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Whitepunk [10]
3 years ago
11

A high-mass star has more "fuel" than a low-mass star. which lives longer? why?

Physics
1 answer:
forsale [732]3 years ago
6 0
Low mass lives long .
Because High mass loss his energy sooner and it will die . so low mass longer live
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Problem 6.056 Air enters a compressor operating at steady state at 15 lbf/in.2, 80°F and exits at 275°F. Stray heat transfer and
vova2212 [387]

To solve this process it is necessary to consider the concepts related to the relations between pressure and temperature in an adiabatic process.

By definition the relationship between pressure and temperature is given by

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

Here

P = Pressure

T = Temperature

\gamma =The ratio of specific heats. For air normally is 1.4.

Our values are given as,

P_1 = 15lb/in^2\\T_1= 80\°F = 299.817K\\T_2 =400\°F = 408.15K

Therefore replacing we have,

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

(\frac{P_2}{15})=(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}

Solving for P_2,

P_2 = 15*(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}

P_2 = 44.15Lbf/in^2

Therefore the maximum theoretical pressure at the exit is 44.15Lbf/in^2

5 0
3 years ago
Which statement best describes the difference between acceleration and velocity?
Sloan [31]
C. is the answer because acceleration is the change in velocity in time while velocity is speed with a direction
8 0
3 years ago
A speedboat initially at rest accelerates uniformly at 4.0 m/s (squared) for 7.0 s. How fast is the boat moving after 7.0 s?
zaharov [31]
Well if the boat initially at rest accelerates at uniformly at 4.0 m/s (squared) then it would be best to muitlply it so 4.0 squared equals 2 by multiplying that by 7.0 your answer would be 14 s
8 0
3 years ago
Number these from least (1) to most (5) inertia.
worty [1.4K]

Answer:

 1. A feather

2. A baseball

3. a small car

4. A truck

5. A large train

Explanation:

5 0
2 years ago
Read 2 more answers
The smallest detail visible with ground-based solar telescopes is about 1 arc second. How large a region (in km) does this repre
mestny [16]

Answer:

x = 727.5 km

Explanation:

With the conditions given using trigonometry, we can find the tangent

       tan θ = CO / CA

With CO the opposite leg and CE is the adjacent leg which is the distance from the Tierral to Sun

   

        D =150 10⁶ km (1000m / 1 km)

        D = 150 10⁹ m.

We must take the given angle to radians.

       1º  = 3600 arc s  

       π rad = 180º

       θ = 1 arc s (1º / 3600 s arc) (pi rad / 180º) =

       θ = 4.85 10⁻⁶ rad

That angle is extremely small, so we can approximate the tangent to the angle

     

       θ = x / D

       x = θ D

       x = 4.85 10-6  150 109

       x = 727.5 103 m

       x = 727.5 km

4 0
3 years ago
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