Surface temperature to Atmospheric temp.

AI Thread Summary
The surface temperature of a city is 20 degrees Celsius, and the problem requires calculating the temperature at 2 km above the surface, given that the temperature decreases at half the average rate of the troposphere. The average rate of temperature decrease in the troposphere is 10 degrees Celsius per km, so half of that would be 5 degrees Celsius per km. Therefore, at 2 km, the temperature would decrease by 10 degrees Celsius, resulting in a temperature of 10 degrees Celsius at that altitude. The calculations provided are confirmed to be correct based on the given parameters.
droseph
Messages
1
Reaction score
0

Homework Statement


The surface temperature of a city is 20 degrees celsius. What is the temperature at 2 km above the surface if the temperature decreases at "half the average rate" of the troposphere? Show your calculations.


Homework Equations


None really, it may be just simple math.


The Attempt at a Solution


rate = 20 degrees / 2 km = 10 degrees loss / km and it says half of that so it would be 5 degrees loss per km. My answer would be 10 degrees celsius @ 2 kms.

I don't know if that's right? Thanks for helping
 
Physics news on Phys.org
Hi droseph,

droseph said:

Homework Statement


The surface temperature of a city is 20 degrees celsius. What is the temperature at 2 km above the surface if the temperature decreases at "half the average rate" of the troposphere? Show your calculations.


Homework Equations


None really, it may be just simple math.


The Attempt at a Solution


rate = 20 degrees / 2 km = 10 degrees loss / km and it says half of that so it would be 5 degrees loss per km. My answer would be 10 degrees celsius @ 2 kms.

I don't know if that's right? Thanks for helping

If that rate for the troposphere (10 degrees/km) is what you should be using, then the rest looks right to me.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Replies
3
Views
792
Replies
7
Views
2K
Replies
6
Views
2K
Replies
4
Views
2K
Replies
4
Views
1K
Replies
12
Views
2K
Back
Top