- #1

goonking

- 434

- 3

## Homework Statement

## Homework Equations

P

_{1}= P

_{atm}+ ρ

_{water}vg

P

_{2}= P

_{atm}+ ρ

_{mer.}vg

## The Attempt at a Solution

so both sides of the tubes are open to atmosphere pressure and are in equilibrium.

I can get P

_{1}=P

_{2}

the 1 atm from each side cancels out.

g (gravity) cancels out

I'm left with volume (or distance in this case) of water times density of water = distance of mercury times density of mercury:

(1000 kg/m

^{3}) (0.12m) = distance of mercury ( 13600 kg/m

^{3})

solving for distance of mercury gives 0.088m or 0.88 cm

is this correct?