Suppose the temperature of the atmosphere decreased from a surface virtual temperature of 15.8°C with a lapse rate of 7.4 K/km. Find the average virtual temperature of a layer extending from the surface to 7250 m.
ii. Find the scale height of an atmosphere with an average virtual temperature of −11°C.
iii. Using your result from (ii), find the pressure at a height of 7250 m above MSL in air with the above mean virtual temperature and a MSL pressure of 1025 hPa.
I don't expect you to solve them for me. If you could just show me the process for working them out then I would be very grateful. Particularly where I use the (1+0.6078) for the virtual temp and where I convert to kelvin.
2007-05-05 22:46:26 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
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2007-05-05 22:58:58 · answer #1 · answered by Anonymous · 0⤊ 0⤋
first you must know that a degree K is the same that a Cesius degree
answer of the first i) 7250m =7.25km
each km of height , temperature decreases 7.4° . so for 7.25km,
it decreases 7.4*7.25 =53.65°
and the final temperature is 15.8-53.65 =-37.85
answer to ii the degrease to reach -11 is -11-15.8 = -26.8°
so height in km -26.8/-7.4 =3.62 km
For iii I do not know the formula!
2007-05-06 06:00:25 · answer #2 · answered by maussy 7 · 0⤊ 0⤋
don't repeat yourself, don't repeat yourself...
2007-05-06 05:56:13 · answer #3 · answered by Anonymous · 0⤊ 0⤋