Question 1

A large stone is resting on the bottom of the swimming pool.The normal force of the bottom of the pool on the stone is equal to the:

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Question 2

What volume of water is displaced by a submerged 2.5-kg cylinder made of solid aluminum? (aluminum density = 2.7 × 10³ kg/m³ and water density = 1.0 × 10³ kg/m³)

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Question 3

If the column of mercury in a barometer stands at 73.8 cm,what is the atmospheric pressure? (The density of mercury is 13.6 × 10³ kg/m³ and g = 9.80 m/s²)

Accepted Answer

Atmospheric pressure = height of mercury column × density of mercury × acceleration due to gravity= 73.8 × 10^-2 m × 13.6 × 10³ kg/m³ × 9.80 m/s²= 98.361 × 10³ N/m²= 0.984 × 10⁵ N/m²

Question 4

What is the total force on the bottom of a 4.0-m-diameter by 1.5-m-deep round wading pool due to the weight of the air and the weight of the water? (Note the pressure contribution from the atmosphere is 1.0 × 10⁵ N/m²,the density of water is 1 000 kg/m³,and g = 9.8 m/s². )

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Question 5

A ping-pong ball has an average density of 0.0840 g/cm³ and a diameter of 3.80 cm.What force would be required to keep the ball completely submerged under water?

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Question 6

How deep under the surface of a lake would the pressure be double that at the surface? (1 atm = 1.01 × 10⁵ Pa)

Accepted Answer

The pressure in a liquid increases with depth according to the formula P = ρgh, where P is the pressure, ρ is the density of the liquid, g is the acceleration due to gravity, and h is the depth below the surface. Double the pressure at the surface means that the pressure at depth will be P = 2P0 = 2ρgh, where P0 is the pressure at the surface. Setting this equal to the pressure formula for the surface, P = P0 = ρgh, we can solve for h: h = 2P0 / ρg = 2(1 atm)/(1000 kg/m3 × 9.8 m/s2) = 10.3 m. Therefore, the answer is C.

Question 7

A cube of wood of density 0.78 g/cm³ is 11 cm on a side.When placed in water,what height of the block will float above the surface? (water density = 1.00 g/cm³)

Accepted Answer

To find the height of the cube that will float, we need to use the principle of buoyancy:

Buoyant force = weight of displaced water

The weight of the cube is:
Weight = Density x Volume x Gravity
Volume = Side^3 = 11^3 = 1331 cm^3
Weight = 0.78 g/cm^3 x 1331 cm^3 x 9.8 m/s^2 = 10,225.46 g

The weight of the displaced water is:
Weight = Density x Volume x Gravity
Volume = Area of base x Height = 11 cm x height
Weight = 1 g/cm^3 x (11 cm x height) x 9.8 m/s^2 = 107.8 x height g

Since the cube floats, the buoyant force is equal to the weight of the cube, so:

10,225.46 g = 107.8 x height g
height = 10,225.46 g / (107.8 g/cm^3 x 11 cm) = 2.4 cm

Therefore, the height of the cube that will float is 2.4 cm, which corresponds to choice C.

Question 8

A solid rock,suspended in air by a spring scale,has a measured mass of 8.50 kg.When the rock is submerged in water,the scale reads 3.50 kg.What is the density of the rock? (water density = 1 000 kg/m³)

Accepted Answer

The buoyant force on the submerged rock is equal to the weight of the water displaced by the rock. This buoyant force reduces the apparent weight of the rock measured by the scale from its actual weight.

The weight of the water displaced by the rock is:

w = mg - F_B = 8.50 kg x 9.81 m/s^2 - 3.50 kg x 9.81 m/s^2 = 48.97 N

where g is the acceleration due to gravity and F_B is the buoyant force.

The volume of water displaced by the rock is equal to the volume of the rock. The density of the water is 1000 kg/m^3. Therefore, the volume of the rock is:

V = m_water / rho_water = w / rho_water x g = 48.97 N / (1000 kg/m^3 x 9.81 m/s^2) = 0.00499 m^3

The density of the rock is:

rho_rock = m_rock / V = 8.50 kg / 0.00499 m^3 = 1703.41 kg/m^3

Rounding to two significant digits, we get 1.7 × 10^3 kg/m^3 or 1.7 × 10^15 g/cm^3. The correct option is D.

