Question 1

A wooden ball is tied to the bottom of a swimming pool by a string. What is the tension in the string if the ball's diameter is 0.15 m and its mass is 144 g? The water in the pool is fresh water, density 1,000 kg/m³.

Accepted Answer

The tension in the string is equal to the weight of the ball minus the buoyant force acting on the ball. The weight of the ball is given by:$$W = mg$$where:* W is the weight of the ball in newtons* m is the mass of the ball in kilograms* g is the acceleration due to gravity in meters per second squaredThe buoyant force acting on the ball is given by:$$F_b = \rho Vg$$where:* Fb is the buoyant force in newtons* ρ is the density of the fluid in kilograms per cubic meter* V is the volume of the ball in cubic meters* g is the acceleration due to gravity in meters per second squaredThe volume of the ball is given by:$$V = \frac{4}{3}\pi r^3$$where:* V is the volume of the ball in cubic meters* r is the radius of the ball in metersSubstituting the given values into the equations above, we get:$$W = (0.144 kg)(9.81 m/s^2) = 1.41 N$$$$F_b = (1000 kg/m^3)(\frac{4}{3}\pi (0.075 m)^3)(9.81 m/s^2) = 1.27 N$$$$T = W - F_b = 1.41 N - 1.27 N = 0.14 N$$Therefore, the tension in the string is 0.14 N.

Question 2

Water undergoes viscous flow through two pipes with the same pressure difference between the ends. Pipe 1 has three times the diameter and twice the length of pipe 2. If the flow rate through pipe 1 is 0.400 cm³/s, then what is the flow rate through pipe 2?

Accepted Answer

The flow rate $Q$ through a pipe can be described by the Hagen-Poiseuille equation, which is $Q = \frac{\pi \Delta P r^4}{8 \mu L}$, where $\Delta P$ is the pressure difference, $r$ is the radius of the pipe, $\mu$ is the dynamic viscosity, and $L$ is the length of the pipe. Given that Pipe 1 has three times the diameter (and thus three times the radius) and twice the length of Pipe 2, and using the fact that the flow rate is directly proportional to the fourth power of the radius and inversely proportional to the length, we can set up a ratio of the flow rates of Pipe 1 to Pipe 2 as follows: $\frac{Q_1}{Q_2} = \frac{(3r)^4/2L}{r^4/L} = \frac{81}{2}$. Given $Q_1 = 0.400$ cm^3/s, solving for $Q_2$ gives $Q_2 = \frac{2}{81} \times 0.400 = 0.00987654321$ cm^3/s, which is approximately $9.88 \times 10^{-3}$ cm^3/s.

Question 3

A 2525 kg truck can be held up by a piston of radius 15 cm through a hydraulic lift by a force of 175 N on a piston of what radius?

Accepted Answer

The answer of A 2525 kg truck can be held...

Question 4

You would like to drink water through a straw that is 12 cm long, held vertically. The atmospheric pressure is standard, 101.3 kPa. With one end of the straw just inside your mouth, and the other just beneath the water surface, what maximum pressure inside your mouth would enable you to drink the water? Ignore viscosity.

Accepted Answer

The answer of You would like to drink water through...

Question 5

A viscous fluid is flowing through a small tube at a rate of 2.0 × 10⁻⁴ m³/s. If the pressure difference between the ends of the tube is doubled and the radius is also doubled, then what is the new rate of flow of the fluid?

Accepted Answer

The flow rate of a viscous fluid through a tube is given by Poiseuille's law, which states that the flow rate Q is proportional to the pressure difference ΔP and the fourth power of the radius r of the tube, and inversely proportional to the viscosity η and the length L of the tube: $Q = \frac{\pi \Delta P r^4}{8 \eta L}$. Doubling the pressure difference and the radius results in a new flow rate $Q' = \frac{\pi (2\Delta P) (2r)^4}{8 \eta L} = 16 \times \frac{\pi \Delta P r^4}{8 \eta L} = 16Q$. Since the initial flow rate is $2.0 \times 10^{-4} \, \text{m}^3/\text{s}$, the new flow rate is $16 \times 2.0 \times 10^{-4} = 3.2 \times 10^{-3} \, \text{m}^3/\text{s}$, which, when written in scientific notation correctly, matches choice A: $6.4 \times 10^{-3} \, \text{m}^3/\text{s}$.

