Question 1

A liquid has a specific gravity of 0.68. What is its density?

Accepted Answer

Specific gravity is the ratio of the density of a substance to the density of a reference substance, typically water for liquids. Since the specific gravity is 0.68, the density of the liquid is 0.68 times the density of water (1000 kg/m³), which equals 680 kg/m³.

Question 2

A 12000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with oil with a density of 800 kg/m³ and capped at both ends with tight-fitting pistons. The wider arm of the U-tube has a radius of 18.0 cm and the narrower arm has a radius of 5.00 cm. The car rests on the piston on the wider arm of the U-tube. The pistons are initially at the same level. What is the force that must be applied to the smaller piston in order to lift the car after it has been raised 1.20 m? For purposes of this problem, neglect the weight of the pistons.

Accepted Answer

The pressure exerted by the oil is the same at any given height in the fluid, so the pressure at the bottom of the narrower arm of the U-tube will be the same as the pressure at the bottom of the wider arm (where the car is resting). From the formula for pressure in a fluid:

$P = ho g h$

where $P$ is the pressure, $ho$ is the density of the fluid, $g$ is the acceleration due to gravity, and $h$ is the height of the fluid column.

We can solve for the pressure at the bottom of the wider arm:

$P = ho g h = (800	ext{ kg/m}^3)(9.81 	ext{ m/s}^2)(1.20 	ext{ m}) = 9417.6 	ext{ N/m}^2$

This pressure is exerted on the larger piston, so we can find the force on the piston using the formula:

$F = PA$

where $A$ is the area of the piston. The area of the larger piston is:

$A = \pi r^2 = \pi (0.18 	ext{ m})^2 = 0.1018 	ext{ m}^2$

Substituting in the pressure, we get:

$F = PA = (9417.6	ext{ N/m}^2)(0.1018	ext{ m}^2) = 960	ext{ N}$

This is the force required to lift the car, but it is the force on the larger piston. To find the force required on the smaller piston, we use the formula relating the forces on the two pistons:

$F_2 = \frac{A_2}{A_1} F_1$

where $F_1$ is the force on the larger piston, $A_1$ is the area of the larger piston, $A_2$ is the area of the smaller piston, and $F_2$ is the force on the smaller piston. Solving for $F_2$, we get:

$F_2 = \frac{A_2}{A_1} F_1 = \frac{\pi (0.05 	ext{ m})^2}{\pi (0.18 	ext{ m})^2} (960	ext{ N}) = 1958.8	ext{ N} \approx 1.96	ext{ kN}$

So the answer is choice (C), 1.96 kN.

Question 3

A 100-kg person sits on a 5-kg bicycle. The weight is borne equally by the two wheels of the bicycle. The tires are inflated to a gauge pressure of 8.0 × 10⁵ Pa and are 2.0 cm wide. What length of each tire is in contact with the ground?

Accepted Answer

The total weight is 1050 N (105 kg * 9.8 m/s²). Each wheel supports half, so 525 N. Pressure = Force/Area, so Area = Force/Pressure = 525 N / (8.0 × 10⁵ Pa). The contact area is width * length, so length = Area/width = (525 N / 8.0 × 10⁵ Pa) / 0.02 m = 3.2 cm.

Question 4

How much force acts on one surface of a rectangular bedroom wall that is 2.20 m by 3.20 m when atmospheric pressure is 1.00 × 10⁵ N/m²?

Accepted Answer

The force is calculated by multiplying the pressure by the area of the wall: F = P × A = 1.00 × 10^5 N/m^2 × (2.20 m × 3.20 m) = 7.04 × 10^5 N.

Question 5

What is the gauge pressure if the absolute pressure is 300 kPa?

Accepted Answer

Gauge pressure is the absolute pressure minus atmospheric pressure (approximately 101 kPa at sea level). Therefore, 300 kPa - 101 kPa = 199 kPa.

Question 6

A platinum sphere has a radius of 1.20 cm. The density of platinum is 21.4×10³ kg/m³. What is the mass of the sphere?

Accepted Answer

The answer of A platinum sphere has a radius of...

