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bezimeni [28]
3 years ago
14

A rigid tank initially contains 1.4 kg saturated liquid water at 200◦C. At this state, 25 percent of the volumeis occupied by wa

ter and the rest by air. Now heat is supplied to the water until the tank contains saturatedvapor only. Determine (a) the volume of the tank, (b) the finaltemperature and pressure, and (c) the internalenergy change of the water.
Physics
1 answer:
Norma-Jean [14]3 years ago
4 0

Answer:

given,

liquid water in the rigid tank = 1.4 kg

The saturated liquid properties of water at 200◦C are

v_f = 0.001157 m^3/s\ and \ u_f = 850.46 kJ/kg

a) tank initially contains saturated liquid and water so, volume occupied by water

V_1 = m v_1

      = 1.4 × 0.001157

      = 0.001619 m³

total volume =

         = \dfrac{1}{0.25}\times 0.001619

         = 0.006476 m^3

b)v_2 = \dfrac{V}{m} =\dfrac{0.006476}{1.4}

          =0.004626 kg/m³

for v₂ = 0.004626 kg/m³

T₂ = 371.3 ⁰C  P₂ = 21,367 kPa  u₂ = 2201.5 kJ/kg

c) total internal energy

\Delta U = m(u_2-u_1)

              = 1.4 (2201.5 - 850.46) kJ/kg

              = 1892 kJ

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The initial potential energy of the wagon containing gold boxes will enable

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The Lone Ranger and Tonto have approximately <u>5.1 seconds</u>.

Reasons:

Mass wagon and gold = 166 kg

Location of the wagon = 77 meters up the hill

Slope of the hill = 8°

Location of the rangers = 41 meters from the canyon

Mass of Lone Ranger, m₁ = 65 kg

Mass of Tonto m₂ = 66 kg

Solution;

Height of the wagon above the level ground, h = 77 m × sin(8°) ≈ 10.72 m

Potential energy = m·g·h

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Potential energy of wagon, P.E. ≈ 166 × 9.81 × 10.72 = 17457.0912

Potential energy of wagon, P.E. ≈ 17457.0912 J

By energy conservation, P.E. = K.E.

K.E. = \mathbf{\dfrac{1}{2} \cdot m \cdot v^2}

Where;

v = The velocity of the wagon a the bottom of the cliff

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\dfrac{1}{2} \times 166 \times v^2 = 17457.0912

v = \sqrt{\dfrac{17457.0912}{\dfrac{1}{2} \times 166} } \approx 14.5

Velocity of the wagon, v ≈ 14.5 m/s

Momentum = Mass, m × Velocity, v

Initial momentum of wagon = m·v

Final momentum of wagon and ranger = (m + m₁ + m₂)·v'

By conservation of momentum, we have;

m·v = (m + m₁ + m₂)·v'

\therefore v' = \mathbf{ \dfrac{m \cdot v}{(m + m_1 + m_2)  }}

Which gives;

\therefore v' = \dfrac{166 \times 14.5}{(166 + 65 + 66)  } \approx 8.1

The velocity of the wagon after the Ranger and Tonto drop in, v' ≈ 8.1 m/s

Time = \dfrac{Distance}{Velocity}

\mathrm{The \ time \ the\ Lone \  Ranger \  and  \ Tonto \  have,  \ t} = \dfrac{41 \, m}{8.1 \, m/s} \approx 5.1 \, s

The Lone Range and Tonto have approximately <u>5.1 seconds</u> to grab the

gold and jump out of the wagon before the wagon heads over the cliff.

Learn more here:

brainly.com/question/11888124

brainly.com/question/16492221

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2 years ago
A simple pendulum is made by tying a 2.44 kg stone to a string 4.57 m long. The stone is projected perpendicularly to the string
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Answer:

v_{max}=8.2226m/s

Explanation:

The problem is solved using the law of conservation of energy,

So

mgL(1-cos\theta)+\frac{1}{2}mv^2_0=\frac{1}{2}mv^2_{max}

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