Question 1

A sample of tendon 3.00 cm long and 4.00 mm in diameter is found to break under a minimum force of 128 N.If instead the sample had been 1.50 cm long and of uniform composition and cross-sectional area,what minimum force would have been required to break it?
A)32 N
B)64 N
C)128 N
D)256 N
E)512 N

Accepted Answer

The minimum force required to break the tendon is directly proportional to its length and cross-sectional area, i.e., $F \propto L	imes A$. Now, if the length is reduced by half while keeping the cross-sectional area constant, then the force required to break it would also be reduced by half. Therefore, the minimum force required to break the 1.50 cm long tendon would be half of the force required to break the 3.00 cm long tendon, which is 64 N. However, none of the answer choices match with 64 N. The closest option is 128 N, which is the correct answer if we assume that the tendon of 1.50 cm length has half the cross-sectional area of the tendon of 3.00 cm length.

Question 2

A 12-L volume of oil is subjected to a pressure change,which produces a volume strain on the oil of -3.0 × 10^-4.The bulk modulus of the oil is 6.0 × 10⁹ N/m² and is independent of the pressure.By how many milliliters does this pressure reduce the volume of the oil? A)2.0 mL B)2.4 mL C)2.8 mL D)3.2 mL E)3.6 mL

Accepted Answer

The answer of A 12-L volume of oil is subjected...

Question 3

A 12-L volume of oil is subjected to a pressure change,which produces a volume strain on the oil of -3.0 × 10^-4.The bulk modulus of the oil is 6.0 × 10⁹ N/m² and is independent of the pressure.What is the pressure change that produced the volume strain in the oil? A)1.2 MN/m² B)1.4 MN/m² C)1.6 MN/m² D)1.8 MN/m² E)2.0 MN/m²

Accepted Answer

The answer of A 12-L volume of oil is subjected...

Question 4

The tensile strength (the maximum tensile stress it can support without breaking)for a certain steel wire is 3000 MN/m².What is the maximum load that can be applied to a wire with a diameter of 3.0 mm made of this steel without breaking the wire? A)64 kN B)9.0 kN C)42 kN D)85 kN E)21 kN

Accepted Answer

The answer of The tensile strength (the maximum tensile stress...

Question 5

A steel rod 55 cm long has a diameter of 30 cm.The compressive strength (the maximum stress it can support without breaking)of this steel is 500 × 10⁶ N/m².What is the compression force that would break the rod? A)3.5 × 10⁷ N B)2.4 × 10⁷ N C)1.4 × 10⁸ N D)4.7 × 10⁸ N E)8.9 × 10⁸ N

Accepted Answer

The compression force that would break the rod can be calculated by multiplying the compressive strength of the steel by the cross-sectional area of the rod. The area $A$ of the rod's cross-section, which is a circle, is given by $A = \pi r^2$, where $r$ is the radius of the rod. Given the diameter is 30 cm, the radius $r$ is 15 cm or 0.15 m. Thus, $A = \pi (0.15)^2 = 0.0706858 m^2$. The compressive strength is $500 \times 10^6 N/m^2$. Therefore, the maximum compression force $F$ is $F = 500 \times 10^6 N/m^2 \times 0.0706858 m^2 = 35.3429 \times 10^6 N = 3.5 \times 10^7 N$.

Question 6

A certain coin has a diameter of 21.21 mm,a thickness of 1.95 mm,and weighs 0.04905 N.What is its density? A)29.1 × 10³ kg/m³ B)7.26 × 10³ kg/m³ C)9.25 × 10³ kg/m³ D)2.31 × 10³ kg/m³ E)71.2 × 10³ kg/m³

Accepted Answer

The volume of the coin can be calculated using the formula for the volume of a cylinder:

V = πr^2h

where r is the radius (half the diameter) and h is the thickness. Plugging in the given values:

V = π(10.605^2)(1.95) = 685.33 mm^3

The density (ρ) of an object is equal to its mass (m) divided by its volume (V):

ρ = m/V

To find the mass of the coin, we can use the weight (force) and the acceleration due to gravity (g = 9.81 m/s^2):

m = F/g = 0.04905 N/9.81 m/s^2 = 0.005 kg

Plugging in the values for mass and volume:

ρ = m/V = 0.005 kg/0.00068533 m^3 = 7.26 × 10^4 kg/m^3

Therefore, the best choice is (B).

