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dimulka [17.4K]
3 years ago
6

Pls help me… I missed the whole lesson and I have no clue what to do-

Physics
1 answer:
galben [10]3 years ago
8 0

Answer:

1) 50 facing towards the right

2) 150 facing right

3) 200 facing right

4) 0- no direction

5) 50- facing left

6) 50 facing right

Explanation:

forces in opposite directions and equal magnitudes counteract each other. in number 2 they face the same direction so they would just be added. in number 4 they oppose each other so would be subtracted

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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
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Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of c_p is determined via the expression:

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

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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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<span>E=hc/wav. len
E = (6.62 x 10^-34 x 3 x 10^8)/0.0275 x 10^-9
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