- #1

mirandasatterley

- 62

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The problem is;

“Mercury is poured into a U-tube. The left arm of the tube has cross-sectional area A1 of 10.0cm^2, and the right arm has a cross-sectional area A2 of 5.0cm^2. 100 grams of water are then poured into the right arm. (a) Determine the length of the water column in the right arm of the U-tube. (b) Given that the density of mercury is 13.6g/cm^3, what distance h does the mercury rise in the left arm?

For part (a) of this problem;

m(water) = d(water) V(water)

m(water) = d(water) A2 h(water)

h(water) = 20 cm

For part (b);

We use the point where the water and mercury meet in the right arm and the point horizontal to that in the left arm. And we get an equation;

P1 = P(atm) + d(Hg) g (h1 +h2), where hi is the height on the diagram from the horizontal line to the top of the mercury and h2 is the height from the horizontal line where the water and mercury meet to the horizontal line on the diagram.

P2 = P(atm) + d(water) g h (water)

P1 = P2

d(Hg) g (h1 + h2) = d(water) g h(water)

d(Hg) (h1 + h2) = d(water) h(water)

(h1 + h2) = (d(water) h(water))/ d(Hg)

= 0.0147 m,

But I only want to know h1. Any ideas how I could approach this?

“Mercury is poured into a U-tube. The left arm of the tube has cross-sectional area A1 of 10.0cm^2, and the right arm has a cross-sectional area A2 of 5.0cm^2. 100 grams of water are then poured into the right arm. (a) Determine the length of the water column in the right arm of the U-tube. (b) Given that the density of mercury is 13.6g/cm^3, what distance h does the mercury rise in the left arm?

For part (a) of this problem;

m(water) = d(water) V(water)

m(water) = d(water) A2 h(water)

h(water) = 20 cm

For part (b);

We use the point where the water and mercury meet in the right arm and the point horizontal to that in the left arm. And we get an equation;

P1 = P(atm) + d(Hg) g (h1 +h2), where hi is the height on the diagram from the horizontal line to the top of the mercury and h2 is the height from the horizontal line where the water and mercury meet to the horizontal line on the diagram.

P2 = P(atm) + d(water) g h (water)

P1 = P2

d(Hg) g (h1 + h2) = d(water) g h(water)

d(Hg) (h1 + h2) = d(water) h(water)

(h1 + h2) = (d(water) h(water))/ d(Hg)

= 0.0147 m,

But I only want to know h1. Any ideas how I could approach this?