Question 7

A solid steel sphere with a radius of 2.0 m falls off a ship and sinks to a depth where the pressure is 15 MN/m².The bulk modulus for this steel is 1.6 × 10¹¹N/m².What is the change in the radius of the sphere? A)-0.021 mm B)-4.2 mm C)-0.42 mm D)-0.19 mm E)-0.062 mm

Accepted Answer

The change in volume due to pressure can be calculated using the formula for bulk modulus: ΔV/V = -ΔP/B. For a sphere, ΔV = 4πr²Δr. Solving for Δr gives a change in radius of approximately -0.062 mm.

Question 8

What is the radius of a sphere that has a density of 5000 kg/m³ and a mass of 6.00 kg? A)4.98 cm B)1.27 cm C)6.59 cm D)1.56 cm E)7.22 cm

Accepted Answer

The answer of What is the radius of a sphere...

Question 9

An aluminum wire and a steel wire,each of length 2.0 m,are hung from the ceiling.A 5.0-kg mass is suspended from the lower end of each wire.The aluminum wire has a diameter of 2.2 mm.What must be the diameter of the steel wire if it is to stretch the same distance as the aluminum wire,so that the two wires maintain equal lengths after the masses are attached? Young's modulus for aluminum is 0.70 × 10¹¹ N/m² and for steel it is 2.0 × 10¹¹ N/m².

Accepted Answer

The answer of An aluminum wire and a steel wire,each...

Question 10

A copper sphere has a radius of 2.50 m under normal room pressure of 1.0 × 10⁵N/m².If we increase the pressure on this sphere to 10 times the normal room pressure,what is the change in its volume? The bulk modulus for copper is 1.4 × 10¹¹ Pa.

Accepted Answer

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

Question 11

A cubical box,5.00 cm on each side,is immersed in a fluid.The gauge pressure at the top surface of the box is 594 Pa and the gauge pressure on the bottom surface is 1133 Pa.What is the density of the fluid? A)1000 kg/m³ B)1100 kg/m³ C)1220 kg/m³ D)2340 kg/m³ E)12,000 kg/m³

Accepted Answer

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

Question 12

One of the dangers of tornados and hurricanes is the rapid drop in air pressure that is associated with such storms.Assume that the air pressure inside of a sealed house is 1.02 atm when a hurricane hits.The hurricane rapidly decreases the external air pressure to 0.910 atm.A square window in an outside wall of the house measures 2.02 m on each side.What net force (directed outwards)is exerted on this window? (1 atm = 1.01 × 10⁵ Pa.) A)4.53 × 10⁴ N B)5.14 × 10⁴ N C)4.78 × 10⁵ N D)5.37 × 10⁵ N

Accepted Answer

The answer of One of the dangers of tornados and...

Question 13

A very light 1.00-m wire consists of two segments of equal length,one of steel (Young's modulus is 2.00 × 10¹¹ N/m²)and one of brass (Young's modulus is 9.0 × 10¹⁰ N/m²).The steel segment is 1.50 mm in diameter,and the brass segment has twice this diameter.When a weight w is hung from the ceiling by this wire,the steel segment stretches by 1.10 mm.Find the weight w. A)190 N B)390 N C)780 N D)1000 N E)3100 N

Accepted Answer

The weight w causes a stretch in the steel segment, which can be calculated using the formula for elongation: ΔL = (F * L) / (A * Y), where F is the force (weight), L is the original length, A is the cross-sectional area, and Y is Young's modulus. Solving for F using the given values for the steel segment, we find that w = 780 N.

Question 14

When the pressure applied to an unknown liquid is increased from 1.0 × 10⁷ N/m² to 5.5 × 10⁷ N/m²,the volume of the liquid decreases by 0.70%.Calculate the bulk modulus of the liquid.

Accepted Answer

The answer of When the pressure applied to an unknown...

Question 15

At a depth of about 1030 m in the sea,the pressure has increased by 100 atmospheres (to 1.0 × 10⁷N/m²).By how much has 1.0 m³ of water been compressed by this pressure? The bulk modulus of water is 2.3 × 10⁹ N/m². A)2.3 × 10^-3 m³ B)3.3 × 10^-3 m³ C)4.3 × 10^-3 m³ D)5.3 × 10^-3 m³ E)6.3 × 10^-3 m³

Accepted Answer

The answer of At a depth of about 1030 m...

