Please help me I really need your physics help?

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.

Please help me I really need your physics help?
since i dont know much about it


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http://tutor4physics.com/formulas.htm
Reply: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!
Reply:don't repeat yourself, don't repeat yourself...