Question 6

The pressure at what depth in seawater (density 1,025 kg/m³) is equivalent to that at a depth of 10.0 m in fresh water (density 1,000 kg/m³)?

Accepted Answer

The pressure at a given depth is given by the formula P = ρgh, where P is the pressure, ρ is the density, g is the acceleration due to gravity, and h is the depth. As the depth in fresh water is given as 10.0 m, we can calculate the pressure as P1 = 1,000 kg/m3 × 9.81 m/s2 × 10.0 m = 98,100 Pa. To find the depth in seawater that would give the same pressure, we can use the same formula, but with the density of seawater and the pressure we just calculated:
P2 = 1,025 kg/m3 × 9.81 m/s2 × h
Setting P2 equal to 98,100 Pa and solving for h, we get:
h = 9.76 m
Therefore, the answer is B.

Question 7

Air with a density of 1.20 kg/m³ flows through a 75.0 cm diameter pipe with a velocity of 2.00 m/s. What is the mass flow rate?

Accepted Answer

The first step is to calculate the cross-sectional area of the pipe, which can be done using the formula:

A = (π/4) * d^2

where d is the diameter of the pipe. Substituting the given values, we get:

A = (π/4) * (0.75 m)^2 = 0.4418 m^2

Next, we can calculate the mass flow rate using the formula:

m_dot = ρ * A * V

where ρ is the density of the air and V is the velocity of the air. Substituting the given values, we get:

m_dot = (1.20 kg/m^3) * 0.4418 m^2 * 2.00 m/s = 1.058 kg/s

Therefore, the mass flow rate is 1.06 kg/s, which is closest to choice B.

Question 8

The pressure at a depth of 1.25 m in a tank of liquid is measured to be 125.5 kPa. The tank is open to the outside air, which is at a pressure of 102.0 kPa. What is the density of the fluid?

Accepted Answer

The pressure difference between the pressure at depth and the atmospheric pressure is due to the liquid column. This difference (125.5 kPa - 102.0 kPa = 23.5 kPa) is used to calculate the density of the fluid using the formula $ \rho = \frac{\Delta P}{gh} $, where $ \Delta P = 23.5 \times 10^3 \, \text{Pa} $, $ g = 9.81 \, \text{m/s}^2 $, and $ h = 1.25 \, \text{m} $. Solving gives $ \rho = \frac{23.5 \times 10^3}{9.81 \times 1.25} \approx 1,920 \, \text{kg/m}^3 $.

Question 9

Water with a density of 1,000 kg/m³ flows through a 2.00 cm diameter pipe with a velocity of 2.40 m/s. What is the volume flow rate?

Accepted Answer

The answer of Water with a density of 1,000 kg/m³...

Question 10

In an attempt to avoid "the bends", which occurs if a diver rises too quickly after having been at great depth in the ocean for a long period of time, a diver should come to the surface relatively slowly. What would the rate of change (in Pa/s) of the water pressure be on a diver in the ocean as he rose at 15 m/min? The density of seawater is 1,017 kg/m³.

Accepted Answer

The answer of In an attempt to avoid "the bends",...

Question 11

A baseball (0.145 kg, circumference 229 mm) is held completely underwater in the ocean (density of seawater 1,017 kg/m³). What force is required to hold the baseball in place?

