Question #15661

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

Expert's answer

1 m = 3.28 ft

10 ft = 304.88 cm^{3}

v = 3.14 * 1/2/2 * 1/2/2 * 304.88 = 60 cm^{3}

density = 1,3520 g/cm^{3}

m = 80.9 g = 0.0809 kg

p = 0.0809 * 9.8 = 0.7927 N

1 atm = 101352 N/m^{3}

s = 3.14 * 0.25*0.25 = 0.196 cm^{3} = 0.00196 m^{3}

0.0809/0.00196 = 42.27 atm

42.27*760 = 32125.2 torr

10 ft = 304.88 cm

v = 3.14 * 1/2/2 * 1/2/2 * 304.88 = 60 cm

density = 1,3520 g/cm

m = 80.9 g = 0.0809 kg

p = 0.0809 * 9.8 = 0.7927 N

1 atm = 101352 N/m

s = 3.14 * 0.25*0.25 = 0.196 cm

0.0809/0.00196 = 42.27 atm

42.27*760 = 32125.2 torr

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