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Svet_ta [14]
3 years ago
10

What is one of the “don’ts” in drawing dimension lines? they should never be labeled they should never be stacked they should ne

ver cross each other they should never have only one measurement value
Engineering
2 answers:
AnnZ [28]3 years ago
7 0

Answer:

they should never cross each other

Explanation:

Strike441 [17]3 years ago
5 0

Answer:

What is one of the “don’ts” in drawing dimension lines? they should never be labeled they should never be stacked they should never cross each other they should never have only one measurement value

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One kilogram of air, initially at 5 bar, 350 K, and 3 kg of carbon dioxide (CO2), initially at 2 bar, 450 K, are confined to opp
pentagon [3]

Answer:

Check the explanation

Explanation:

Energy alance of 2 closed systems: Heat from CO2 equals the heat that is added to air in

m_{a} c_{v,a}(T_{eq} -T_{a,i)} =m_{co2} c_{v,co2} (T_{co2,i} -T_{eq)}

1x0.723x(T_{eq} -350)=3x0.780x(450-T_{eq} ) ⇒T_{eq} = 426.4 °K

The initail volumes of the gases can be determined by the ideal gas equation of state,

V_{a,i}  = \frac{mRT_{a,i} }{P_{a,i} }=  \frac{1x (8.314 28.97 kJ kg • °K)x 350°K}{5 bar x 100KPa bar} = 0.201m^{3}

The equilibrium pressure of the gases can also be obtained by the ideal gas equation

P_{eq=\frac{(m_{a}R_{a}T_{eq})+(m_{a}R_{a}T_{eq} ) }{(V_{a,eq}+V_{CO2,eq)} } =\frac{(m_{a}R_{a}T_{eq})+(m_{a}R_{a}T_{eq} ) }{(V_{a,i}+V_{CO2,i)} }

P_{eq}= 1x(8.314 28.97)x426.4+3x(8.314 44)x426.4

                             (0.201+1.275)

= 246.67 KPa = 2.47 bar

6 0
3 years ago
What test should be performed on abrasive wheels
Svet_ta [14]

Answer:

before wheel is put on it should be looked at for damage and a sound or ring test should be done to check for cracks, to test the wheel it should be tapped with a non metallic instrument (I looked it up)

3 0
3 years ago
Read 2 more answers
What’s another name for a service overcurrent device?
alexandr402 [8]
Overcurrent protective devices, or OCPDs
4 0
3 years ago
A piston-cylinder assembly has initially a volume of 0.3 m3 of air at 25 °C. Mass of the air is 1 kg. Weights are put on the pis
kogti [31]

Answer:

n=2.32

w= -213.9 KW

Explanation:

V_1=0.3m^3,T_1=298 K

V_2=0.1m^3,T_1=1273 K

Mass of air=1 kg

For polytropic process  pv^n=C ,n is the polytropic constant.

  Tv^{n-1}=C

  T_1v^{n-1}_1=T_2v^{n-1}_2

298\times .3^{n-1}_1=1273\times .1^{n-1}_2

n=2.32

Work in polytropic process given as

       w=\dfrac{P_1V_1-P_2V_2}{n-1}

      w=mR\dfrac{T_1-T_2}{n-1}

Now by putting the values

w=1\times 0.287\dfrac{289-1273}{2.32-1}

w= -213.9 KW

Negative sign indicates that work is given to the system or work is done on the system.

For T_V diagram

  We can easily observe that when piston cylinder reach on new position then volume reduces and temperature increases,so we can say that this is compression process.

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3 years ago
1. ELECTRICAL SHOCK
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