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Elena-2011 [213]
4 years ago
7

Problem 6.056 Air enters a compressor operating at steady state at 15 lbf/in.2, 80°F and exits at 275°F. Stray heat transfer and

kinetic and potential energy effects are negligible. Assuming the ideal gas model for the air, determine the maximum theoretical pressure at the exit, in lbf/in.2 p2,max = lbf/in.2
Physics
1 answer:
vova2212 [387]4 years ago
5 0

To solve this process it is necessary to consider the concepts related to the relations between pressure and temperature in an adiabatic process.

By definition the relationship between pressure and temperature is given by

(\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}

Here

P = Pressure

T = Temperature

\gamma =The ratio of specific heats. For air normally is 1.4.

Our values are given as,

P_1 = 15lb/in^2\\T_1= 80\°F = 299.817K\\T_2 =400\°F = 408.15K

Therefore replacing we have,

(\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}

(\frac{P_2}{15})=(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}

Solving for P_2,

P_2 = 15*(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}

P_2 = 44.15Lbf/in^2

Therefore the maximum theoretical pressure at the exit is 44.15Lbf/in^2

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A 6.5×10−2 kg arrow hits the target at 25 m/s and penetrates 3.8 cm before stopping.
White raven [17]
1) We use:
2as = v² - u², with v = 0,
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2 x 0.038 x a = -(25)²
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B) The arrows force is the same but in the opposite directioin.
534.2 N

C) a = -8.22 x 10³ m/s²
s = -(65)²/(2 x -8.22 x 10³)
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8 0
3 years ago
When you rub a glass rod with a piece of silk, the glass rod becomes positively charged?
Lynna [10]
Glass loses electrons easily and silk takes them. Because electrons are negatively charged, the glass rod becomes positive and the silk becomes negative.
3 0
4 years ago
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A meter stick balances horizontally on a knife-edge at the 51 cm mark. With two nickels stacked over the 6.0 cm mark, the stick
Oliga [24]

Answer:

65g

Explanation:

Two main conditions for equilibrium are:

I. The resultant force must be equal to zero. That is, sum of the forces acting in one direction about a point must be equal to the sum of the forces acting in the opposite direction about the same point.

II. The resultant moment must be equal to zero. That is, sum of the moments in one direction about a point must be equal to the sum of the moments in another direction about the same point.

For the above question,

the 51cm mark is the point where the resultant weight of the meter stick lies,

the pivot or point is the 45cm mark where the stick balanced when 2 nickels ( total mass (5.0g x 2) 10g were placed at the 6cm mark.

Using the conversion factor:

1000g(1kg) = 10N, we can convert mass to weight, calculate the weight of the meter stick then reconvert to mass.

That is,

mass of 2 nickels = 10g = 10/1000 = 0.01N.

Moment = Force x distance from line of force to pivot of rotation

Applying the principle of equilibrium,

Moment of left side = Moment of right side

0.01 x (45-6) = W x (51-45)

Where W = weight of the meter stick

W x 6 = 0.01 x 39

W x 6 = 0.39

W = 0.39/6

W= 0.065N

Therefore, mass of meter stick = 0.065 x 1000 = 65g.

4 0
3 years ago
The carpal bones in the hands are an example of __________.
Olenka [21]
The carpal bones in the hands are an example of __________.
Answer: gliding joints

<span>A gliding joint means a freely moving joint in which the articulations allow only gliding motions</span>
6 0
3 years ago
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what is the rotational kinetic energy of the earth? use the moment of inertia you calculated in part a rather than the actual mo
Ivenika [448]

The Earth's rotational kinetic energy is the kinetic Energy that the Earth

has due to rotation.

The rotational kinetic energy of the Earth is approximately <u>3.331 × 10³⁶ J</u>

Reasons:

<em>The parameters required for the question are;  </em>

<em>Mass of the Earth, M = </em><em>5.97 × 10²⁴ kg</em>

<em>Radius of the Earth, R = </em><em>6.38 × 10⁶ m</em>

<em>The rotational period of the Earth, T = </em><em>24.0 hrs</em><em>.</em>

The \ moment  \ of \  inertia \  of \  uniform \  sphere \  is \ I =   \mathbf{\dfrac{2}{5} \cdot M \cdot R^2}

Which gives;

\mathbf{I_{Earth}} =   \dfrac{2}{5} \times 5.97 \times 10 ^{24} \cdot \left(6.38 \times 10^6 \right)^2 = 9.7202107 \times 10^{37}

\mathrm{The \ rotational \  kinetic  \ energy \  is} \   E_{rotational} = \mathbf{\dfrac{1}{2} \cdot I \cdot \omega^2}

\mathrm{The \ angular \ speed, \ \omega} = \mathbf{\dfrac{2 \dcdot \pi}{T}}

Therefore;

\omega = \dfrac{2 \cdot \pi}{24}  = \dfrac{\pi}{24}

Which gives;

\mathbf{E_{rotational}} = \dfrac{1}{2} \times  9.7202107 \times 10^{37} \times  \left(  \dfrac{\pi}{12} \right)^2 = 3.331 \times 10^{36}

The rotational kinetic energy of the Earth, E_{rotational} = <u>3.331 × 10³⁶ Joules</u>

Learn more here:

brainly.com/question/13623190

<em>The moment of inertia from part A  of the question (obtained online) is that of the Earth approximated to a perfect sphere</em>.

<em>Mass of the Earth, M = 5.97 × 10²⁴ kg</em>

<em>Radius of the Earth, R = 6.38 × 10⁶ m</em>

<em>The rotational period of the Earth, T = 24.0 hrs</em>

3 0
3 years ago
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