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WITCHER [35]
3 years ago
14

A 15-ft3 tank contains oxygen initially at 14.7 psia and 80°F. A paddle wheel within the tank is rotated until the pressure insi

de rises to 20 psia. During the process 20 Btu of heat is lost to the surroundings. Determine the paddle-wheel work done. Neglect the energy stored in the paddle wheel. The gas constant and molar mass of oxygen are R = 0.3353 psia·ft3/lbm·R and M = 32 lbm/lbmol. The specific heat of oxygen at the average temperature of Tavg = (735 + 540)/2 = 638 R is cv ,avg= 0.160 Btu/lbm·R.
Physics
1 answer:
Nikolay [14]3 years ago
6 0

Explanation:

Equation for energy balance will be as follows.

         \Delta E_{system} = E_{in} - E_{out}

        \Delta U = W_{in} - Q_{out}

Hence,    W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})

Therefore, we will calculate the final temperature as follows.

            \frac{P_{1}V}{T_{1}} = \frac{P_{2}V}{T_{2}}

       T_{2} = \frac{20 psia}{14.7}(638 R)

                   = 868.03 R

Now, we will calculate the mass as follows.

             m = \frac{P_{1}V}{RT_{1}}

                 = \frac{14.7 psia \times 15 ft^{3}}{0.3353 psi ft^{3}/lbm R \times 638 R}

                 = 1.031 lbm

Hence,

        W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})

Putting the values into the above equation as follows.

            W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})

    W_{in} = 20 Btu + 1.031 lbm (\frac{0.160 Btu}{lbm R})(735 - 540)R

            W_{in} = 655.2 Btu

Thus, we can conclude that work done by paddle wheel is 655.2 Btu.

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Why do raindrops fall with a constant speed during the later stages of their descents?
BartSMP [9]
Firstly they have a acceleration downwards due the force downwards due they gravitational field acting on it's mass.
as it falls it gains speed, and as it gains speed the air Resistance which is a upward force actin on the drop increases, eventually the rain drop's upward and downward forces are balanced and hence there is no RESULTANT force therefore no acceleration, so the drops falls in constant speed (terminal verlocity is a better term)

Are you wondering that why is the raindrop still moving given that the forces are balanced? If so according to Newton's 1st law an object will keep moving or Remain at rest until a RESULTANT force acts on it.
6 0
3 years ago
What does it mean to say something is proportional?
mamaluj [8]

Answer: When quantities have the same relative size. In other words they have the same ratio corresponding in size or amount to something else.

Explanation: Example: "A rope's length and weight are in proportion."

Example: "The punishment should be proportional to the crime"

4 0
2 years ago
A high-speed flywheel in a motor is spinning at 450 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and
alexira [117]

Answer:

A) \omega_f=17.503\ rad.s^{-1}

B) t=55.6822\ s

C) \theta=1312\ rad

Explanation:

Given:

  • mass of flywheel, m=40\ kg
  • diameter of flywheel, d=0.72\ m
  • rotational speed of flywheel, N_i=450\ rpm \Rightarrow \omega_i=\frac{450\times 2\pi}{60} =15\pi\ rad.s^{-1}
  • duration for which the power is off, t_0=35\ s
  • no. of revolutions made during the power is off, \theta=180\times 2\pi=360\pi\ rad

<u>Using equation of motion:</u>

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

360\pi=15\pi\times 35+\frac{1}{2} \times \alpha\times35^2

\alpha=-0.8463\ rad.s^{-2}

Negative sign denotes deceleration.

A)

Now using the equation:

\omega_f=\omega_i+\alpha.t

\omega_f=15\pi-0.8463\times 35

\omega_f=17.503\ rad.s^{-1} is the angular velocity of the flywheel when the power comes back.

B)

Here:

\omega_f=0\ rad.s^{-1}

Now using the equation:

\omega_f=\omega_i+\alpha.t

0=15\pi-0.8463\times t

t=55.6822\ s is the time after which the flywheel stops.

C)

Using the equation of motion:

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

\theta=15\pi\times 55.68225-0.5\times 0.8463\times 55.68225^2

\theta=1312\ rad revolutions are made before stopping.

3 0
3 years ago
3.
Ivan

Answer:

car B will be 30 Km ahead of car A.

Explanation:

We'll begin by calculating the distance travelled by each car. This is illustrated below:

For car A:

Speed = 40 km/h

Time = 3 hours

Distance =?

Speed = distance / time

40 = distance / 3

Cross multiply

Distance = 40 × 3

Distance = 120 Km

For car B:

Speed = 50 km/h

Time = 3 hours

Distance =?

Speed = distance / time

50 = distance / 3

Cross multiply

Distance = 50 × 3

Distance = 150 Km

Finally, we shall determine the distance between car B an car A. This can be obtained as follow:

Distance travelled by car B (D₆) = 150 Km

Distance travelled by car A (Dₐ) = 120 Km

Distance apart =?

Distance apart = D₆ – Dₐ

Distance apart = 150 – 120

Distance apart = 30 Km

Therefore, car B will be 30 Km ahead of car A.

7 0
3 years ago
Pleasse help
GaryK [48]

Hi,

The correct answer is letter B.

The last group contains noble gases, while both along the top and along the bottom the elements on the right are non-metals.

3 0
3 years ago
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