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ZanzabumX [31]
3 years ago
8

Many scientific discoveries are made by chance. While studying one phenomenon, a scientist may discover a different one. Such wa

s the case when William Herschel discovered the planet Uranus in 1781. Herschel was charting faint stars in the sky, when he discovered something that was not a star. How can astronomers tell the difference between a star and a planet
Physics
1 answer:
Vedmedyk [2.9K]3 years ago
3 0

Answer:

One simple explanation is that planets move, while stars remained fixed in the sky. You can observe a planet's orbit, but a star will stay in the same position.

Explanation:

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Explain how smoking damages the respiratory system and name two conditions that can develop when a person smokes. ribe the locat
kipiarov [429]

Answer

Explanation:

Smoking can cause lung disease by damaging your airways and the small air sacs (alveoli) found in your lungs. Lung diseases caused by smoking include COPD, which includes emphysema and chronic bronchitis. Cigarette smoking causes most cases of lung cancer.

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3 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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Which kind of air front forms when a warm air mass meets the area of a cooler air mass?
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Which of the following changes would be a physical change to a substance?(1 point)
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