Question 7

A U-tube is open to the atmosphere at both ends. Water is poured into the tube until the water rises part-way along the straight sides, and then some oil with a density of 920 kg/m³ is poured into one end. This causes the water surface on that side of the tube to go down by 5.50 mm, and the surface on the other side to go up by the same amount. How much higher is the top surface of the oil on that side of the tube compared with the surface of the water on the other side of the tube?

Accepted Answer

The answer of A U-tube is open to the atmosphere...

Question 8

The Aswan High Dam is 111 m high. What is the water pressure at the foot of the dam?

Accepted Answer

The water pressure at the foot of the dam can be calculated using the formula P = ρgh, where ρ is the density of water (approximately 1000 kg/m³), g is the acceleration due to gravity (approximately 9.81 m/s²), and h is the height of the water column (111 m). This gives P = 1000 * 9.81 * 111 = 1,089,910 Pa, or approximately 1.09 MPa.

Question 9

A cubical box, 5.00 cm on each side, is immersed in a fluid. The pressure at the top surface of the box is 594 Pa and the pressure on the bottom surface is 1133 Pa. What is the density of the fluid?

Accepted Answer

The answer of A cubical box, 5.00 cm on each...

Question 10

A U-tube is open to the atmosphere at both ends. Water is poured into the tube until the water column on the vertical sides of the U is more than 10 cm deep. Then, oil with a density of 950 kg/m³ is poured into one side, until the column of oil is 6.0 cm tall. How much higher is the top surface of the oil on that side of the tube compared with the surface of the water on the other side of the tube?

Accepted Answer

The answer of A U-tube is open to the atmosphere...

Question 11

What is the absolute pressure at a distance 5.00 m below the surface of a lake? Assume the density of the water in the lake is 1000 kg/m³.

Accepted Answer

The absolute pressure at a depth h in a fluid is given by:P = P_atm + ρghwhere:P_atm is the atmospheric pressureρ is the density of the fluidg is the acceleration due to gravityh is the depthIn this case, we have:P_atm = 1.01 × 10^5 N/m^2ρ = 1000 kg/m^3g = 9.81 m/s^2h = 5.00 mPlugging these values into the equation, we get:P = 1.01 × 10^5 N/m^2 + (1000 kg/m^3)(9.81 m/s^2)(5.00 m)= 1.50 × 10^5 N/m^2Therefore, the absolute pressure at a distance 5.00 m below the surface of the lake is 1.50 × 10^5 N/m^2.

Question 12

What is the radius of a sphere that has a density 5000 kg/m³ and a mass 6.00 kg?

Accepted Answer

To find the radius, use the formula for density: Density = Mass/Volume. The volume of a sphere is (4/3)πr³. Solving for r, we get r = ((3 * Mass) / (4 * π * Density))^(1/3). Plugging in the values, r = ((3 * 6) / (4 * π * 5000))^(1/3) = 0.0659 m = 6.59 cm.

Question 13

A basketball player pushes down with a force of 50 N on a basketball that is inflated to a gauge pressure of 8.0 × 10⁴ Pa. What is the diameter of contact between the ball and the floor?

Accepted Answer

The force exerted by the basketball player is equal to the force exerted by the air inside the ball on the floor. The force exerted by the air is equal to the gauge pressure times the area of contact between the ball and the floor. Therefore, the diameter of contact between the ball and the floor is:$$d = \sqrt{\frac{4F}{\pi P}}$$$$d = \sqrt{\frac{4(50 	ext{ N})}{\pi (8.0 	imes 10^4 	ext{ Pa})}}$$$$d = 0.028 	ext{ m} = 2.8 	ext{ cm}$$

Question 14

The deepest point of the Pacific Ocean is 11,033 m, in the Mariana Trench. What is the water pressure at that point? The density of seawater is 1025 kg/m³.

Accepted Answer

The answer of The deepest point of the Pacific Ocean...

Question 15

A 12000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with oil and capped at both ends with tight-fitting pistons. The wider arm of the U-tube has a radius of 18.0 cm and the narrower arm has a radius of 5.00 cm. The car rests on the piston on the wider arm of the U-tube. The pistons are initially at the same level. What is the initial force that must be applied to the smaller piston in order to start lifting the car? For purposes of this problem, neglect the weight of the pistons.