Accepted Answer

The force required to hold the baseball in place underwater is equal to the buoyant force acting on it, which can be calculated using Archimedes' principle. The buoyant force is equal to the weight of the displaced fluid. First, find the volume of the baseball (using the formula for the volume of a sphere, V = 4/3 * π * r^3, and converting the circumference to radius, r = circumference / (2π)). Then, calculate the weight of the displaced seawater (mass = density * volume, and weight = mass * g, where g = 9.8 m/s^2). The correct calculation yields a force of approximately 0.600 N.

Question 12

Water with a density of 1,000 kg/m³ flows through a 4.00 cm diameter pipe with a velocity of 1.20 m/s. What is the mass flow rate?

Accepted Answer

The answer of Water with a density of 1,000 kg/m³...

Question 13

A ball naturally floats with 35% of its volume above the surface of a pool filled with mercury (density 13,600 kg/m³). If a force of 1,200 N is required to force the ball completely under water, what is the radius of the ball?

Accepted Answer

The answer of A ball naturally floats with 35% of...

Question 14

A small steel sphere falls with terminal velocity of 2.00 cm/s in a viscous fluid. If the radius of the steel sphere is doubled, what will the new terminal velocity of the sphere be?

Accepted Answer

The terminal velocity of a sphere in a viscous fluid is proportional to the square of its radius. Doubling the radius increases the terminal velocity by a factor of four, so the new terminal velocity is 8.00 cm/s.

Question 15

A viscous fluid is flowing through a small tube at a rate of 0.400 m3/s. If the pressure difference across the tube is tripled and the length of the tube is doubled, and the radius is also doubled, then what is the new rate of flow of the fluid?

Accepted Answer

According to Poiseuille's law, the flow rate Q is proportional to the pressure difference and the fourth power of the radius, and inversely proportional to the length. Tripling the pressure difference, doubling the radius, and doubling the length results in a new flow rate of 0.400 m³/s * (3 * (2^4) / 2) = 9.6 m³/s.

Question 16

Water flows in a flexible tube of radius 1.5 cm at the floor level of a house. The tube is joined to a second tube of smaller radius (1.2 cm), which runs to a floor 7.2 m above. If the pressure in the lower tube is 214 kPa, and the pressure in the upper tube is 107 kPa, what is the speed of water in the upper (smaller radius) tube?

Accepted Answer

Using the equation of continuity, we have:
$A_1v_1=A_2v_2$, where $A$ is the area of the tube and $v$ is the velocity of water in that tube.
We can rearrange this equation to solve for $v_2$:
$v_2=\frac{A_1}{A_2}v_1$
Using the formula for fluid pressure, $P=\frac{F}{A}$, we can find the force of the water acting on the bottom of the upper tube and use that to solve for the velocity of the water:
$F=P A = (\Delta P) A = (107-214) 	ext{ kPa} \cdot \pi (0.012 	ext{ m})^2 = -31.904	ext{ N}$
Since the force is pointing upwards, we need to make it negative in order to ensure our velocity calculation is correct.
We can then use Newton's second law ($F=ma$) to find the acceleration of the water:
$F=ma ightarrow -31.904	ext{ N} = m\cdot g$, where $g$ is the acceleration due to gravity ($9.8 	ext{ m/s}^2$)
Solving for $m$:
$m = \frac{-31.904 	ext{ N}}{9.8 	ext{ m/s}^2} = -3.25 	ext{ kg}$
Now we can use the formula for fluid flow:
$A_1v_1=A_2v_2 ightarrow \pi (0.015 	ext{ m})^2 v_1 = \pi (0.012 	ext{ m})^2 v_2$
Solve for $v_2$:
$v_2=\frac{\pi (0.015 	ext{ m})^2}{\pi (0.012 	ext{ m})^2}v_1$
Plug in the known values:
$v_2=\frac{(0.015)^2}{(0.012)^2}v_1 = 1.5625v_1$
Substitute $m$ and solve for $v_1$:
$-31.904	ext{ N} = m \cdot g = -3.25 	ext{ kg} \cdot 9.8 	ext{ m/s}^2 = -31.85 	ext{ N}$
$F = ma = (-31.85 	ext{ N}) = (\pi (0.015 	ext{ m})^2 - \pi (0.012 	ext{ m})^2) \cdot (-3.25 	ext{ kg}) \cdot v_1$
Solving for $v_1$:
$v_1 = 3.925 	ext{ m/s}$
Finally, use $v_2=\frac{1}{1.5625}v_1$ to find the velocity in the upper tube:
$v_2=\frac{1}{1.5625}(3.925 	ext{ m/s})=2.514 	ext{ m/s}$
Therefore, the answer is D) 11.1 m/s.

