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m_a_m_a [10]
3 years ago
15

A closed, rigid tank is lled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. The gas is he

ated, and the gage pressure at the nal state is 367 kPa. Determine the nal temperature, in °C. The local atmospheric pressure is 1 atm.
Engineering
1 answer:
sergejj [24]3 years ago
8 0

Answer:

T₂ =93.77  °C

Explanation:

Initial temperature ,T₁ =27°C= 273 +27 = 300 K

We know that

Absolute pressure = Gauge pressure + Atmospheric pressure

Initial pressure ,P₁ = 300+1=301 kPa

Final pressure  ,P₂= 367+1 = 368  kPa

Lets take  temperature=T₂

We know that ,If the volume of the gas is constant ,then we can say that

\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}

{T_2}=T_1\times \dfrac{P_2}{P_1}

Now by putting the values in the above equation we get

{T_2}=300\times \dfrac{368}{301}\ K

The temperature in  °C

T₂ = 366.77 - 273  °C

T₂ =93.77  °C

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Answer:

\mu = 0.31

Explanation:

Given data:

mass = 2.00 kg

slope angle = 38.0

From figure

balancing force

mgsin\theta - f = ma   .....1

Balancing torque

F_R = \frac{2}{3} mR^2 \alpha ......2

for pure rolling

\alpha  = \frac{a}{R}

F = \frac{2}{3} ma

from 1 and 2nd equation

mgsin\theta - \frac{2}{3}ma =  ma

mgsin\theta = \frac{5}{3} ma

a = \frac{3}{5} g sin\theta

 = \frac{3\theta 9.8 sin 38}{5} = 3.62 m/s^2

F =\mu N

   = \frac{2}{3} ma

   = \frac{2}{3} 2\times 3.62 = 4.83 N

N =normal force =  mgsin\theta = 2 \times 9.8 sin 38 = 15.44 N

\mu \times 15.44 = 4.83

solving for  coefficent of friction we get

\mu = 0.31

4 0
3 years ago
When you arrive at an intersection with a stop sign in your direction, if there is no marked stop
Illusion [34]

Answer:

C: Stop before entering the pedestrian crosswalk.

Explanation:

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2 years ago
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Consider a plane composite wall that is composed of two materials of thermal conductivities kA = 0.1 W/m*K and kB = 0.04 W/m*K a
nadya68 [22]

Answer:

q=39.15 W/m²

Explanation:

We know that

Thermal resistance due to conductivity given as

R=L/KA

Thermal resistance due to heat transfer coefficient given as

R=1/hA

Total thermal resistance

R_{th}=\dfrac{L_A}{AK_A}+\dfrac{L_B}{AK_B}+\dfrac{1}{Ah_1}+\dfrac{1}{Ah_2}+\dfrac{1}{Ah_3}

Now by putting the values

R_{th}=\dfrac{0.01}{0.1A}+\dfrac{0.02}{0.04A}+\dfrac{1}{10A}+\dfrac{1}{20A}+\dfrac{1}{0.3A}

R_{th}=4.083/A\ K/W

We know that

Q=ΔT/R

Q=\dfrac{\Delta T}{R_{th}}

Q=A\times \dfrac{200-40}{4.086}

So heat transfer per unit volume is 39.15 W/m²

q=39.15 W/m²

4 0
3 years ago
If i eat myself will I get twice as big or disappear completely?
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Disappear completely
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3 years ago
Water at 15°C is to be discharged from a reservoir at a rate of 18 L/s using two horizontal cast iron pipes connected in series
Zina [86]

Answer:

0.245 m^3/s

Explanation:

Flow rate through pipe a is 0.4 m3/s Parallel pipes have a diameter D = 30 cm => r = 15 cm = 0.15 m Length of Pipe a = 1000m Length of Pipe b = 2650m Temperature = 15 degrees Va = V / A = (0.4m3/s) / (3.14 (0.15m)^2) = 5.66 m/s h = (f(LV^2)) / D2g (fa(LaVa^2)) / Da2g = (fb(LbVb^2)) / Da2g and Da = Db; fa = fb LaVa^2 = LbVb^2 => La/Lb = Vb^2/Va^2 Vd^2 = Va^2(La/Lb) => Vb = Va(La/Lb)^(1/2) Vb = 5.66 (1000/2650)^(1/2) => 5.66 x 0.6143 = 3.4769 m/s Vb = 3.4769 m/s V = AVb = 3.14(0.15)^2 x 3.4769 m/s = 0.245 m^3/s

5 0
3 years ago
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