Question 9

The bottom of a flat-bottomed aluminum boat has area = 4.2 m² and mass = 60 kg.If two fishermen and their fishing gear with total mass of 315 kg are placed in the boat,how much lower will the boat ride in the water? (H₂O density = 1.0 × 10³ kg/m³)

Accepted Answer

The weight of the boat can be calculated as: 
mass = density x volume
volume = area x depth
mass = density x area x depth
60 kg = (1.0 x 10^3 kg/m^3) x (4.2 m^2) x depth
depth = 0.357 m

When the fishermen and gear are added, the total weight of the boat and contents is:
mass = density x volume
volume = area x depth_f
mass = density x area x depth_f
315 kg + 60 kg = (1.0 x 10^3 kg/m^3) x (4.2 m^2) x depth_f
depth_f = 0.432 m

The difference in depth is:
0.432 m - 0.357 m = 0.075 m

Therefore, the boat will ride 0.075 m lower in the water. The answer is choice C.

Question 10

Dams at two different locations are needed to form a lake.When the lake is filled,the water level will be at the top of both dams.The Dam #2 is twice as high and twice as wide as Dam #1.How much greater is the force of the water on Dam #2 than the force on Dam #1? (Ignore atmospheric pressure;it is pushing on both sides of the dams. )

Accepted Answer

The force of the water on a dam depends on the height and width of the dam. 
Since Dam #2 is twice as high and twice as wide as Dam #1, the force of the water on Dam #2 will be:
2 x 2 = 4 times greater than the force of water on Dam #1. 
Therefore, the correct answer is C) 8.

Question 11

The water behind Grand Coulee Dam is 1 100 m wide and 150 m deep.Find the hydrostatic force on the back of the dam.(Hint: the total force = average pressure × area)

Accepted Answer

The hydrostatic force is calculated using the formula: Force = average pressure × area. The average pressure is (density of water × gravity × depth/2), and the area is width × depth. Substituting the values, the force is approximately 12.13 × 10¹⁰ N.

Question 12

Atmospheric pressure is 1.0 × 10⁵ N/m²,and the density of air is 1.29 kg/m³.If the density of air is constant as you get higher and higher,calculate the height of the atmosphere needed to produce this pressure.

Accepted Answer

The pressure at a depth h in a fluid is given by P = ρgh, where ρ is the density of the fluid and g is the acceleration due to gravity. In this case, the pressure is 1.0 × 10⁵ N/m², the density of air is 1.29 kg/m³, and the acceleration due to gravity is 9.8 m/s². Solving for h, we get h = P/ρg = (1.0 × 10⁵ N/m²)/(1.29 kg/m³)(9.8 m/s²) = 7900 m.

Question 13

Legend says that Archimedes,in determining whether or not the king's crown was made of pure gold,measured its volume by the water displacement method.If the density of gold is 19.3 g/cm³,and the crown's mass is 600 g,what volume would be necessary to prove that it is pure gold?

Accepted Answer

To determine the volume of the crown, Archimedes measured its water displacement. The volume of the displaced water is equal to the volume of the crown. We can use the formula:

density = mass/volume

Solving for the volume, we get:

volume = mass/density

Substituting the given values, we get:

volume = 600 g / 19.3 g/cm3
volume = 31.1 cm3

Therefore, the necessary volume to prove that the crown is pure gold is 31.1 cm3. 
The correct answer is A.

Question 14

By what factor is the total pressure greater at a depth of 810 m in water than at the surface where pressure is one atmosphere? (water density = 1.0 × 10³ kg/m³,1 atmosphere pressure = 1.01 × 10⁵ N/m²,and g = 9.8 m/s²)

Accepted Answer

The total pressure at a depth of 810 m in water is the sum of the atmospheric pressure and the pressure due to the water column. The pressure due to the water column is given by:$$P = ho g h$$where:* P is the pressure (N/m^2)* ρ is the density of the water (kg/m^3)* g is the acceleration due to gravity (m/s^2)* h is the depth of the water (m)Substituting the given values into the equation, we get:$$P = (1.0 	imes 10^3 	ext{ kg/m}^3)(9.8 	ext{ m/s}^2)(810 	ext{ m}) = 7.9 	imes 10^6 	ext{ N/m}^2$$The total pressure at a depth of 810 m in water is therefore:$$P_{total} = P_{atm} + P_{water} = 1.01 	imes 10^5 	ext{ N/m}^2 + 7.9 	imes 10^6 	ext{ N/m}^2 = 8.0 	imes 10^6 	ext{ N/m}^2$$The total pressure at a depth of 810 m in water is therefore 79 times greater than the pressure at the surface.

