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3 years ago

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Two pounds of water vapor at 30 psia fill the 4-ft3 left chamber of a partitioned system. The right chamber has twice the volume

of the left and is initially evacuated. Determine the pressure of the water after the partition has been removed and enough heat has been transferred so that the temperature of the water is 40oF.

1 answer:

fomenos3 years ago

6 0

Answer:

3.38atm

Explanation:

Using data from the steam table we have that

Moles of water vapour = 907.19 / 18

= 50.4 moles

So

p1 = 30 psi = 30 x 0.68 = 2.04 atm

v1 = 4ft³= 113.2 L

Then from

PV= nRT

Then to find T we use

T1 = p1 V1 / n R

= 2.04 x 113.2 / 50.4 x 0.0821

= 55.8 K

Then to find volume two

v2 = 2v1 + v1

So

3 v1 = 339.6 K

The pressure two we use

P2 = n R T2 / V2

= 50.4 x 0.0821 x 277.6 / 339.6

So we have

= 3.38 atm =

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A record is dropped vertically onto a freely rotating (undriven) turntable. Frictional forces act to bring the record and turnta

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Answer:

The loss of initial Kinetic energy = 37.88 %

Explanation:

Given:

Rotational inertia of the turntable = $I_t$

Rotational inertia ( $I_r$ ) of the record = $0.61\times I_t$

According to the question:

<em>Frictional forces act to bring the record and turntable to a common angular speed.</em>

So,angular momentum will be conserved as it is an inelastic collision.

Considering the initial and final angular velocity of the turn table as $\omega _i\ ,\ \omega_f$ respectively.

Note :

Angular momentum $(L)$ = Product of moment of inertia $(I)$ and angular velocity $(\omega)$ .

Lets say,

⇒ initial angular momentum = final angular momentum

⇒ $L_i=L_f$

⇒ $(I_t)\times \omega_i = (I_t+I_r)\times \omega_f$

⇒ $\omega _f=\frac{I_t}{I_t+I_r} \times (\omega_i)$ ...equation (i)

Now we will find the ratio of the Kinetic energies.

⇒ $K_i=\frac{I_t\times \omega_i^2}{2}$ ⇒ $K_f=\frac{(I_r+I_t)\times \omega_f^2}{2}$

Their ratios:

⇒ $\frac{K_f}{K_i} =\frac{\frac{(I_t+I_r)\times \omega_f^2}{2} }{\frac{I_t\times \omega_i^2}{2} }$

⇒ $\frac{K_f}{K_i} = {\frac{(I_t+I_r)\times \omega_f^2}{2} } \times {\frac{2}{I_t\times \omega_i^2}}$

Plugging the values of $\omega _f^2$ as $\omega _f^2 =(\frac{I_t}{I_t+I_r} \times \omega_i\ )^2$ from equation (i) in the ratios of the Kinetic energies.

⇒ $\frac{K_f}{K_i} =\frac{(I_t+I_r)\times \frac{(I_t)^2}{(I_t+I_r)^2} \times \omega_i^2}{I_t\times \omega_i^2} =\frac{(I_t)^2}{(I_t+I_r)}\times \frac{1}{I_t}=\frac{I_t}{I_t+I_r}$

Now,

The Kinetic energy lost in fraction can be written as:

⇒ $\frac{K_f-K_i}{K_i}$

Now re-arranging the terms.

$\frac{K_f-K_i}{K_i} =(\frac{K_f}{K_i} -1)= \frac{I_t}{I_t+I_r} -1=\frac{I_t-I_t-I_r}{I_t+I_r} =\frac{-I_r}{(I_t+I_r)}$

Plugging the values of $I_r$ and $I_t$ .

⇒ $\frac{K_f}{K_i} = \frac{-0.61I_t}{0.61I_t+I_t} =\frac{-0.61}{1.61} =-0.3788$

To find the percentage we have to multiply it with $100$ and here negative means for loss of Kinetic energy.

⇒ $\frac{K_f}{K_i} = =-0.3788\times 100= 37.88$

So the percentage of the initial Kinetic energy lost is 37.88

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zhannawk [14.2K]

Answer:

See the answers below

Explanation:

To solve this problem we must use the definition of power and work in physics.

a)

The function of the conveyor belt is to carry the boxes from an initial point that is at low altitude to an end point that is at high altitude. In this way the conveyor belt prints a speed to the box to be able to raise it to the required vertical distance.

Since we have a velocity at the beginning and then we place the box at a high position, where then the box remains at rest, we can say that it converts kinetic energy to potential energy.

b)

Power is defined as the relationship of work over time. Therefore we have:

$P=W/t$

where:

P = power = 3.8 [kW] = 3800 [W]

W = work [J]

t = time = 14 [s]

$W=P*t\\W=3800*14\\W= 53200[J] = 53.2[kJ]$

c)

Since the given time is equal to the given time at Point b, we can use the same work calculated.

We know that work is defined as the product of force by the distance traveled.

$W =F*d$

So, the force is equal to:

$F=W/d\\F=53200/5.3\\F=10037.73[N]$

Now we know that force is defined as the product of mass by gravitation acceleration.

$F =m*g$

where:

F = force or weight = 10037.73 [N]

g = gravity acceleration = 9.81 [m/s²]

m = mass [kg]

$m=F/g\\m = 10037.73/9.81\\m = 1023.2 [kg]$

d)

This part can be solved by means of the energy conservation theorem, where the potential energy is transformed into kinetic energy or vice versa.

$E_{pot}=m*g*h = E_{kin}=0.5*m*v^{2}$

where:

h = elevation = 4.7 [m]

v = velocity [m/s]

$m*g*h=0.5*m*v^{2}\\g*h=0.5*v^{2} \\v=\sqrt{\frac{g*h}{0.5} } \\v=\sqrt{\frac{9.81*4.7}{0.5} } \\v = 9.6 [m/s]$

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3 years ago