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3 years ago

15

What did the romans adopt from the Greek representation of the human art form

1 answer:

alexdok [17]3 years ago

5 0

Answer:

The Romans adopted the Greek's use of realistic features when making human art.

Explanation:

i took the test

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10–25. The 45° strain rosette is mounted on the surface of a shell. The following readings are obtained for each gage: ε a = −20

vazorg [7]

Answer:

The answer is 380.32×10^-6

Refer below for the explanation.

Explanation:

Refer to the picture for brief explanation.

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2 years ago

What might cause a shotgun to explode?

Tanzania [10]

Answer:

When you pull the trigger to shoot a shotshell from a shotgun or a cartridge from a rifle or handgun, the firing pin strikes the primer in the base of the cartridge or shotshell. This causes the primer to explode. The spark from the primer ignites the gunpowder, which burns rapidly and converts to a gas.

Explanation:

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2 years ago

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engineering uses data from pareto charts to analyse motor faults caused during their production. Explain one advantage of using

lions [1.4K]

The advantage of a pareto chart is to make sure they have all of their tools

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3 years ago

A specific internal combustion engine has a displacement volume VD of 5.6 liters. The processes within each cylinder of the engi

Kisachek [45]

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

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3 years ago

A counter-flow double pipe heat exchanger is heat heat water from 20 degrees Celsius to 80 degrees Celsius at the rate of 1.2 kg

lakkis [162]

Answer:

$L=107.6m$

Explanation:

Cold water in: $m_{c}=1.2kg/s, C_{c}=4.18kJ/kg\°C, T_{c,in}=20\°C, T_{c,out}=80\°C$

Hot water in: $m_{h}=2kg/s, C_{h}=4.18kJ/kg\°C, T_{h,in}=160\°C, T_{h,out}=?\°C$

$D=1.5cm=0.015m, U=649W/m^{2}K, LMTD=?\°C, A_{s}=?m^{2},L=?m$

Step 1: Determine the rate of heat transfer in the heat exchanger

$Q=m_{c}C_{c}(T_{c,out}-T_{c,in})$

$Q=1.2*4.18*(80-20)$

$Q=1.2*4.18*(80-20)$

$Q=300.96kW$

Step 2: Determine outlet temperature of hot water

$Q=m_{h}C_{h}(T_{h,in}-T_{h,out})$

$300.96=2*4.18*(160-T_{h,out})$

$T_{h,out}=124\°C$

Step 3: Determine the Logarithmic Mean Temperature Difference (LMTD)

$dT_{1}=T_{h,in}-T_{c,out}$

$dT_{1}=160-80$

$dT_{1}=80\°C$

$dT_{2}=T_{h,out}-T_{c,in}$

$dT_{2}=124-20$

$dT_{2}=104\°C$

$LMTD = \frac{dT_{2}-dT_{1}}{ln(\frac{dT_{2}}{dT_{1}})}$

$LMTD = \frac{104-80}{ln(\frac{104}{80})}$

$LMTD = \frac{24}{ln(1.3)}$

$LMTD = 91.48\°C$

Step 4: Determine required surface area of heat exchanger

$Q=UA_{s}LMTD$

$300.96*10^{3}=649*A_{s}*91.48$

$A_{s}=5.07m^{2}$

Step 5: Determine length of heat exchanger

$A_{s}=piDL$

$5.07=pi*0.015*L$

$L=107.57m$

7 0

2 years ago