Question 15

The bottom of a flat-bottomed aluminum boat has an area of 4.4 m² and the boat's mass is 62 kg.When set afloat in water,how far below the water surface is the boat bottom? (water density = 1.0 × 10³ kg/m³)

Accepted Answer

The buoyant force on the boat is equal to the weight of the water displaced by the boat. The volume of water displaced is equal to the area of the bottom of the boat times the distance that the boat is submerged. The density of water is 1.0 × 10³ kg/m³, so the mass of the water displaced is equal to the volume of water displaced times the density of water. The mass of the water displaced is equal to the mass of the boat, so the volume of water displaced is equal to the mass of the boat divided by the density of water. The distance that the boat is submerged is equal to the volume of water displaced divided by the area of the bottom of the boat.$$V = \frac{m}{ ho} = \frac{62 ext{ kg}}{1.0 imes 10^3 ext{ kg/m}^3} = 0.062 ext{ m}^3$$$$d = \frac{V}{A} = \frac{0.062 ext{ m}^3}{4.4 ext{ m}^2} = 0.014 ext{ m}$$

Question 16

A heavily loaded boat is floating in a pond.The boat sinks because of a leak.What happens to the surface level of the pond?

Accepted Answer

The boat is heavily loaded, which means it displaces a large volume of water. When it sinks, the volume of water displaced by the boat is replaced by the water that was on top of the boat, causing the overall volume of water in the pond to decrease. This means the surface level of the pond will go down.

Question 17

As ice floats in water,about 10% of the ice floats above the surface of the water.If we float some ice in a glass of water,what will happen to the water level as the ice melts?

Accepted Answer

The water level remains unchanged because the volume of water displaced by the ice is equal to the volume of water the ice turns into when it melts. This is due to the principle of buoyancy and the fact that ice expands to occupy a larger volume than the water it was made from, which compensates for the part of the ice that floats above the water level.

Question 18

A 13 500-N car on a hydraulic lift rests on a cylinder with a piston of radius 0.10 m.If a connecting cylinder with a piston of 0.050-m radius is driven by compressed air,what force must be applied to this smaller piston in order to lift the car?

Accepted Answer

Using Pascal's principle, the force applied to the smaller piston is calculated by the ratio of the areas of the pistons: F1/A1 = F2/A2. Solving for F1 gives F1 = (A1/A2) * F2 = (π*(0.05)^2/π*(0.10)^2) * 13,500 N = 3,375 N.

Question 19

In a large tank of liquid,the hydrostatic pressure at a given depth is a function of:

Accepted Answer

Hydrostatic pressure depends on both the depth and the density of the liquid. Surface area is not a factor in determining hydrostatic pressure. Therefore, choices A and C are both valid, making choice D the correct answer.

Question 20

A blimp is filled with 430 m³ of helium.How big a payload can the balloon lift? (The density of air is 1.29 kg/m³;the density of helium is 0.18 kg/m³. )

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Quiz 9: Solids and Fluids

A large stone is resting on the bottom of the swimming pool.The normal force of the bottom of the pool on the stone is equal to the:

What volume of water is displaced by a submerged 2.5-kg cylinder made of solid aluminum? (aluminum density = 2.7 × 103 kg/m3 and water density = 1.0 × 103 kg/m3)

If the column of mercury in a barometer stands at 73.8 cm,what is the atmospheric pressure? (The density of mercury is 13.6 × 103 kg/m3 and g = 9.80 m/s2)

What is the total force on the bottom of a 4.0-m-diameter by 1.5-m-deep round wading pool due to the weight of the air and the weight of the water? (Note the pressure contribution from the atmosphere is 1.0 × 105 N/m2,the density of water is 1 000 kg/m3,and g = 9.8 m/s2. )

A ping-pong ball has an average density of 0.0840 g/cm3 and a diameter of 3.80 cm.What force would be required to keep the ball completely submerged under water?

How deep under the surface of a lake would the pressure be double that at the surface? (1 atm = 1.01 × 105 Pa)

A cube of wood of density 0.78 g/cm3 is 11 cm on a side.When placed in water,what height of the block will float above the surface? (water density = 1.00 g/cm3)

A solid rock,suspended in air by a spring scale,has a measured mass of 8.50 kg.When the rock is submerged in water,the scale reads 3.50 kg.What is the density of the rock? (water density = 1 000 kg/m3)

The bottom of a flat-bottomed aluminum boat has area = 4.2 m2 and mass = 60 kg.If two fishermen and their fishing gear with total mass of 315 kg are placed in the boat,how much lower will the boat ride in the water? (H2O density = 1.0 × 103 kg/m3)