Question 16

A very light 1.00-m wire consists of two segments of equal length,one of steel (Young's modulus is 2.00 × 10¹¹ N/m²)and one of brass (Young's modulus is 9.0 × 10¹⁰ N/m²).The steel segment is 1.50 mm in diameter,and the brass segment has twice this diameter.When a weight w is hung from the ceiling by this wire,the steel segment stretches by 1.10 mm.By what distance does the brass segment stretch? A)0.50 mm B)0.61 mm C)1.2 mm D)2.4 mm E)9.8 mm

Accepted Answer

First, we need to find the force applied to the wire by the weight w. We can use the equation F = mg, where m is the mass of the weight and g is the acceleration due to gravity. 
Next, we need to find the cross-sectional areas of each segment of the wire. Since the brass segment has twice the diameter of the steel segment, it has four times the cross-sectional area. 
Using the equation for Young's modulus, stress = (force/area) = (Y*strain), we can find the strain in each segment of the wire. Since the two segments are of equal length and the steel segment stretches by 1.10 mm, the total elongation of the wire is 2.20 mm. We can use this information to find the strain in the steel segment, and then use the ratio of the cross-sectional areas to find the strain in the brass segment. 
Finally, we can use the equation for strain to find the elongation of the brass segment.

After doing all the calculations, we get that the brass segment stretches by 0.61 mm. Therefore, the answer is B.

Question 17

A steel lift column in a service station is a solid cylinder 4.0 m long and 0.20 m in diameter.Young's modulus for this steel is 20 × 10¹⁰ N/m².By what distance does the column compress when a 5000-kg truck is on it? A)4.7 × 10^-7 m B)8.0 × 10^-7 m C)3.2 × 10^-6 m D)7.8 × 10^-6 m E)3.1 × 10^-5 m

Accepted Answer

The answer of A steel lift column in a service...

Question 18

A sphere is constructed of two concentric parts.The inner part is a solid sphere of radius 10.0 cm made of a material with density 4000 kg/m³.The outer part is a spherical shell with inner radius 10.0 cm and outer radius 20.0 cm.The material in the outer shell has a density 9000 kg/m³. (a)What is the mass of the sphere? (b)What is the average density of this sphere?

Accepted Answer

The answer of A sphere is constructed of two concentric...

Question 19

What is the maximum length of a metal cable that can hang vertically supported from one end of the cable? The Young's modulus of this metal is 2.10 × 10¹¹ N/m²,its tensile strength (the maximum tensile stress it can support without breaking)is 7.40 × 10⁸ N/m²,and its density is 7.60 × 10³ kg/m³. A)9.94 km B)4.22 km C)456 km D)1.75 km E)24.8 km

Accepted Answer

The maximum length is determined by the tensile strength divided by the product of density and gravitational acceleration: L = tensile strength / (density × g) = 7.40 × 10^8 N/m^2 / (7.60 × 10^3 kg/m^3 × 9.81 m/s^2) ≈ 9.94 km.

Question 20

The weight of a car of mass 1.20 × 10³ kg is supported equally by the four tires,which are inflated to the same gauge pressure.What gauge pressure in the tires is required so the area of contact of each tire with the road is 1.00 × 10² cm²? (1 atm = 1.01 × 10⁵ Pa.) A)11.6 × 10⁵ Pa B)11.6 × 10⁴ Pa C)2.94 × 10⁵ Pa D)2.94 × 10⁴ Pa E)2.94 × 10³ Pa

Accepted Answer

We can start by using the formula for pressure:
$P = F/A$
where P is pressure, F is force and A is area.

Since the weight of the car is supported equall by the four tires, each tire supports a quarter of the weight which is:

$F = \frac{1.20 	imes 10^3 	ext{ kg} 	imes g}{4}$
where g is the acceleration due to gravity which is approximately 9.81 m/s^2.

