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3 years ago

3 MAOP stands for which of the following?

Engineering

1 answer:

raketka [301]3 years ago

3 0

Answer:

MAOP Master of Arts in Organizational Psychology (various universities)

MAOP Maximum Allowable Operating Pressure

MAOP Mid-Atlantic Association of Oracle Professionals

MAOP Mid Atlantic Oncology Program

MAOP Maryland Association of Osteopathic Physicians

MAOP Master Air Operations Planner

MAOP Meharry Association of Office Personnel

MAOP Managers' Annual Operating Plan

MAOP Military Assistance Officer Program (US DoD)

Explanation:

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The heat flux through a 1-mm thick layer of skin is 1.05 x 104 W/m2. The temperature at the inside surface is 37°C and the tempe

miss Akunina [59]

Answer:

a) Thermal conductivity of skin: $k_{skin}=1.5W/mK$

b) Temperature of interface: $T_{interface}=35.6\°C$

Heat flux through skin: $\frac{Q}{A}=2100W/m^2$

Explanation:

$k=\frac{QL}{A(T_{2}-T_{1})}$

Where: $k$ is thermal conductivity of a material, $\frac{Q}{A}$ is heat flux through a material, $L$ is the thickness of the material, $T_{1}$ is the temperature on the first side and $T_{2}$ is the temperature on the second side

$k_{skin}=\frac{QL}{A(T_{2}-T_{1})}$

$k_{skin}=\frac{Q}{A}*\frac{L}{(T_{2}-T_{1})}$

$k_{skin}=1.05*10^{4}*\frac{1*10^{-3}}{(37-30)}$

$k_{skin}=1.5W/mK$

$k_{insulation}=\frac{k_{skin}}{2}$

$k_{insulation}=\frac{1.5}{2}$

$k_{insulation}=0.75W/mK$

The heat flux between both surfaces is constant, assuming the temperature is maintained at each surface.

$\frac{Q}{A}=\frac{k(T_{2}-T_{1})}{L}$

$\frac{k_{skin}(T_{skin}-T_{interface})}{L_{skin}}=\frac{k_{insulation}(T_{interface}-T_{insulation})}{L_{insulation}}$

$\frac{1.5*(37-T_{interface})}{0.001}=\frac{0.75*(T_{interface}-30)}{0.002}$

$55500-1500T_{interface}=375T_{interface}-11250$

$1875T_{interface}=66750$

$T_{interface}=35.6\°C$

$\frac{Q}{A}=\frac{k_{skin}(T_{skin}-T_{interface})}{L_{skin}}$

$\frac{Q}{A}=\frac{1.5*(37-35.6)}{0.001}$

$\frac{Q}{A}=2100W/m^2$

3 0

3 years ago

Technician A says that latent heat is hidden heat and cannot be measured on a thermometer. Technician B says that latent heat is

AysviL [449]

Answer: C

Both A and B are correct

Explanation:

Latent heat is the hidden heat.

Latent heat is the heat energy required to change one state of matter to another state of matter without change in temperature. For example, solid state to liquid state, or liquid state to gaseous state.

Thermometer can not detect the latent heat. That is why it is called hidden heat.

If Technician A says that latent heat is hidden heat and cannot be measured on a thermometer. And Technician B says that latent heat is hidden heat that is required for a change of state of matter, then we can therefore conclude that both Technician A and Technician B are correct.

5 0

3 years ago

Solve the inequality. Then graph your solution.<br> -9v – 10 < 7y +6

Cerrena [4.2K]

16

if you add 9+10 you get 18 - 7+6

5 0

3 years ago

with a digital system, if you have measured incorrectly and use too low of a kvp for adequate penetration, what do you need to d

Lubov Fominskaja [6]

The x-ray beam's penetrating power is regulated by kVp (beam quality). Every time an exposure is conducted, the x-rays need to be powerful (enough) to sufficiently penetrate through the target area.

<h3>How does kVp impact the exposure to digital receptors?</h3>

The radiation's penetration power and exposure to the image receptor both increase as the kVp value is raised.

<h3>Exposure to the image receptor is enhanced with an increase in kVp, right?</h3>

Due to an increase in photon quantity and penetrability, exposure at the image receptor rises by a factor of five of the change in kVp, doubling the intensity at the detector with a 15% change in kVp.

To know more about kVp visit:-

brainly.com/question/17095191

#SPJ4

5 0

1 year ago

An air-conditioning system operating on the reversed Carnot cycle is required to transfer heat from a house at a rate of 755 kJ/

Lyrx [107]

Answer:

There is 0.466 KW required to operate this air-conditioning system

Explanation:

<u>Step 1:</u> Data given

Heat transfer rate of the house = Ql = 755 kJ/min

House temperature = Th = 24°C = 24 +273 = 297 Kelvin

Outdoor temperature = To = 35 °C = 35 + 273 = 308 Kelvin

<u>Step 2: </u> Calculate the coefficient of performance o reversed carnot air-conditioner working between the specified temperature limits.

COPr,c = 1 / ((To/Th) - 1)

COPr,c = 1 /(( 308/297) - 1)

COPr,c = 1/ 0.037

COPr,c = 27

<u>Step 3:</u> The power input cna be given as followed:

Wnet,in = Ql / COPr,max

Wnet, in = 755 / 27

Wnet,in = 27.963 kJ/min

Win = 27.963 * 1 KW/60kJ/min = 0.466 KW

There is 0.466 KW required to operate this air-conditioning system

3 0

3 years ago