The escape velocity of any object from earth is 11.2 km/s. (a) express this speed in m/s and km/h. (b) at what temperature would oxygen molecules (molecular mass is equal to 32.0 g/mol) have an average velocity vrms equal to earth’s escape velocity of 11.1 km/s

a ) 11.1 *10^3 m/s = 39.96 Km/h

b) T_{o2} =1.58*10^5 K

Explanation:

a)= 11.1 km/s =11.1 *10^3 m/s = 39.96 Km/h

b)

M_O2 = 32.00 g/mol =32.0*10^{-3} kg/mol

gas constant R = 8.31 j/mol.K

So,

multiply each side by M_{o2}, so we have

solving for temperature T_{o2}

In the question given,

T_{o2} =1.58*10^5 K

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form