Ask question

Ask question

All categories

3 years ago

9

Air is contained in a vertical piston–cylinder assembly such that the piston is in static equilibrium. The atmosphere exerts a p

ressure of 101 kPa on top of the 0.5-m-diameter piston. The gage pressure of the air inside the cylinder is 1.2 kPa. The local acceleration of gravity is g = 9.81 m/s2 . Subsequently, a weight is placed on top of the piston causing the piston to fall until reaching a new static equilibrium position. At this position, the gage pressure of the air inside the cylinder is 2.8 kPa. Determine (a) the mass of the piston, in kg, and (b) the mass of the added weight, in kg

1 answer:

oee [108]3 years ago

5 0

Answer:

a) 24 kg

b) 32 kg

Explanation:

The gauge pressure is of the gas is equal to the weight of the piston divided by its area:

p = P / A

p = m * g / (π/4 * d^2)

Rearranging

p * (π/4 * d^2) = m * g

m = p * (π/4 * d^2) / g

m = 1200 * (π/4 * 0.5^2) / 9.81 = 24 kg

After the weight is added the gauge pressure is 2.8kPa

The mass of piston plus addded weight is

m2 = 2800 * (π/4 * 0.5^2) / 9.81 = 56 kg

56 - 24 = 32 kg

The mass of the added weight is 32 kg.

You might be interested in

How are eras different from decades?

ivolga24 [154]

Answer:

decades are 10 years, while eras do not have set periods of time

5 0

3 years ago

A small family home in Tucson, Arizona has a rooftop area of 2667 square feet, and it is possible to capture rain falling on abo

Kazeer [188]

Answer:

volume = 53.747 m3 = 14198.138 gal

weight = 526652 N = 118396.08 lbf

Explanation:

We know that volume of water

$volume = A'\times H$

where A' = 61% of A

$= 0.61\times 2667 = 1626.87 sq ft$

$volume = 1626.87 \times (\frac{14}{12} ft)$

=1898.015 ft^3

$in\ m^3 = \frac{ 1898.015}{35.315} = 53.7457 m^3$

$in\ gallon = 1898.015 \times 7.481 = 14198.138 gallon$

$weight = \rho Vg$

$= 1000\times 53.74\times 9.8$

=526652 N

$In\ lbf = \frac{526652}{4.448} = 118396.08 lbf$

7 0

3 years ago

Explain why surface temperature increases when two bodies are rubbed against each other. What is the significance of temperature

Ronch [10]

Answer:

The surface temperature increases when two bodies are rubbed against each other due to friction.

Explanation:

No object has a perfectly even surface. So, when two bodies with uneven surfaces are rubbed against each other, they experience friction.

Friction is a resistance experienced by the two bodies when they are moved against each other.

The friction between the two surfaces, converts the kinetic energy of the movement to the thermal energy.

Thus, resulting in rise in the surface temperature of the two bodies.

Therefore, when two bodies are rubbed against each other, the surface temperature increases due to friction.

7 0

3 years ago

Problem 3: Soil Classification using the AASHTO and USCS Systems

nataly862011 [7]

<u>Solution:</u>

$Given\\ \(\quad W=3000 Ib , \quad m=\frac{W}{g}=\frac{3000}{322} \ slug =93.1677 slug\)\\K_{e q}=2160 lbs / wp =2100 \frac{ lbs }{10} \frac{ x 12}{1 ft }=(2160 \times 12) lb / ft$$$

a) The natural frequency

$\begin{aligned}&\left(\omega_{n}\right)=\sqrt{\frac{K_{e q}}{m}}\\&=\sqrt{\frac{2160 \times 12}{93.1677}}\\&\omega_{n}=16.68 \text { rad } | s\\&\omega_{n}=\frac{2 \pi}{T}\\&16.68=\frac{2 \pi}{T}\\&T=0.3767 s\end{aligned}$

b)

$Given, \(t=10 s , \quad y(t)=6 in = A\)\\\(y(t)=A \cos \left(\omega_{n} t+\phi\right) \rightarrow 0\)\\\(6=6 \cos (16.68 \times 10+\phi)\)\\\(1=\cos (166.8+\phi)\)\\\(166.8+\phi=0\)\\\phi=-166.8\)\\At \(t=0, \quad y(0)=6 \cos (16.68 \times 0-166.8)\) {y(0)}=-5.74 in$

5 0

3 years ago

(35-39) A student travels on a school bus in the middle of winter from home to school. The school bus temperature is 68.0° F. Th

arlik [135]

Answer:

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

Explanation:

From Heat Transfer we determine that heat transfer rate due to electromagnetic radiation ( $\dot Q$ ), measured in BTU per hour, is represented by this formula:

$\dot Q = \epsilon\cdot A\cdot \sigma \cdot (T_{s}^{4}-T_{b}^{4})$ (1)

Where:

$\epsilon$ - Emissivity, dimensionless.

$A$ - Surface area of the student, measured in square feet.

$\sigma$ - Stefan-Boltzmann constant, measured in BTU per hour-square feet-quartic Rankine.

$T_{s}$ - Temperature of the student, measured in Rankine.

$T_{b}$ - Temperature of the bus, measured in Rankine.

If we know that $\epsilon = 0.90$ , $A = 16.188\,ft^{2}$ , $\sigma = 1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}}$ , $T_{s} = 554.07\,R$ and $T_{b} = 527.67\,R$ , then the heat transfer rate due to electromagnetic radiation is:

$\dot Q = (0.90)\cdot (16.188\,ft^{2})\cdot \left(1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}} \right)\cdot [(554.07\,R)^{4}-(527.67\,R)^{4}]$

$\dot Q = 417.492\,\frac{BTU}{h}$

Under the consideration of steady heat transfer we find that the net energy transfer from the student's body during the 20 min-ride to school is:

$Q = \dot Q \cdot \Delta t$ (2)

Where $\Delta t$ is the heat transfer time, measured in hours.

If we know that $\dot Q = 417.492\,\frac{BTU}{h}$ and $\Delta t = \frac{1}{3}\,h$ , then the net energy transfer is:

$Q = \left(417.492\,\frac{BTU}{h} \right)\cdot \left(\frac{1}{3}\,h \right)$

$Q = 139.164\,BTU$

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

7 0

3 years ago