A 0.170 kg cube of ice (frozen water) is floating in glycerin. The glycerin is in a tall cylinder that has inside radius 5.50 cm. The level of the glycerin is well below the top of the cylinder. If the ice completely melts, by what distance does the height of liquid in the cylinder change? Express your answer with the appropriate units. Enter positive value if the surface of the water is above the original level of the glycerin before the ice melted and negative value if the surface of the water is below the original level of the glycerin.

Mercury is poured into a U-tube as shown in Figure P15.17a. The left arm of the tube has cross-sectional area A1 of 10.0 cm2, and the right arm has a cross-sectional area A2 of 5.00 cm2. One hundred grams of water are then poured into the right arm as shown in Figure P15.17b. (a) Determine the length of the water column in the right arm of the U-tube. (b) Given that the density of mercury is 13.6 g/cm3, what distance h does the mercury rise in the left arm?

A uniform wooden board of length L and mass M is hinged at the top of a vertical wall of a container partially filled with a certain liquid (Fig. P15.81). (If there were no liquid in the container, the board would hang straight down.) Three-fifths of the length of the board is submerged in the liquid when the board is in equilibrium. Find the ratio of the densities of the liquid and the board.

(a) Calculate the absolute pressure at an ocean depth of 1 000 m. Assume the density of seawater is 1 030 kg/m3 and the air above exerts a pressure of 101.3 kPa. (b) At this depth, what is the buoyant force on a spherical submarine having a diameter of 5.00 m?

A spherical submersible 2.00 m in radius, armed with multiple cameras, descends under water in a region of the Atlantic Ocean known for shipwrecks and finds its first shipwreck at a depth of 1.75 103 m. Seawater has density 1.03 103 kg/m3, and the air pressure at the oceans surface is 1.013 105 Pa. a. What is the absolute pressure at the depth of the shipwreck? b. What is the buoyant force on the submersible at the depth of the shipwreck?

A manometer is shown in Figure P15.36. Rank the pressures at the five locations indicated from highest to lowest. Indicate equal pressures, if any. FIGURE P15.36

A 10.0-kg block of metal measuring 12.0 cm by 10.0 cm by 10.0 cm is suspended from a scale and immersed in water as shown in Figure P15.24b. The 12.0-cm dimension is vertical, and the top of the block is 5.00 cm below the surface of the water. (a) What are the magnitudes of the forces acting on the top and on the bottom of the block due to the surrounding water? (b) What is the reading of the spring scale? (c) Show that the buoyant force equals the difference between the forces at the top and bottom of the block.

A rock with a mass of 540 g in air is found to have an apparent mass of 342 g when submerged in water. (a) What mass of water is displaced? (b) What is the volume of the rock? (c) What is its average density? Is this consistent with the value for granite?

(a) The pressure inside an alveolus with a 2.00104 -m radius is 1.40103 Pa, due to its fluid-lined walls. Assuming the alveolus acts like a spherical bubble, what is the surface tension of the fluid? (b) Identify the likely fluid. (You may need to extrapolate between values in Table 11.3.)

A fluid flows through a horizontal pipe that widens, making a 45 angle with the y axis (Fig. P15.48). The thin part of the pipe has radius R, and the fluids speed in the thin part of the pipe is v0. The origin of the coordinate system is at the point where the pipe begins to widen. The pipes cross section is circular. a. Find an expression for the speed v(x) of the fluid as a function of position for x 0 b. Plot your result: v(x) versus x. FIGURE P15.48 (a) The continuity equation (Eq. 15.21) relates the cross-sectional area to the speed of the fluid traveling through the pipe. A0v0 = A(x)v(x) v(x)=A0v0A(x) The cross sectional area is the area of a circle whose radius is y(x). The widening pan of the pipe is a straight line with slope of 1 and intercept y(0) = R. y(x) = mx + b = x + R A(x) = [y(x)]2 = (x + R)2 Plug this into the formula for the velocity. Plug this into the formula for the velocity. v(x)=A0v0(x+R)2