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2 years ago

12

Who else hates this because i do

2 answers:

Arisa [49]2 years ago

8 0

Answer:

uh

Explanation:

┓┏┓┏┓┃

┛┗┛┗┛┃＼○／

┓┏┓┏┓┃ /

┛┗┛┗┛┃ノ)

┓┏┓┏┓┃

┛┗┛┗┛┃

┓┏┓┏┓┃

┛┗┛┗┛┃

┓┏┓┏┓┃

┃┃┃┃┃┃

┻┻┻┻┻┻

ella [17]2 years ago

5 0

The first one is obviously weird

Anddd the second one looks nicee from the front

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Is my paper's main idea, or thesis, clearly stated early on (within the first paragraph, ideally)?

Burka [1]

I dont know is your papers main idea stated clearly?

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2 years ago

Five kilograms of air at 427°C and 600 kPa are contained in a piston–cylinder device. The air expands adiabatically until the pr

son4ous [18]

Answer:

The entropy change of the air is $0.240kJ/kgK$

Explanation:

$T_{1} =427+273K,T_{1} =700K\\P_{1} =600kPa\\P_{2} =100kPa$

$T_{2}$ is unknown

we can apply the following expression to find $T_{2}$

$-w_{out} =mc_{v} (T_{2} -T_{1} )$

$T_{2} =T_{1} -\frac{w_{out } }{mc_{v} }$

now substitute

$T_{2} =700K-\frac{600kJ}{5kg*0.718kJ/kgK} \\T_{2}=533K$

To find entropy change of the air we can apply the ideal gas relationship

Δ $s_{air}$ $=c_{p} ln\frac{T_{2} }{T_{1} } -Rln\frac{P_{2} }{P_{1} }$

Δ $s_{air} =$ $1.005*ln(\frac{533}{700})-0.287* in(\frac{100}{600} )$

Δ $s_{air} =0.240kJ/kgK$

4 0

2 years ago

A 3 m aluminum pole is kept at a residential site for construction

Aliun [14]

Answer:

I don't know sorry

Explanation:

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3 years ago

Water is pumped steadily through a 0.10-m diameter pipe from one closed pressurized tank to another tank. The pump adds 4.0 kW o

jekas [21]

Complete Question

Complete Question is attached below.

Answer:

$V'=5m/s$

Explanation:

From the question we are told that:

Diameter $d=0.10m$

Power $P=4.0kW$

Head loss $\mu=10m$

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$\frac{300*10^3}{\rho g}+35+Hm=\frac{500*10^3}{\rho g}+15+10$

$H_m=(\frac{200*10^3}{1000*9.8}-10)$

$H_m=10.39m$

Generally the equation for Power is mathematically given by

$P=\rho gQH_m$

Therefore

$Q=\frac{P}{\rho g H_m}$

$Q=\frac{4*10^4}{1000*9.81*10.9}$

$Q=0.03935m^3/sec$

Since

$Q=AV'$

Where

$A=\pi r^2\\A=3.142 (0.05)^2$

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Therefore

$V'=\frac{0.03935m^3/sec}{7.85*10^{-3}}$

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2 years ago

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Elenna [48]

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2 years ago

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