Question 1

A hot air balloon along with its cargo has a mass of 4.0 x 10⁵ kg, and it holds 7.0 x 10⁵ m³ of hot air. It is floating at a constant height in air with a density of 1.29 kg/m³. What is the density of the hot air in the balloon?

Accepted Answer

First, we need to calculate the weight of the displaced air for the balloon to be floating at a constant height. This is given by:
weight of displaced air = density of air x volume of displaced air
weight of displaced air = 1.29 kg/m^3 x 7.0 x 10^15 m^3
weight of displaced air = 9.03 x 10^15 kg

Since the weight of the balloon and its cargo is 4.0 x 10^15 kg, this means that the weight of the displaced air is supporting the weight of the balloon and its cargo.

Now, we can use the fact that the weight of the hot air inside the balloon must be equal to the weight of the displaced air to find the density of the hot air:
weight of hot air = density of hot air x volume of hot air
weight of hot air = density of air x volume of displaced air
density of hot air x volume of hot air = density of air x volume of displaced air
density of hot air = (density of air x volume of displaced air) / volume of hot air
density of hot air = (1.29 kg/m^3 x 7.0 x 10^15 m^3) / (7.0 x 10^13 m^3)
density of hot air = 0.72 kg/m^3

Therefore, the answer is B) 0.72 kg/m^3.

Question 2

For a certain patient, the build-up of fatty tissue on the wall of an artery has decreased the arterial radius by 10%. By how much would the pressure provided by the heart have to be increased to maintain a constant volume of blood flow? Model the blood as an ideal incompressible fluid.

Accepted Answer

According to Poiseuille's Law, the volume of blood flow through a vessel is directly proportional to the fourth power of the radius of the vessel. Therefore, if the radius of the artery decreases by 10%, the volume of blood flow through the artery decreases by (0.9)^4 = 0.66 or 34%. To maintain a constant volume of blood flow, the pressure provided by the heart must be increased by the same factor, which is 1/0.66 or approximately 1.52 or 52%. Therefore, the correct answer is (B) 52%.

Question 3

Water flowing through a cylindrical pipe suddenly comes to a section of pipe where the diameter decreases to 86% of its previous value. If the speed of the water in the larger section of the pipe was $32 \mathrm {~m} / \mathrm { s }$ what is its speed in this smaller section if the water behaves like an ideal incompressible fluid?

Accepted Answer

According to the principle of continuity for incompressible fluids, the product of cross-sectional area and velocity remains constant. If the diameter decreases to 86%, the area decreases to $0.86^2$. Therefore, the velocity increases by a factor of $1/0.86^2$, resulting in a speed of approximately 43 m/s in the smaller section.

Question 4

A person who weighs 550 N empties her lungs as much as possible and is then completely immersed in water of density 1000 kg/m³ while suspended from a harness. Her apparent weight is now 21.2 N. What is her density?

Accepted Answer

The answer of A person who weighs 550 N empties...

Question 5

Water, which we can treat as ideal and incompressible, flows at 12 m/s in a horizontal pipe with a pressure of 3.0 × 10⁴ Pa. If the pipe widens to twice its original radius, what is the pressure in the wider section?

Accepted Answer

The answer of Water, which we can treat as ideal...

Question 6

An ideal incompressible fluid flows at 0.252 m/s through a 44-mm diameter cylindrical pipe. The pipe widens to a square cross sectional area that is 5.5 cm on a side. Assume steady flow throughout the system.
(a)What is the speed of the fluid through the square section of pipe?
(b)What is the flow rate in liters per minute?

Accepted Answer

The answer of An ideal incompressible fluid flows at 0.252...

Question 7

Ideal incompressible water flows through a horizontal pipe of cross-sectional area 10.0 cm² at a pressure of 0.250 atm with a volume flow rate of 1.00 × 10^-3 m³/s. At a valve, the effective cross-sectional area of the pipe is reduced to 5.00 cm². What is the pressure at the valve?

