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2 years ago

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How to convert 45.77 cubic inches to cubic centimeters

1 answer:

Paha777 [63]2 years ago

6 0

45✖️77 and then you will get your answer

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Some steamboat restaurants use paper pots for their customers to boil the food themselves. What is the reason for the paper not

Ber [7]

Answer: D. 1,2,3

The paper pots used in steamboat restaurants will not catch fire when in contact with the flame for the heat is conducted quickly to the water because the paper is thin and the burning temperature of the paper is much higher than the boiling point of water. Lastly, the texture of the paper is thick that it can withstand the temperature of the flame.

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3 years ago

The temperature of a sample of water increases from 20°c to 46.6°c as it absorbs 5650 calories of heat what is the mass of the s

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M=energy transferred/ (temperature change*specific heat) M= 5650/(26.6*1.0) M=212g

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3 years ago

Please help!!! what is the main point of paragraph 3?

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2 years ago

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Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is

Nataly [62]

Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of $c_p$ is determined via the expression:

$c_p = \frac{KR}{K_1}$

where ;

R = universal gas constant = $\frac{8.314 \ J}{28.97 \ kg.K}$

k = constant = 1.4

$c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}$

$c_p= 1.004 \ kJ/kg.K$

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

$0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}$ ------ equation(1)

we can rewrite the above equation as :

$0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}$

$T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}$

where:

$T_1 = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K$

$T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}$

$T_2 = 402.36 \ K$

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation: $\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}$

$P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}$

$P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}$

$P_2 = 17.79 \ bar$

Therefore, the exit pressure is 17.79 bar

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2 years ago

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marysya [2.9K]

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3 years ago

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