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3 years ago

Which of the following scenarios represents the best use of a scientific model?

Physics

1 answer:

valkas [14]3 years ago

4 0

Answer:

Predicting weather patterns is the answer :-)

Explanation:

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Air is compressed from 14.7 psia and 60°F to a pressure of 150 psia while being cooled at a rate of 10 Btu/lbm by circulating wa

iren2701 [21]

Answer:

A. 6.36 lbm/s

b. $T_2=341\textdegree F$

Explanation:

a. Given the following information;

#Compressor inlet:

Air pressure, $P_1=14.7psia,T_1=60\textdegree F$

#Compressor outlet:

$Air \ Pressure, P_2=150psia\\\\Air \ volume \ flow \ rate, \dot V_1=5000ft^3/min$

#Cooling rate,

$q_{out}=10Btu/lbm, \dot W=700hp$

# From table A-1E

Gas constant of air $R=0.3704\ psia.ft^3/lbm.R$

Specific enthalpy at $P_1=520R-h_1=124.27Btu/lbm$

Using the mass balance:

$\dot m_{in}-\dot m_{out}=\bigtriangleup \dot m_{sys}=0.0\\\\\dot m_{in}=\dot m_{out}\\\\v_1=\frac{RT_1}{P_1}\\\\v_1=\frac{0.3704\times520}{14.7}\\\\\\v_1=13.1026ft^3/lbm\\\\\dot m=\frac{5000/60}{13.1025}\\\\\dot m=6.36\ lbm/s$

Hence, the mass flow rate of the air is 6.36lbm/s

b The specific enthalpy at the exit is defined as the energy balance on the system:

$\dot m_{in}-\dot m_{out}=\bigtriangleup \dot m_{sys}=0.0\\\\\dot m_{in}=\dot m_{out}\\\\\dot W_{in}+ \dot mh_1=\dot Q_{out}+\dot m_2\\\\h_2=\frac{\dot W_{in}-\dot Q_{out}}{\dot m}+h_1\\\\h_2=\frac{700\times 0.7068-10\times 6.36}{6.36}+124.27\\\\h_2=192.06\ Btu/lbm\\\\#at \ h_2=182.06Btu/lbm, T_2=801R=341\textdegree F$

Hence, the temperature at the compressor exit $T_2=341\textdegree F$

5 0

3 years ago

On a hot day, Dave is relaxing by the pool. He touches the concrete next to the pool and then touches the surface of the water.

Alenkasestr [34]

Answer:

the concrete because the concrete absorbs the suns energy, making it hot

Explanation:

Please give me brainliest

5 0

3 years ago

Read 2 more answers

Hannah has information about an object in circular orbit around Earth.

Elena-2011 [213]

The correct answer is:

the distance of the orbiting object to Earth.

In fact, we know that the gravitational force that keeps the object in circular motion around the Earth is equal to the centripetal force, so we can write:

$G\frac{Mm}{r^2}=m\frac{v^2}{r}$

If we re-arrange the equation, we find an expression for the tangential speed of the object:

$v=\sqrt{\frac{GM}{r}}$

and we see that it depends on 3 quantities: G, M (the mass of the Earth) and r (the distance of the object from the Earth).

3 0

3 years ago

Read 2 more answers

Compressional forces within the crust can produce:

brilliants [131]

tension, compression, and shearing and can i get brainliest plz

4 0

3 years ago

What is the average velocity of atoms in 1.00 mol of neon (a monatomic gas) at 465 K? For m, use 0.0202 kg.

TiliK225 [7]

Answer: 757m/s

Explanation:

Given the following :

Mole of neon gas = 1.00 mol

Temperature = 465k

Mass = 0.0202kg

Using the ideal gas equation. For calculating the average kinetic energy molecule :

0.5(mv^2) = 3/2 nRt

Where ;

M = mass, V = volume. R = gas constant(8.31 jK-1 mol-1, t = temperature in Kelvin, n = number of moles

Plugging our values

0.5(0.0202 × v^2) = 3/2 (1 × 8.31 × 465)

0.0101 v^2 = 5796.225

v^2 = 5796.225 / 0.0101

v^2 = 573883.66

v = √573883.66

v = 757.55109m/s

v = 757m/s

5 0

3 years ago