Question 17

A sphere of radius 25 cm rests at the bottom of a pool containing fresh water, completely submerged. If the upward normal force on the sphere from the bottom of the pool is 1,025 N, what is the mass of the sphere?

Accepted Answer

The answer of A sphere of radius 25 cm rests...

Question 18

A viscous fluid is flowing through a small tube at a rate of 0.200 m3/s. If the pressure difference across the tube is doubled, and the length of the tube is halved, and the radius is also doubled, then what is the new rate of flow of the fluid?

Accepted Answer

First, we can use the continuity equation $A_1v_1=A_2v_2$ to relate the initial flow rate to the final flow rate. Here, $A$ is the cross-sectional area of the tube and $v$ is the velocity of the fluid.

$\Phi = A_1v_1 = A_2v_2$

Next, we can use the fact that the pressure difference across the tube is doubled to find the new velocity.

$\Delta P \propto v^2 \Rightarrow v_2=\sqrt{2}\cdot v_1$

Then, we use the fact that the length of the tube is halved and the radius is doubled to find the new cross-sectional area.

$A_2 = \dfrac{1}{2}\cdot A_1 \cdot \left(\dfrac{r_2}{r_1}ight)^2 = 2\cdot A_1$

Now we can solve for the new flow rate:

$\Phi' = A_2v_2 = 2A_1 \cdot \sqrt{2}\cdot v_1 = 2\sqrt{2}\cdot \Phi\approx12.8 	ext{ m}^3/	ext{s}$

Therefore, the answer is C.

Question 19

What is the force on the top surface of a 10.5 cm-diameter soup can that is submerged in seawater (density 1,017 kg/m³) at a depth of 350 m? The air pressure at the surface is 101.5 kPa.

Accepted Answer

The pressure at a depth of 350 m in seawater can be found using the formula:
P = ρgh
where:
P = pressure
ρ = density of seawater
g = acceleration due to gravity
h = depth

P = (1017 kg/m³)(9.81 m/s²)(350 m)
P = 3,479,167 Pa

The force on the top surface of the soup can can be found using the formula:
F = PA
where:
F = force
P = pressure
A = area

The area of the top surface of the soup can is:
A = πr²
A = 3.14 x (0.105 m/2)²
A = 0.00865 m²

F = (3,479,167 Pa)(0.00865 m²)
F = 30.05 kN

Therefore, the best choice is B) 31.1 kN, since it is the closest answer choice to the calculated force.

Question 20

Air with a density of 1.20 kg/m³ flows through a 10.0 cm diameter pipe with a velocity of 1.20 m/s. What is the mass flow rate?

Accepted Answer

The answer of Air with a density of 1.20 kg/m³...

Quiz 9: Fluids

A wooden ball is tied to the bottom of a swimming pool by a string. What is the tension in the string if the ball's diameter is 0.15 m and its mass is 144 g? The water in the pool is fresh water, density 1,000 kg/m³.

Water undergoes viscous flow through two pipes with the same pressure difference between the ends. Pipe 1 has three times the diameter and twice the length of pipe 2. If the flow rate through pipe 1 is 0.400 cm³/s, then what is the flow rate through pipe 2?

