a column of mercury has an inside diameter of 1/2 cm and is 10 ft tall. what is the down ward force in newtons, atmospheres and torr for this column of mercury
1 m = 3.28 ft
10 ft = 304.88 cm3
v = 3.14 * 1/2/2 * 1/2/2 * 304.88 = 60 cm3
density = 1,3520 g/cm3
m = 80.9 g = 0.0809 kg
p = 0.0809 * 9.8 = 0.7927 N
1 atm = 101352 N/m3
s = 3.14 * 0.25*0.25 = 0.196 cm3 = 0.00196 m3
0.0809/0.00196 = 42.27 atm
42.27*760 = 32125.2 torr