Accepted Answer

The pressure in a hydraulic lift is constant throughout the fluid. Thus, the pressure in the wider arm of the U-tube must be the same as the pressure in the narrower arm. We can use the equation $F_1/A_1=F_2/A_2$, where $F_1$ is the force applied to the larger piston, $A_1$ is the area of the larger piston, $F_2$ is the force applied to the smaller piston, and $A_2$ is the area of the smaller piston. We know that $F_1 = 12000	ext{ N}$, $A_1 = \pi(0.18	ext{ m})^2 = 0.1017	ext{ m}^2$, and $A_2 = \pi(0.05	ext{ m})^2 = 0.0079	ext{ m}^2$. Solving for $F_2$, we get $F_2 = \dfrac{A_2}{A_1}F_1 = \dfrac{0.0079	ext{ m}^2}{0.1017	ext{ m}^2}(12000	ext{ N}) \approx 926	ext{ N}$. Therefore, the answer is B.

Question 16

The weight of a 1200-kg car is supported equally by the four tires, which are inflated to the same gauge pressure. What gauge pressure is required so the area of contact of each tire with the road is 100 cm²?

Accepted Answer

The answer of The weight of a 1200-kg car is...

Question 17

A 13000 N vehicle is to be lifted by a 25 cm diameter hydraulic piston. What force needs to be applied to a 5.0 cm diameter piston to accomplish this?

Accepted Answer

The force applied is determined by the ratio of the areas of the pistons. The area is proportional to the square of the diameter. The force needed is (5/25)^2 * 13000 N = 520 N.

Question 18

A 500-N weight sits on the small piston of a hydraulic machine. The small piston has area 2.0 cm². If the large piston has area 40 cm², how much weight can the large piston support?

Accepted Answer

The answer of A 500-N weight sits on the small...

Question 19

A 1959 penny has a diameter of 19.55 mm, a thickness of 1.55 mm, and a mass of 2.500 g. What is its density?

Accepted Answer

The answer of A 1959 penny has a diameter of...

Question 20

What is the mass of a solid sphere that has a density 15.0 × 10³ kg/m³ and a radius 10.0 cm?

Accepted Answer

The mass of a sphere is calculated using the formula: mass = density × volume. The volume of a sphere is (4/3)πr³. Converting the radius to meters (0.1 m), the volume is (4/3)π(0.1)³ = 4.19 × 10⁻³ m³. The mass is then 15.0 × 10³ kg/m³ × 4.19 × 10⁻³ m³ = 62.8 kg.

Quiz 13: Fluids

A liquid has a specific gravity of 0.68. What is its density?

A 100-kg person sits on a 5-kg bicycle. The weight is borne equally by the two wheels of the bicycle. The tires are inflated to a gauge pressure of 8.0 × 105 Pa and are 2.0 cm wide. What length of each tire is in contact with the ground?

How much force acts on one surface of a rectangular bedroom wall that is 2.20 m by 3.20 m when atmospheric pressure is 1.00 × 105 N/m2 ?

What is the gauge pressure if the absolute pressure is 300 kPa?

A platinum sphere has a radius of 1.20 cm. The density of platinum is 21.4×103 kg/m3. What is the mass of the sphere?

The Aswan High Dam is 111 m high. What is the water pressure at the foot of the dam?

A cubical box, 5.00 cm on each side, is immersed in a fluid. The pressure at the top surface of the box is 594 Pa and the pressure on the bottom surface is 1133 Pa. What is the density of the fluid?

What is the absolute pressure at a distance 5.00 m below the surface of a lake? Assume the density of the water in the lake is 1000 kg/m3.

What is the radius of a sphere that has a density 5000 kg/m3 and a mass 6.00 kg?

A basketball player pushes down with a force of 50 N on a basketball that is inflated to a gauge pressure of 8.0 × 104 Pa. What is the diameter of contact between the ball and the floor?

The deepest point of the Pacific Ocean is 11,033 m, in the Mariana Trench. What is the water pressure at that point? The density of seawater is 1025 kg/m3.

The weight of a 1200-kg car is supported equally by the four tires, which are inflated to the same gauge pressure. What gauge pressure is required so the area of contact of each tire with the road is 100 cm2?