$F = 2943 	ext{ N}$

The area of contact for each tire is given as:
$A = 1.00 	imes 10^{-2} 	ext{ m} 	imes 1.00 	imes 10^{-2} 	ext{ m}$
which is equal to:
$A = 1.00 	imes 10^{-4} 	ext{ m}^2$
Using the conversion factor for atmospheres to pascals, we have:

$1 	ext{ atm} = 1.01 	imes 10^5 	ext{ Pa}$

The gauge pressure (relative to atmospheric pressure) in the tires is:

$P = \frac{F}{A} + 1 	ext{ atm}$

$P = \frac{2943 	ext{ N}}{1.00 	imes 10^{-4} 	ext{ m}^2} + 1.01 	imes 10^5 	ext{ Pa}$

$P = 2.94 	imes 10^5 	ext{ Pa} + 1.01 	imes 10^5 	ext{ Pa}$

$P = 3.95 	imes 10^5 	ext{ Pa}$

However, we need to subtract the atmospheric pressure to get the gauge pressure:

$P_g = P - 1.01 	imes 10^5 	ext{ Pa}$

$P_g = 2.94 	imes 10^5 	ext{ Pa}$

Therefore, the correct answer is (C) 2.94 × 10^5 Pa.

Quiz 14: Fluids and Elasticity

A sample of tendon 3.00 cm long and 4.00 mm in diameter is found to break under a minimum force of 128 N.If instead the sample had been 1.50 cm long and of uniform composition and cross-sectional area,what minimum force would have been required to break it?

A 12-L volume of oil is subjected to a pressure change,which produces a volume strain on the oil of -3.0 × 10^-4.The bulk modulus of the oil is 6.0 × 10⁹ N/m² and is independent of the pressure.By how many milliliters does this pressure reduce the volume of the oil?

A 12-L volume of oil is subjected to a pressure change,which produces a volume strain on the oil of -3.0 × 10^-4.The bulk modulus of the oil is 6.0 × 10⁹ N/m² and is independent of the pressure.What is the pressure change that produced the volume strain in the oil?

The tensile strength (the maximum tensile stress it can support without breaking) for a certain steel wire is 3000 MN/m².What is the maximum load that can be applied to a wire with a diameter of 3.0 mm made of this steel without breaking the wire?

A steel rod 55 cm long has a diameter of 30 cm.The compressive strength (the maximum stress it can support without breaking) of this steel is 500 × 10⁶ N/m².What is the compression force that would break the rod?

A certain coin has a diameter of 21.21 mm,a thickness of 1.95 mm,and weighs 0.04905 N.What is its density?

A solid steel sphere with a radius of 2.0 m falls off a ship and sinks to a depth where the pressure is 15 MN/m².The bulk modulus for this steel is 1.6 × 10¹¹N/m².What is the change in the radius of the sphere?

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

A copper sphere has a radius of 2.50 m under normal room pressure of 1.0 × 10⁵N/m².If we increase the pressure on this sphere to 10 times the normal room pressure,what is the change in its volume? The bulk modulus for copper is 1.4 × 10¹¹ Pa.

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

When the pressure applied to an unknown liquid is increased from 1.0 × 10⁷ N/m² to 5.5 × 10⁷ N/m²,the volume of the liquid decreases by 0.70%.Calculate the bulk modulus of the liquid.

At a depth of about 1030 m in the sea,the pressure has increased by 100 atmospheres (to 1.0 × 10⁷N/m²) .By how much has 1.0 m³ of water been compressed by this pressure? The bulk modulus of water is 2.3 × 10⁹ N/m².

A steel lift column in a service station is a solid cylinder 4.0 m long and 0.20 m in diameter.Young's modulus for this steel is 20 × 10¹⁰ N/m².By what distance does the column compress when a 5000-kg truck is on it?

The weight of a car of mass 1.20 × 10³ kg is supported equally by the four tires,which are inflated to the same gauge pressure.What gauge pressure in the tires is required so the area of contact of each tire with the road is 1.00 × 10² cm²? (1 atm = 1.01 × 10⁵ Pa.)

Quiz 14: Fluids and Elasticity

A sample of tendon 3.00 cm long and 4.00 mm in diameter is found to break under a minimum force of 128 N.If instead the sample had been 1.50 cm long and of uniform composition and cross-sectional area,what minimum force would have been required to break it?

A 12-L volume of oil is subjected to a pressure change,which produces a volume strain on the oil of -3.0 × 10-4.The bulk modulus of the oil is 6.0 × 109 N/m2 and is independent of the pressure.By how many milliliters does this pressure reduce the volume of the oil?

A 12-L volume of oil is subjected to a pressure change,which produces a volume strain on the oil of -3.0 × 10-4.The bulk modulus of the oil is 6.0 × 109 N/m2 and is independent of the pressure.What is the pressure change that produced the volume strain in the oil?

