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DanielleElmas [232]
3 years ago
6

two punds of water vapor at 30 psia fill the 4ft3 left chmaber of a partitioned system. The right chmaber has twice the volume o

f the left and is initially evacuated. Detrmine the pressure of water after the paertiion has been removed nd enough heat has been transfered so that the temperature of the water is 40F.
Physics
1 answer:
tamaranim1 [39]3 years ago
7 0

Answer:

pressure of water will be 49.7 atm

Explanation:

given data

pressure = 30 psi = 2.04 atm

water = 2 pound = 907.18

mole of water vapor = 907.19 /2 = 50.4 mole

volume = 4 ft³ = 113.2 L

temperature = 40 F = 277.59 K

to find out

pressure of water

solution

we will apply here ideal gas condition

that is

PV = nRT  .......................1

put here all value and here R = 0.0821 , T temperature and V volume and P pressure and n is no of mole

and we get here temperature

PV = nRT  

2.04 × 113.2 = 50.4×0.0821×T

solve it and we get

T = 55.8 K

so we have given right chamber has twice the volume of the left chamber i.e

volume = twice of volume + volume

volume = 2(113.2) + 113.2

volume = 339.6 L

so from equation 1 pressure will be

PV = nRT

P(339.6) = 50.4 × ( 0.0821) × (277.59)

P = 3.3822 atm = 49.7 atm

so pressure of water will be 49.7 atm

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Scorpion4ik [409]

Answer:

The height of the cliff is, h = 78.4 m

Explanation:

Given,

The horizontal velocity of the projectile, Vx = 20 m/s

The range of the projectile, s = 80 m

The projectile projected from a height is given by the formula

                            <em> S = Vx [Vy + √(Vy² + 2gh)] / g </em>

Therefore,  

                            h = S²g/2Vx²

Substituting the values

                             h = 80² x 9.8/ (2 x 20²)

                                = 78.4 m

Hence, the height of the cliff is, h = 78.4 m

8 0
3 years ago
An iron railroad rail is 700 ft long when the temperature is 30°C. What is its length when the temperature is – 10°C?
MAVERICK [17]

Answer:

699.67ft

Explanation:

We are given with,

  • α = 1.2×10⁻⁵ / °C
  • L₀ = 700 ft
  • ΔT = -10°C − 30°C = -40°C

Now, We have to find ΔL:

  • ΔL = (1.2×10⁻⁵ / °C) (700 ft) (-40°C)
  • ΔL = −0.336

Rounded to two significant figures, the change in length is −0.33ft.

<u>Therefore, the final length is approximately 700 ft − 0.33 ft = 699.67ft</u>.

6 0
3 years ago
The parallel plates in a capacitor, with a plate area of 8.00 cm2 and an air-filled separation of 2.70 mm, are charged by a 8.70
MrRa [10]

Answer:

a)  ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J,  d)  W = 150 10⁻¹² J

Explanation:

Let's find the capacitance of the capacitor

         C = \epsilon_o \frac{A}{d}

         C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³

         C = 2.62 10⁻¹² F

for the initial data let's look for the accumulated charge on the plates

          C = \frac{Q}{\Delta V}

          Q₀ = C ΔV

           Q₀ = 2.62 10⁻¹² 8.70

           Q₀ = 22.8 10⁻¹² C

a) we look for the capacity for the new distance

          C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³

          C₁ = 1.04 10⁻¹² F

       

          C₁ = Q₀ / ΔV₁

          ΔV₁ = Q₀ / C₁

          ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²

          ΔV₁ = 21.9 V

b) initial stored energy

          U₀ = \frac{Q_o}{ 2C}

          U₀ = (22.8 10⁻¹²)²/(2  2.62 10⁻¹²)

          U₀ = 99.2 10⁻¹² J

c) final stored energy

          U_f = (22.8 10⁻¹²) ² /(2  1.04 10⁻⁻¹²)

          U_f = 249.9 10⁻¹² J

d) the work of separating the plates

as energy is conserved work must be equal to energy change

          W = U_f - U₀

          W = (249.2 - 99.2) 10⁻¹²

          W = 150 10⁻¹² J

note that as the energy increases the work must be supplied to the system

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umka21 [38]
The answer to the question is true
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elena-14-01-66 [18.8K]

Time = 15/-0.5 = 14.5/-0.5 = 14.5s

The answer is 14.5s

Sorry no one would answer it sooner.

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