- #1

MelissaJL

- 50

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## Homework Statement

1) Calculate the absolute pressure at an ocean depth of two kilometers. Assume the density of seawater is 1000 kg/m

^{3}, the density of air is 1.3 kg/m

^{3}, and that the depth of the atmosphere above the ocean is 10 km.

2) The small piston of a hydraulic lift has a cross sectional area of 3.0 cm

^{2}, and its large piston has a cross sectional area of 200 cm

^{2}. What downward force of magnitude F

_{1}must be applied to the small piston for the lift to raise a load whose mass is 1000 kg?

3) The mass of an object is 3 kg. When the object is suspended from a spring and submerged in water, the scale reads 2 N. Find the density of the object.

## Homework Equations

1) P=P

_{o}+ρgh

2)F=mg, F

_{1}= [itex]\frac{(F2)(A1)}{(A2)}[/itex]

3)F=mg, ρ= [itex]\frac{m}{v}[/itex]

## The Attempt at a Solution

1) So this one I thought I should add the density and distance values together into the formula as following and also using the value of air pressure as P

_{o}=1 atm=1.013x10

^{5}kg/m*s

^{2}. I'm not very confident in this one though:

P=P

_{o}+(ρ

_{sea}+ρ

_{air})g(h+y), where h=10 km and y=2 km.

P=1.013x10

^{5}kg/m*s

^{2}+(1001.3 kg/m

^{3})(9.81 m/s

^{2})(12 km)=2.19x10

^{5}kg/m*s

^{2}

2)

F

_{2}=(1000 kg)(9.81 m/s

^{2})=9810 N

F

_{1}=[itex]\frac{(9810 N)(0.03 m)}{(2 m)}[/itex]=147.2 N

3)

Force exerted by gravity on the object:

F=(3 kg)(9.81 m/s

^{2})= 29.43 N

Difference between submerged and not:

29.43 N-2 N= 27.43 N

Volume displaced by liquid and therefore the volume of the object (I'm not sure if this part is right):

[itex]\frac{(27.43 N)}{(9.81 m/s^{2})}[/itex] = 2.796 kg ∴ V= 2.796 m

^{2}

ρ=[itex]\frac{(3 kg)}{(2.796 m^{3})}[/itex] = 1.073 kg/m

^{3}

I'm quite dyslexic and new to putting in formulas on here so I apologize for any typos that may be in here. Thank you.