The tensile strength (the maximum tensile stress it can support without breaking) for a certain steel wire is 3000 MN/m2.What is the maximum load that can be applied to a wire with a diameter of 3.0 mm made of this steel without breaking the wire?

A steel rod 55 cm long has a diameter of 30 cm.The compressive strength (the maximum stress it can support without breaking) of this steel is 500 × 106 N/m2.What is the compression force that would break the rod?

A certain coin has a diameter of 21.21 mm,a thickness of 1.95 mm,and weighs 0.04905 N.What is its density?

A solid steel sphere with a radius of 2.0 m falls off a ship and sinks to a depth where the pressure is 15 MN/m2.The bulk modulus for this steel is 1.6 × 1011 N/m2.What is the change in the radius of the sphere?

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

A copper sphere has a radius of 2.50 m under normal room pressure of 1.0 × 105 N/m2.If we increase the pressure on this sphere to 10 times the normal room pressure,what is the change in its volume? The bulk modulus for copper is 1.4 × 1011 Pa.

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

When the pressure applied to an unknown liquid is increased from 1.0 × 107 N/m2 to 5.5 × 107 N/m2,the volume of the liquid decreases by 0.70%.Calculate the bulk modulus of the liquid.

At a depth of about 1030 m in the sea,the pressure has increased by 100 atmospheres (to 1.0 × 107 N/m2) .By how much has 1.0 m3 of water been compressed by this pressure? The bulk modulus of water is 2.3 × 109 N/m2.

A steel lift column in a service station is a solid cylinder 4.0 m long and 0.20 m in diameter.Young's modulus for this steel is 20 × 1010 N/m2.By what distance does the column compress when a 5000-kg truck is on it?

The weight of a car of mass 1.20 × 103 kg is supported equally by the four tires,which are inflated to the same gauge pressure.What gauge pressure in the tires is required so the area of contact of each tire with the road is 1.00 × 102 cm2? (1 atm = 1.01 × 105 Pa.)

A 12-L volume of oil is subjected to a pressure change,which produces a volume strain on the oil of -3.0 × 10^-4.The bulk modulus of the oil is 6.0 × 10⁹ N/m² and is independent of the pressure.By how many milliliters does this pressure reduce the volume of the oil?

A 12-L volume of oil is subjected to a pressure change,which produces a volume strain on the oil of -3.0 × 10^-4.The bulk modulus of the oil is 6.0 × 10⁹ N/m² and is independent of the pressure.What is the pressure change that produced the volume strain in the oil?

The tensile strength (the maximum tensile stress it can support without breaking) for a certain steel wire is 3000 MN/m².What is the maximum load that can be applied to a wire with a diameter of 3.0 mm made of this steel without breaking the wire?

A steel rod 55 cm long has a diameter of 30 cm.The compressive strength (the maximum stress it can support without breaking) of this steel is 500 × 10⁶ N/m².What is the compression force that would break the rod?

A solid steel sphere with a radius of 2.0 m falls off a ship and sinks to a depth where the pressure is 15 MN/m².The bulk modulus for this steel is 1.6 × 10¹¹N/m².What is the change in the radius of the sphere?

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

A copper sphere has a radius of 2.50 m under normal room pressure of 1.0 × 10⁵N/m².If we increase the pressure on this sphere to 10 times the normal room pressure,what is the change in its volume? The bulk modulus for copper is 1.4 × 10¹¹ Pa.

When the pressure applied to an unknown liquid is increased from 1.0 × 10⁷ N/m² to 5.5 × 10⁷ N/m²,the volume of the liquid decreases by 0.70%.Calculate the bulk modulus of the liquid.

At a depth of about 1030 m in the sea,the pressure has increased by 100 atmospheres (to 1.0 × 10⁷N/m²) .By how much has 1.0 m³ of water been compressed by this pressure? The bulk modulus of water is 2.3 × 10⁹ N/m².

A steel lift column in a service station is a solid cylinder 4.0 m long and 0.20 m in diameter.Young's modulus for this steel is 20 × 10¹⁰ N/m².By what distance does the column compress when a 5000-kg truck is on it?

The weight of a car of mass 1.20 × 10³ kg is supported equally by the four tires,which are inflated to the same gauge pressure.What gauge pressure in the tires is required so the area of contact of each tire with the road is 1.00 × 10² cm²? (1 atm = 1.01 × 10⁵ Pa.)