plane in West direction = 350 N

Frictional force on the runway = 100 N in east

force exerted by the wind = 100 N in east

net force and direction = ?

consider west to be positive and east be negative.

when airplane will be moving there will be frictional as well as wind resistance will be acting in opposite direction of airplane

Net force = 350 N - 100 N - 100 N

                 = 150 N

as our answer comes out to be positive so the airplane will be moving in West

hence, the correct answer is option C

8 0
3 years ago
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