Question

A U-tube contains liquid of unknown density. An oil of density 770 kg/m³ is poured into one arm of the tube until the oil column is 13.5 cm high. The oil-air interface is then 4.5 cm above the liquid level in the other arm of the U-tube. Find the density of the liquid.

A raft is made of 11 logs lashed together. Each is 38.0 cm in diameter and has a length of 7.00 m. How many people (whole number) can the raft hold before they start getting their feet wet, assuming the average person has a mass of 68.0 kg? Do not neglect the weight of the logs. Assume the density of wood is 600 kg/m³

A block of mass 0.50 kg and density 2300 kg/m³ is completely submerged under the water and in a static equilibrium on top of a spring (k = 70.0 N/m) that is fixed to the bottom of the container as shown in the figure. How much is the spring compressed? (the density of water is 1000 kg/m³).
cm

A solid cylinder (radius = 13.0 cm, height = 16.0 cm) has a mass of 7.22 kg. This cylinder is floating in water. Then oil (density = 725 kg/m³) is poured on top of the water until the situation shown in the drawing results. How much of the volume of the cylinder is in the water?
m³

A lead block is suspended by means of a string from the underside of a 5.00 kg block of wood of density of 748 kg/m³. If the upper surface of the wood is just level with the water, what is the tension in the string?(The density of lead is 11340 kg/m³)

A spherically shaped balloon has a radius of 9.5 m, and is filled with helium. How much extra mass a cargo can lift, assuming that the skin and the cargo of the balloon have a mass of 900 kg? Neglect the buoyant force on the cargo volume itself.(The density of air is 1.29 kg/m³ and the density of helium is 0.179 kg/m³)

The tank shown in the figure is very large and open to the atmosphere at the top. Find the speed of the liquid leaving the pipe at point (1) if the height h = 32.0 m.

An air-plane has an effective wing surface area of 13.0 m² that is generating the lift force. In level flight the air speed over the top of the wings is 68.0 m/s, while the air speed beneath the wings is 39.5 m/s. What is the weight of the plane?(The density of air is 1.29 kg/m³)

Water at a pressure of 3.50 atm at street level flows into an office building at a speed of 0.55 m/s through a pipe 6.40 cm in diameter. The pipes taper down to 2.50 cm in diameter by the top floor, 29.0 m above. Calculate the water pressure in such a pipe on the top floor.

Water flows straight down from an open faucet. The cross-sectional area of the faucet is 1.20 × 10^-4 m², and the speed of the water is 0.86 m/s as it leaves the faucet. Ignoring air resistance, find the cross-sectional area of the water stream at a point 0.12 m below the faucet.

Answer 1

1) p = p0 + ρgh

Let pi = pressure at interface of the two liquids. pA = atmospheric pressure. ρ1 = density of oil. ρ2 = density of unknown liquid. h1 = 13.5 cm. h2 = 4.5 cm.

At the interface between the oil and liquid the pressure will be the same as the pressure 13.5 – 4.5= 9 cm below the surface:

pA + ρ2g(h1 - h2) = pA + ρ1gh1

ρ2 = ρ1h1 / (h1 - h2) = [770kg/m^3][13.5/9] = 1155 kg/m^3

2) Volume of 11 logs = ((pi r^2) x L) x 11, where r is 0.19 metre, = 8.733 m^3.
Mass of 11 logs = (8.733 x 600) = 5239.8 kg.
When just submerged, the logs will displace 8.733 m^3 of water, so assuming fresh water, it has a mass of 8733 kg.
(8733 – 5239.8)/68 = 51.37 people.

4) Without the drawing, I will assume that the axis of revolution of the cylinder is pointing up. That makes the volume of the submerged cylinder proportional to the height submerged.

There is a buoyant force due to the oil, and another buoyant force due to the water. Their sum is equal to the weight of the cylinder.

Bo + Bw = W
po*g*Vo + pw*g*Vw = pc*g*Vc
po*Vo + pw*Vw = pc*Vc
po*(Vo/Vc) + pw*(Vw/Vc) = pc

where:
Vo, Vw, Vc = volume of cylinder in oil, in water, total volume of cyl
po, pw, pc = density of oil = 725, water = 1000, cylinder=849.93

Further, the portions in oil and water make up the whole, so we have

(Vo/Vc) + (Vw/Vc) = 1

We now have a system of equations, and the solution is

(Vo/Vc) = (pc - pw)/(po - pw)= 0.546
(Vw/Vc) = (po - pc)/(po - pw)= 0.454

Now
0.546*0.13 = 0.071 m is in the oil.