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A 2-m3insulated rigid tank contains 3.2 kg of carbon dioxide at 120 kPa.Paddle-wheel work is done on the system until the pressu

AleksandrR [38]

Answer:

The change in entropy is found to be 0.85244 KJ/k

Explanation:

In order to solve this question, we first need to find the ration of temperature for both state 1 and state 2. For that, we can use Charles' law. Because the volume of the tank is constant.

P1/T1 = P2/T2

T2/T1 = P2/P1

T2/T1 = 180 KPa/120KPa

T2/T1 = 1.5

Now, the change in entropy is given as:

ΔS = m(s2 - s1)

where,

s2 = Cv ln(T2/T1)

s1 = R ln(V2/V1)

ΔS = change in entropy

m = mass of CO2 = 3.2 kg

Therefore,

ΔS = m[Cv ln(T2/T1) - R ln(V2/V1)]

Since, V1 = V2, therefore,

ΔS = mCv ln(T2/T1)

Cv at 300 k for carbondioxide is 0.657 KJ/Kg.K

Therefore,

ΔS = (3.2 kg)(0.657 KJ/kg.k) ln(1.5)

3 0

2 years ago

If im 14 and your 14 what does that equal

Karo-lina-s [1.5K]

Answer:

hmmmmmmmmmmmmmmmmmmmmmmm

Explanation:

4 0

2 years ago

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While playing a game of catch on the quadrangle, you throw a ball at an initial velocity of 17.6 m/s (approximately 39.4 mi/hr),

MAXImum [283]

Answer:

a) The y-component of velocity just before the ball hits the ground is -14.860 meters per second.

b) The ball is in the air during approximately 2.890 seconds.

c) The horizontal distance covered by the ball is 32.695 meters.

d) The magnitude of the velocity of the ball just before it hits the ground is approximately 18.676 meters per second.

e) The angle of the total velocity of the ball just before it hits the ground is approximately 52.717º below the horizontal.

Explanation:

a) The ball experiments a parabolic motion, which is a combination of horizontal motion at constant velocity and vertical motion at constant acceleration. First, we calculate the time taken by the ball to hit the ground:

$y = y_{o} + (v_{o}\cdot \sin \theta) \cdot t+\frac{1}{2}\cdot g\cdot t^{2}$ (1)

Where:

$y_{o}$ , $y$ - Initial and final vertical position, measured in meters.

$v_{o}$ - Initial speed, measured in meters per second.

$\theta$ - Launch angle, measured in sexagesimal degrees.

$g$ - Gravitational acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

If we know that $y_{o} = 2\,m$ , $y = 0\,m$ , $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ and $g = -9.807\,\frac{m}{s^{2}}$ , then the time taken by the ball is:

$-4.904\cdot t^{2}+13.482\cdot t +2 = 0$ (2)

This second order polynomial can be solved by Quadratic Formula:

$t_{1} \approx 2.890\,s$ and $t_{2} \approx -0.141\,s$

Only the first root offers a solution that is physically reasonable. That is, $t \approx 2.890\,s$ .

The vertical velocity of the ball is calculated by this expression:

$v_{y} = v_{o}\cdot \sin \theta +g\cdot t$ (3)

Where:

$v_{o,y}$ , $v_{y}$ - Initial and final vertical velocity, measured in meters per second.

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ , $g = -9.807\,\frac{m}{s^{2}}$ and $t \approx 2.890\,s$ , then the final vertical velocity is:

$v_{y} = -14.860\,\frac{m}{s}$

The y-component of velocity just before the ball hits the ground is -14.860 meters per second.

b) From a) we understand that ball is in the air during approximately 2.890 seconds.

c) The horizontal distance covered by the ball ( $x$ ) is determined by the following expression:

$x = (v_{o}\cdot \cos \theta)\cdot t$ (4)

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ and $t \approx 2.890\,s$ , then the distance covered by the ball is:

$x = 32.695\,m$

The horizontal distance covered by the ball is 32.695 meters.

d) The magnitude of the velocity of the ball just before hitting the ground ( $v$ ), measured in meters per second, is determined by the following Pythagorean identity:

$v = \sqrt{(v_{o}\cdot \cos \theta )^{2}+v_{y}^{2}}$ (5)

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ and $v_{y} = -14.860\,\frac{m}{s}$ , then the magnitude of the velocity of the ball is:

$v \approx 18.676\,\frac{m}{s}$ .

The magnitude of the velocity of the ball just before it hits the ground is approximately 18.676 meters per second.

e) The angle of the total velocity of the ball just before it hits the ground is defined by the following trigonometric relationship:

$\tan \theta = \frac{v_{y}}{v_{o}\cdot \cos \theta_{o}}$

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta_{o} = 50^{\circ}$ and $v_{y} = -14.860\,\frac{m}{s}$ , the angle of the total velocity of the ball just before hitting the ground is:

$\theta \approx -52.717^{\circ}$

The angle of the total velocity of the ball just before it hits the ground is approximately 52.717º below the horizontal.

3 0

2 years ago

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A insulated vessel s has two compartments separated by a membreane. On one side is 1kg of steam at 400 degC and 200 bar. The oth

Lilit [14]

Answer:

See explaination

Explanation:

See attachment for the detailed step by step solution of the given problem.

5 0

2 years ago

Four common causes of product failure are poor design, poor construction, poorly communicated operating instructions, and ______

Inga [223]

Answer:

B. operator error or misuse

Explanation:

A product is a failure if it is not able to achieve the anticipated life cycle as expected by the organization.

In such a case, there is a withdrawal of the product from the market as a result of its ultimate failure of a product to achieve profitability.

Four common causes of product failure are poor design, poor construction, poorly communicated operating instructions, and <u>operator error or misuse</u>

3 0

3 years ago