Accepted Answer

The volume flow rate is conserved in an incompressible fluid, so we have:

$A_1 v_1 = A_2 v_2$

where $A_1$ and $v_1$ are the cross-sectional area and velocity before the valve, and $A_2$ and $v_2$ are the cross-sectional area and velocity after the valve. We can rearrange this equation to solve for $v_2$:

$v_2 = \frac{A_1}{A_2} v_1$

We also know that the fluid is incompressible, so the mass flow rate is conserved:

$ho_1 A_1 v_1 = ho_2 A_2 v_2$

where $ho_1$ and $ho_2$ are the densities before and after the valve. We can rearrange this equation to solve for $ho_2$:

$ho_2 = \frac{ho_1 A_1 v_1}{A_2 v_2}$

Since the density of water is (approximately) constant, we can use this equation to find the pressure after the valve. The pressure difference across the valve is given by:

$\Delta P = ho_2 g h$

where $g$ is the acceleration due to gravity and $h$ is the difference in height between the two points. In this case, the two points are at the same height, so $\Delta P = 0$. Therefore we have:

$P_1 + \frac{1}{2} ho_1 v_1^2 = P_2 + \frac{1}{2} ho_2 v_2^2$

Solving for $P_2$ and substituting in the given values, we get:

$P_2 = P_1 + \frac{1}{2} (ho_1 v_1^2 - ho_2 v_2^2)$

$P_2 = 0.250 \, 	ext{atm} + \frac{1}{2} \left[\left(1000 \frac{	ext{kg}}{	ext{m}^3}ight) \left(10^{-2} \, 	ext{m}^2ight) \left(1.00 	imes 10^{-3} \frac{	ext{m}}{	ext{s}}ight)^2 - \left(1000 \frac{	ext{kg}}{	ext{m}^3}ight) \left(5.00 	imes 10^{-4} \, 	ext{m}^2ight) \left(\frac{10^{-2} \, 	ext{m}}{2.00 	imes 10^{-3} \, 	ext{m}/	ext{s}}ight)^2ight]$

$P_2 = 0.235 \, 	ext{atm}$

Therefore the answer is D.

Question 8

A rectangular wooden block measures 10.0 cm × 4.00 cm × 2.00 cm. When the block is placed in water, it floats horizontally, with its bottom face 1.65 cm below the surface. What is the density of the wood? The density of water is 1.00 g/cm³.

Accepted Answer

The volume of the wooden block is 10.0 cm × 4.00 cm × 2.00 cm = 80.0 cm^3. When the block is placed in water, it displaces a volume of water equal to its own volume, which is 80.0 cm^3. The buoyant force on the block is equal to the weight of the displaced water, which is 80.0 cm^3 × 1.00 g/cm^3 = 80.0 g. For the block to float with its bottom face 1.65 cm below the surface, the weight of the block must be balanced by the buoyant force plus the weight of the water above the bottom face, which is (10.0 cm × 4.00 cm × 1.65 cm) × 1.00 g/cm^3 = 66.0 g. Therefore, the weight of the block is 80.0 g - 66.0 g = 14.0 g. The density of the wood is equal to its mass divided by its volume, which is 14.0 g ÷ 80.0 cm^3 = 0.175 g/cm^3. Therefore, the answer is A, which is equivalent to 0.825 g/cm^3.

Question 9

In a section of horizontal pipe with a diameter of 3.0 cm, the pressure is 100 kPa and water is flowing with a speed of 1.5 m/s. The pipe narrows to 2.0 cm. What is the pressure in the narrower region? Treat the water as an ideal incompressible fluid.

