AP Physics 2 - Fluids | Practice Hub

Seawater density: $1.03\frac{g}{cm^3}$

A baseball has a mass of and a circumference of . Determine what percentage of a baseball will be submerged when it is floating in seawater after a home run?

None of these

Explanation

$F_{net}=F_{gravity}+F_{buoyant}$

$0=\rho_{seawater}*V_{submerged}*g-m*g$

Solving for $V_{submerged}$

$\frac{m_{ball}}{\rho_{seawater}}=V_{submerged}$

Plugging in values

$\frac{145}{1.03}=V_{submerged}$

$V_{submerged}=141cm^3$

If the circumference is , it is

$C=2\pi r=23cm$

$V_{ball}=\frac{4}{3}\pi r^3$

$V_{ball}=\frac{4}{3}\pi *3.66^3$

$V_{ball}=205cm^3$

$\frac{V_{submerged}}{V_{ball}}=\frac{141cm^3}{205cm^3}=69%$

Seawater density: $1.03\frac{g}{cm^3}$

A baseball has a mass of and a circumference of . Determine what percentage of a baseball will be submerged when it is floating in seawater after a home run?

None of these

Explanation

$F_{net}=F_{gravity}+F_{buoyant}$

$0=\rho_{seawater}*V_{submerged}*g-m*g$

Solving for $V_{submerged}$

$\frac{m_{ball}}{\rho_{seawater}}=V_{submerged}$

Plugging in values

$\frac{145}{1.03}=V_{submerged}$

$V_{submerged}=141cm^3$

If the circumference is , it is

$C=2\pi r=23cm$

$V_{ball}=\frac{4}{3}\pi r^3$

$V_{ball}=\frac{4}{3}\pi *3.66^3$

$V_{ball}=205cm^3$

$\frac{V_{submerged}}{V_{ball}}=\frac{141cm^3}{205cm^3}=69%$

Seawater density: $1.03\frac{g}{cm^3}$

A baseball has a mass of and a circumference of . Determine what percentage of a baseball will be submerged when it is floating in seawater after a home run?

None of these

Explanation

$F_{net}=F_{gravity}+F_{buoyant}$

$0=\rho_{seawater}*V_{submerged}*g-m*g$

Solving for $V_{submerged}$

$\frac{m_{ball}}{\rho_{seawater}}=V_{submerged}$

Plugging in values

$\frac{145}{1.03}=V_{submerged}$

$V_{submerged}=141cm^3$

If the circumference is , it is

$C=2\pi r=23cm$

$V_{ball}=\frac{4}{3}\pi r^3$

$V_{ball}=\frac{4}{3}\pi *3.66^3$

$V_{ball}=205cm^3$

$\frac{V_{submerged}}{V_{ball}}=\frac{141cm^3}{205cm^3}=69%$

Pipe has radius $R_{1}$ , and pipe has radius $R_{2}$ . The two pipes are connected. In order for the speed of water in pipe to be times as great as the speed in pipe , what must be $\frac{R_{2}}{R_{1}}$ ?

$\frac{\sqrt{2}}{4}$

$2\sqrt{2}$

$\frac{1}{8}$

$\frac{1}{4}$

Explanation

The continuity equation says that the cross sectional area of the pipe multiplied by velocity must be constant. Let be the water speed in pipe .

$\pi R_{1}^2v=\pi R_{2}^2(8v)$

$\frac{R_{2}^2}{R_{1}^2}=\frac{1}{8}$

$\frac{R_{2}}{R_{1}}=\frac{\sqrt{2}}{4}$

$\frac{\sqrt{2}}{4}$

$2\sqrt{2}$

$\frac{1}{8}$

$\frac{1}{4}$

Explanation

The continuity equation says that the cross sectional area of the pipe multiplied by velocity must be constant. Let be the water speed in pipe .

$\pi R_{1}^2v=\pi R_{2}^2(8v)$

$\frac{R_{2}^2}{R_{1}^2}=\frac{1}{8}$

$\frac{R_{2}}{R_{1}}=\frac{\sqrt{2}}{4}$

$\frac{\sqrt{2}}{4}$

$2\sqrt{2}$

$\frac{1}{8}$

$\frac{1}{4}$

Explanation

The continuity equation says that the cross sectional area of the pipe multiplied by velocity must be constant. Let be the water speed in pipe .

$\pi R_{1}^2v=\pi R_{2}^2(8v)$

$\frac{R_{2}^2}{R_{1}^2}=\frac{1}{8}$

$\frac{R_{2}}{R_{1}}=\frac{\sqrt{2}}{4}$

What is the net force on a ball of mass and volume of when it is submerged under water?

$g = 10\frac{m}{s^2}$

$\rho_{water} = 1.0\frac{g}{ml}$

Explanation

The buoyant force on the ball is simply the weight of water displaced by the ball:

$F_b = V_{ball}\cdot\rho_{water}\cdot g = (0.2m^3)(1000\frac{kg}{m^3})(10\frac{m}{s^2}) = 2000 N$

The force of gravity on the ball is:

$F_g = mg = 15kg (10\frac{m}{s^2})= 150 N$

These forces oppose each other, so we can say:

$F_{net} = F_b - F_g = 2000N - 150N = 1850N$

What is the net force on a ball of mass and volume of when it is submerged under water?

$g = 10\frac{m}{s^2}$

$\rho_{water} = 1.0\frac{g}{ml}$

Explanation

The buoyant force on the ball is simply the weight of water displaced by the ball:

$F_b = V_{ball}\cdot\rho_{water}\cdot g = (0.2m^3)(1000\frac{kg}{m^3})(10\frac{m}{s^2}) = 2000 N$

The force of gravity on the ball is:

$F_g = mg = 15kg (10\frac{m}{s^2})= 150 N$

These forces oppose each other, so we can say:

$F_{net} = F_b - F_g = 2000N - 150N = 1850N$

What is the net force on a ball of mass and volume of when it is submerged under water?

$g = 10\frac{m}{s^2}$

$\rho_{water} = 1.0\frac{g}{ml}$

Explanation

The buoyant force on the ball is simply the weight of water displaced by the ball:

$F_b = V_{ball}\cdot\rho_{water}\cdot g = (0.2m^3)(1000\frac{kg}{m^3})(10\frac{m}{s^2}) = 2000 N$

The force of gravity on the ball is:

$F_g = mg = 15kg (10\frac{m}{s^2})= 150 N$

These forces oppose each other, so we can say:

$F_{net} = F_b - F_g = 2000N - 150N = 1850N$

Suppose that a barometer is filled with water, as shown in the diagram below. If this barometer's column has a height of , what would be the height of a barometer that used mercury instead of water?

Note: Mercury has a specific gravity of .

Vt physics barometer problem

Explanation

For this question, we're asked to see how the height of a barometer's column will change once the liquid changes from water to mercury.

Barometers are a measure of atmospheric pressure. As the ambient pressure pushes down on the surface of the liquid, this pressure is transmitted throughout the liquid, such that it causes the liquid level in the water to rise. As the ambient pressure increases, the column in the barometer climbs.

We'll need to use the equation for pressure as it relates to density and height.

$P=\rho gh$

Since the ambient pressure is the same in both instances, we can set these terms equal to each other. That is to say, it doesn't matter which liquid is being used in the column, because both of them will feel the same pressure from the atmosphere.

$P_{w}=P_{m}$