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Rina8888 [55]
3 years ago
9

A piston-cylinder device initially contains 0.6 kg of water with a volume of 0.1 m3 . The mass of the piston is such that it mai

ntains a constant pressure of 1000 kPa. The cylinder is connected through a valve to a supply line that carries steam at 5 MPa and 500 o C. Now the valve is opened and steam is allowed to flow slowly into the cylinder until the volume of the cylinder doubles and the temperature in the cylinder reaches 280 o C, at which point the valve is closed. The pressure remains constant during the process. Determine:
Physics
1 answer:
Sloan [31]3 years ago
3 0

Answer: Hello the missing piece of your question is attached

question : Determine mass of steam that has entered ( in kg )

answer : 0.206 kg

Explanation:

V1 = 0.1 m^3 ,

v' = V1 / m1 = 0.1 / 0.6 = 0.167 m^3/kg

V2 = 0.2 m^3

using the steam tables

at ; P = 1000 kPa, v' = 0.167 m^3/kg

U1 = 2321  KJ/kg

at ; P = 1000 kPa , T2 = 280°C

v'2= 0.2481 m^3kg

U2 = 2760.6

at ; P = 5MPa ,  T = 500°C

h1 = 3434.7 KJ/Kg

calculate final mass ( m2 )

M2 = V2 / v'2

      = 0.2 / 0.2481 =  0.806 kg

therefore the mass added =  m2 - m1

                                            = 0.806 - 0.6 =  0.206 kg

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Answer: Probably clean

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Hope this helps your soap problems! :b

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3 years ago
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How does charging by conduction occur?
lutik1710 [3]

Answer : (A) A charged object is brought near a neutral object without touching it.

Explanation:

Bodies can be charged by the method of conduction. By conduction the body acquires the same charge as on the charging body.

Charging the body can be understood by the following example of charging a paper cylinder by conduction. Make a paper cylinder by rolling a strip of paper on a pencil and then gently pulling out the pencil. Suspend the paper cylinder by a string tied to its center, now touch the paper cylinder with a glass rod rubbed with silk so it has a positive charge. Remove the class rod and the again bring it near to the paper cylinder.

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8 0
3 years ago
Need help pls. It is for acellus
Artist 52 [7]

Answer:

The height of the water column = 1.62405\overline{30} × 10⁻¹ m

Explanation:

The air cavity in the Coke bottle = 0.220 m deep

The fundamental (frequency) it plays when water is added to shorten the column and it is blown across the top, f = 528 Hz

The given speed of sound in air, v = 343 m/s

We note that the air cavity in the coke bottle is equivalent to a tube closed at one end

The fundamental frequency for a tube closed at one end, 'f', is given as follows;

f = v/(4·L) = v/λ

Where;

L = The height of the water column

λ = The wavelength of the wave

∴ 4·L = v/f = (343 m/s)/(528 Hz) = 0.6496\overline{21} m

∴ L = 0.6496\overline{21} m/4 = 0.162405\overline{30} m

The height of the water column = 1.62405\overline{30} × 10⁻¹ m.

4 0
3 years ago
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21 of 35 Review A friend of yours is loudly singing a single note at 401 HzHz while racing toward you at 23.0 m/sm/s on a day wh
Ipatiy [6.2K]

Answer:

The frequency of the sound note as heard = 429 Hz

Explanation:

The frequency of sound waves as heard from a distance for a sound wave coming towards one at v₀ m/s and whose real frequency is f₀ is given by

+f = f₀/[1 - (v₀/v)]

+f = frequency of sound as heard from the distance away = ?

f₀ = real frequency of sound = 401 Hz

v₀ = velocity at which the sound source is moving towards the reference point = 23.0 m/s

v = velocity of sound waves = 343 m/s

+f = 401/[1 - (23/343)]

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5 0
3 years ago
Two parallel plates that are initially uncharged are separated by 1.7 mm, have only air between them, and each have surface area
yaroslaw [1]

Answer:

5.63\cdot 10^{-6} C

Explanation:

The capacitor of a parallel-plate capacitor is given by:

C=\epsilon_0 \frac{A}{d}

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A is the area of each plate

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\epsilon_0 is the vacuum permittivity

The energy stored in a capacitor instead is given by

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where

Q is the charge stored in each plate

Substituting the expression we found for C inside the last formula,

U=\frac{1}{2}\frac{Q^2 d}{\epsilon_0 A}

And re-arranging it

Q=\sqrt{\frac{2U\epsilon_0 A}{d}}

Now if we substitute

d=1.7 mm=0.0017 m\\A=16 cm^2 = 16\cdot 10^{-4} m^2\\U = 1.9 J

We find the charge stored on the capacitor:

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7 0
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