Accepted Answer

To find the pressure in the narrower region, we can use the Bernoulli's equation, which in one of its forms can be written as $P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$, where $P_1$ and $P_2$ are the pressures at points 1 and 2, $v_1$ and $v_2$ are the velocities at points 1 and 2, and $\rho$ is the density of the fluid (water in this case). The equation assumes the fluid is incompressible and flow is steady.Given:- $P_1 = 100$ kPa- $v_1 = 1.5$ m/s- Diameter at point 1, $D_1 = 3.0$ cm or 0.03 m- Diameter at point 2, $D_2 = 2.0$ cm or 0.02 mFirst, we find the velocities using the continuity equation $A_1v_1 = A_2v_2$, where $A_1$ and $A_2$ are the cross-sectional areas at points 1 and 2, respectively. The area of a circle is given by $A = \pi r^2$, so we can find $v_2$ as follows:$$A_1 = \pi (0.015)^2$$$$A_2 = \pi (0.01)^2$$$$v_2 = v_1 \times \frac{A_1}{A_2} = 1.5 \times \frac{\pi (0.015)^2}{\pi (0.01)^2} = 1.5 \times \frac{(0.015)^2}{(0.01)^2} = 1.5 \times 2.25 = 3.375 \, \text{m/s}$$Now, applying Bernoulli's equation to solve for $P_2$:$$100 \times 10^3 + \frac{1}{2} \times 1000 \times (1.5)^2 = P_2 + \frac{1}{2} \times 1000 \times (3.375)^2$$Solving for $P_2$:$$P_2 = 100 \times 10^3 + \frac{1}{2} \times 1000 \times (1.5)^2 - \frac{1}{2} \times 1000 \times (3.375)^2$$$$P_2 = 100000 + 1125 - 5681.25 = 100000 - 4556.25 = 95443.75 \, \text{Pa} = 95.44 \, \text{kPa}$$Rounding to the nearest kPa gives us 95 kPa, which corresponds to choice A.

Question 10

Fluid flows at 2.0 m/s through a pipe of diameter 3.0 cm. What is the volume flow rate of the fluid?

Accepted Answer

The volume flow rate (Q) of a fluid is given by Q = A × v, where A is the cross-sectional area of the pipe and v is the velocity of the fluid.

The diameter of the pipe is given as 3.0 cm, which means the radius (r) is 1.5 cm (or 0.015 m). The cross-sectional area of the pipe can be calculated as:

A = πr^2 = π × (0.015 m)^2 = 7.07 × 10^-7 m^2

The velocity of the fluid is given as 2.0 m/s. Using these values, we can calculate the volume flow rate as:

Q = A × v = 7.07 × 10^-7 m^2 × 2.0 m/s = 1.4 × 10^-3 m^3/s

Therefore, the best choice is (A) 1.4 × 10^-3 m^3/s.

Question 11

An incompressible ideal fluid flows steadily through a pipe that has a change in diameter. The fluid speed at a location where the pipe diameter is 8.0 cm is 1.28 m/s. What is the fluid speed at a location where the diameter has narrowed to 4.0 cm?

Accepted Answer

The answer of An incompressible ideal fluid flows steadily through...

Question 12

Consider a very small hole in the bottom of a tank that is $17.0 \mathrm {~cm}$ in diameter and filled with water to a height of $90.0 \mathrm {~cm} .$ Find the speed at which the water exits the tank through the hole.

Accepted Answer

The speed of water exiting the tank can be found using Torricelli's law: $ v = \sqrt{2gh} $, where $ g = 9.8 \, \text{m/s}^2 $ and $ h = 0.9 \, \text{m} $. Calculating gives $ v = \sqrt{2 \times 9.8 \times 0.9} \approx 4.20 \, \text{m/s} $.

Question 13

Ideal incompressible fluid flows through a 4.0-cm-diameter pipe at 1.0 m/s. There is a 2.0 cm diameter constriction in the line. What is the speed of the fluid in this constriction?

Accepted Answer

The answer of Ideal incompressible fluid flows through a 4.0-cm-diameter...

Question 14

The bathyscaphe Trieste consists of a thin-walled but very strong cylindrical hull, 15.2 m long and 3.66 m in diameter, filled with gasoline with a density of 880 kg/m³ to make it buoyant. If it is totally submerged in seawater with a density of 1025 kg/m³ and neither rising nor sinking, what is the weight of the Trieste craft?

Accepted Answer

The weight of the craft can be found using the formula:
Weight = Volume x Density x Gravity
The volume of the craft can be found using the formula for the volume of a cylinder: 
Volume = πr2h = π(1.83 m)2(15.2 m) ≈ 150.67 m3

Substituting the given densities and gravity, we get:
Weight = 150.67 m3 x (1025 kg/m3 - 880 kg/m3) x 9.81 m/s2
Weight ≈ 227.11 kN

Therefore, the weight of the Trieste craft is 227 kN.

Question 15

A pump uses a piston of 15 cm diameter that moves at 2.0 cm/s as it pushes a fluid through a pipe. What is the speed of the fluid when it enters a portion of the pipe that is 3.0 mm in diameter? Treat the fluid as ideal and incompressible.