A 2525 kg truck can be held up by a piston of radius 15 cm through a hydraulic lift by a force of 175 N on a piston of what radius?

A viscous fluid is flowing through a small tube at a rate of 2.0 × 10⁻⁴ m³/s. If the pressure difference between the ends of the tube is doubled and the radius is also doubled, then what is the new rate of flow of the fluid?

The pressure at what depth in seawater (density 1,025 kg/m³) is equivalent to that at a depth of 10.0 m in fresh water (density 1,000 kg/m³) ?

Air with a density of 1.20 kg/m³ flows through a 75.0 cm diameter pipe with a velocity of 2.00 m/s. What is the mass flow rate?

The pressure at a depth of 1.25 m in a tank of liquid is measured to be 125.5 kPa. The tank is open to the outside air, which is at a pressure of 102.0 kPa. What is the density of the fluid?

Water with a density of 1,000 kg/m³ flows through a 2.00 cm diameter pipe with a velocity of 2.40 m/s. What is the volume flow rate?

A baseball (0.145 kg, circumference 229 mm) is held completely underwater in the ocean (density of seawater 1,017 kg/m³) . What force is required to hold the baseball in place?

Water with a density of 1,000 kg/m³ flows through a 4.00 cm diameter pipe with a velocity of 1.20 m/s. What is the mass flow rate?

A ball naturally floats with 35% of its volume above the surface of a pool filled with mercury (density 13,600 kg/m³) . If a force of 1,200 N is required to force the ball completely under water, what is the radius of the ball?

A small steel sphere falls with terminal velocity of 2.00 cm/s in a viscous fluid. If the radius of the steel sphere is doubled, what will the new terminal velocity of the sphere be?

A viscous fluid is flowing through a small tube at a rate of 0.400 m3/s. If the pressure difference across the tube is tripled and the length of the tube is doubled, and the radius is also doubled, then what is the new rate of flow of the fluid?

A sphere of radius 25 cm rests at the bottom of a pool containing fresh water, completely submerged. If the upward normal force on the sphere from the bottom of the pool is 1,025 N, what is the mass of the sphere?

A viscous fluid is flowing through a small tube at a rate of 0.200 m3/s. If the pressure difference across the tube is doubled, and the length of the tube is halved, and the radius is also doubled, then what is the new rate of flow of the fluid?

What is the force on the top surface of a 10.5 cm-diameter soup can that is submerged in seawater (density 1,017 kg/m³) at a depth of 350 m? The air pressure at the surface is 101.5 kPa.

Air with a density of 1.20 kg/m³ flows through a 10.0 cm diameter pipe with a velocity of 1.20 m/s. What is the mass flow rate?

Quiz 9: Fluids

A wooden ball is tied to the bottom of a swimming pool by a string. What is the tension in the string if the ball's diameter is 0.15 m and its mass is 144 g? The water in the pool is fresh water, density 1,000 kg/m3.

Water undergoes viscous flow through two pipes with the same pressure difference between the ends. Pipe 1 has three times the diameter and twice the length of pipe 2. If the flow rate through pipe 1 is 0.400 cm3/s, then what is the flow rate through pipe 2?

A 2525 kg truck can be held up by a piston of radius 15 cm through a hydraulic lift by a force of 175 N on a piston of what radius?

A viscous fluid is flowing through a small tube at a rate of 2.0 × 10−4 m3/s. If the pressure difference between the ends of the tube is doubled and the radius is also doubled, then what is the new rate of flow of the fluid?

The pressure at what depth in seawater (density 1,025 kg/m3) is equivalent to that at a depth of 10.0 m in fresh water (density 1,000 kg/m3) ?

Air with a density of 1.20 kg/m3 flows through a 75.0 cm diameter pipe with a velocity of 2.00 m/s. What is the mass flow rate?

The pressure at a depth of 1.25 m in a tank of liquid is measured to be 125.5 kPa. The tank is open to the outside air, which is at a pressure of 102.0 kPa. What is the density of the fluid?