Accepted Answer

The speed of the fluid increases as it moves from a wider to a narrower section of the pipe due to the conservation of mass principle for an incompressible fluid, which can be described by the equation $A_1v_1 = A_2v_2$, where $A_1$ and $A_2$ are the cross-sectional areas of the wider and narrower sections, respectively, and $v_1$ and $v_2$ are the fluid velocities in these sections. Given the diameters, the areas can be calculated using $A = \pi r^2$, leading to a significant increase in velocity when the fluid moves into the narrower section. Calculating with the given numbers, the velocity in the narrower section is found to be 50 m/s.

Question 16

Water is flowing in a drainage channel of rectangular cross-section. The width of the channel is 15 m, the depth of water is 8.0 m, and the speed of the flow is 2.5 m/s. What is the mass flow rate of the water? The density of water is 1000 kg/m³.

Accepted Answer

The mass flow rate formula is given by:
Mass flow rate = density x cross-sectional area x velocity
The cross-sectional area can be found by multiplying the width by the depth of the channel:
Cross-sectional area = 15 m x 8.0 m = 120 m^2
Plugging in the given values, we get:
Mass flow rate = 1000 kg/m^3 x 120 m^2 x 2.5 m/s = 3.0 x 10^5 kg/s
Therefore, the answer is (C) 3.0 x 10^5 kg/s.

Question 17

A hole of radius 1.0 mm forms in the bottom of a very large water storage tank that holds water to a depth of 15 m. At what volume rate will water flow out of the hole?

Accepted Answer

The answer of A hole of radius 1.0 mm forms...

Question 18

A cylinder, semicircular in cross-section, is 10 m long and 5.0 m in radius. It is completely submerged, with its flat bottom side horizontal and 8.0 m below the water surface, as shown in the figure. What is the magnitude of the net force on the curved upper surface of the half-cylinder? Neglect atmospheric pressure.

Accepted Answer

The answer of A cylinder, semicircular in cross-section, is 10...

Question 19

A barge with vertical sides is 10 m wide and 60 m long and is floating in fresh water of density 1000 kg/m³. How much deeper into the water does the barge sink when 3.0 x 10⁵ kg of coal are loaded on the barge?

Accepted Answer

The answer of A barge with vertical sides is 10...

Question 20

Ideal incompressible water is flowing in a drainage channel of rectangular cross-section. At one point, the width of the channel is 12 m, the depth of water is 6.0 m, and the speed of the flow is 2.5 m/s. At a point downstream, the width has narrowed to 9.0 m, and the depth of water is 8.0 m. What is the speed of the flow at the second point?

Accepted Answer

The answer of Ideal incompressible water is flowing in a...

Quiz 13: Fluids

A hot air balloon along with its cargo has a mass of 4.0 x 105 kg, and it holds 7.0 x 105 m3 of hot air. It is floating at a constant height in air with a density of 1.29 kg/m3. What is the density of the hot air in the balloon?

For a certain patient, the build-up of fatty tissue on the wall of an artery has decreased the arterial radius by 10%. By how much would the pressure provided by the heart have to be increased to maintain a constant volume of blood flow? Model the blood as an ideal incompressible fluid.

A person who weighs 550 N empties her lungs as much as possible and is then completely immersed in water of density 1000 kg/m3 while suspended from a harness. Her apparent weight is now 21.2 N. What is her density?

Water, which we can treat as ideal and incompressible, flows at 12 m/s in a horizontal pipe with a pressure of 3.0 × 104 Pa. If the pipe widens to twice its original radius, what is the pressure in the wider section?

Ideal incompressible water flows through a horizontal pipe of cross-sectional area 10.0 cm2 at a pressure of 0.250 atm with a volume flow rate of 1.00 × 10-3 m3/s. At a valve, the effective cross-sectional area of the pipe is reduced to 5.00 cm2. What is the pressure at the valve?

A rectangular wooden block measures 10.0 cm × 4.00 cm × 2.00 cm. When the block is placed in water, it floats horizontally, with its bottom face 1.65 cm below the surface. What is the density of the wood? The density of water is 1.00 g/cm3.