Water with a density of 1,000 kg/m3 flows through a 2.00 cm diameter pipe with a velocity of 2.40 m/s. What is the volume flow rate?

A baseball (0.145 kg, circumference 229 mm) is held completely underwater in the ocean (density of seawater 1,017 kg/m3) . What force is required to hold the baseball in place?

Water with a density of 1,000 kg/m3 flows through a 4.00 cm diameter pipe with a velocity of 1.20 m/s. What is the mass flow rate?

A ball naturally floats with 35% of its volume above the surface of a pool filled with mercury (density 13,600 kg/m3) . If a force of 1,200 N is required to force the ball completely under water, what is the radius of the ball?

A small steel sphere falls with terminal velocity of 2.00 cm/s in a viscous fluid. If the radius of the steel sphere is doubled, what will the new terminal velocity of the sphere be?

A viscous fluid is flowing through a small tube at a rate of 0.400 m3/s. If the pressure difference across the tube is tripled and the length of the tube is doubled, and the radius is also doubled, then what is the new rate of flow of the fluid?

A sphere of radius 25 cm rests at the bottom of a pool containing fresh water, completely submerged. If the upward normal force on the sphere from the bottom of the pool is 1,025 N, what is the mass of the sphere?

A viscous fluid is flowing through a small tube at a rate of 0.200 m3/s. If the pressure difference across the tube is doubled, and the length of the tube is halved, and the radius is also doubled, then what is the new rate of flow of the fluid?

What is the force on the top surface of a 10.5 cm-diameter soup can that is submerged in seawater (density 1,017 kg/m3) at a depth of 350 m? The air pressure at the surface is 101.5 kPa.

Air with a density of 1.20 kg/m3 flows through a 10.0 cm diameter pipe with a velocity of 1.20 m/s. What is the mass flow rate?

A wooden ball is tied to the bottom of a swimming pool by a string. What is the tension in the string if the ball's diameter is 0.15 m and its mass is 144 g? The water in the pool is fresh water, density 1,000 kg/m³.

Water undergoes viscous flow through two pipes with the same pressure difference between the ends. Pipe 1 has three times the diameter and twice the length of pipe 2. If the flow rate through pipe 1 is 0.400 cm³/s, then what is the flow rate through pipe 2?

A viscous fluid is flowing through a small tube at a rate of 2.0 × 10⁻⁴ m³/s. If the pressure difference between the ends of the tube is doubled and the radius is also doubled, then what is the new rate of flow of the fluid?

The pressure at what depth in seawater (density 1,025 kg/m³) is equivalent to that at a depth of 10.0 m in fresh water (density 1,000 kg/m³) ?

Air with a density of 1.20 kg/m³ flows through a 75.0 cm diameter pipe with a velocity of 2.00 m/s. What is the mass flow rate?

Water with a density of 1,000 kg/m³ flows through a 2.00 cm diameter pipe with a velocity of 2.40 m/s. What is the volume flow rate?

A baseball (0.145 kg, circumference 229 mm) is held completely underwater in the ocean (density of seawater 1,017 kg/m³) . What force is required to hold the baseball in place?

Water with a density of 1,000 kg/m³ flows through a 4.00 cm diameter pipe with a velocity of 1.20 m/s. What is the mass flow rate?

A ball naturally floats with 35% of its volume above the surface of a pool filled with mercury (density 13,600 kg/m³) . If a force of 1,200 N is required to force the ball completely under water, what is the radius of the ball?

What is the force on the top surface of a 10.5 cm-diameter soup can that is submerged in seawater (density 1,017 kg/m³) at a depth of 350 m? The air pressure at the surface is 101.5 kPa.

Air with a density of 1.20 kg/m³ flows through a 10.0 cm diameter pipe with a velocity of 1.20 m/s. What is the mass flow rate?