Question 1

A cable lifting a 4.00-lb bowling ball stretches by 3.00 mm. If the cable is cut in half and both halves are used together to lift the same bowling ball, each cable will stretch by

Accepted Answer

When the cable is cut in half and both halves are used together, the effective spring constant doubles, reducing the stretch by half. Therefore, each cable will stretch by 1.50 mm / 2 = 0.75 mm.

Question 2

The Young's modulus for aluminum is  . If a 35.0-cm long aluminum bar is shortened by 4.00 mm under a certain load, what is the stress on this bar?

Accepted Answer

B

Question 3

A design project requires you to construct a crane to lift a maximum load of QUOTE 5.00×103kg and with a maximum upward acceleration of 2.00 m/s². What size diameter steel cable should you use if the cable is to have a safety factor of exactly 5?

Accepted Answer

We can use the formula:
Tension = mass x (acceleration + gravity)
where tension is the force exerted by the cable, mass is the maximum load of 5.00 x 10^3 kg, acceleration is the maximum upward acceleration of 2.00 m/s^2, and gravity is the acceleration due to gravity, which is 9.81 m/s^2.

Solving for tension:
Tension = (5.00 x 10^3 kg) x (2.00 m/s^2 + 9.81 m/s^2)
Tension = 5.00 x 10^3 kg x 11.81 m/s^2
Tension = 5.91 x 10^4 N

To find the diameter of the steel cable, we can use the formula:
Tension = (pi/4) x (diameter^2) x tensile strength x safety factor
where pi is approximately 3.14, diameter is the diameter of the cable we want to find, and tensile strength is the maximum stress the cable can withstand without breaking. We can assume a tensile strength of 1.84 x 10^9 N/m^2 for steel cable.

Solving for diameter:
diameter^2 = (4 x tension) / (pi x tensile strength x safety factor)
diameter^2 = (4 x 5.91 x 10^4 N) / (3.14 x 1.84 x 10^9 N/m^2 x 5)
diameter^2 = 6.39 x 10^-4 m^2
diameter = 2.68 x 10^-2 m
diameter = 2.68 cm

Therefore, the best choice is C, a diameter of 2.68 cm.

Question 4

A cable lifting a 4.00-lb bowling ball stretches by 3.00 mm. The same cable lifting a 5.00-lb bowling ball stretches by

Accepted Answer

The stretch in the cable is directly proportional to the weight of the bowling ball. Therefore, we can use the formula: 
stretch = constant x weight

To find the constant, we can use the information given: 
stretch = 3.00 mm when weight = 4.00 lbs

So the constant is: 
constant = stretch/weight = 3.00 mm/4.00 lbs = 0.75 mm/lb

Now we can use this constant to find the stretch when the weight is 5.00 lbs: 
stretch = constant x weight = 0.75 mm/lb x 5.00 lbs = 3.75 mm

Therefore, the answer is choice C, 3.75 mm.

Question 5

Fill In the blank. If you double the radius of a steel cable without changing its length, you would __________ its elastic constant.

Accepted Answer

The elastic constant, or stiffness, of a cable is proportional to its cross-sectional area. Doubling the radius increases the area by a factor of four (since area is proportional to the square of the radius), thus quadrupling the elastic constant.

Question 6

A 3.00-m long wire with a diameter of 1.50 mm has Young's modulus of 9.00 x10¹⁰ N/m². What mass has to be attached to the wire so that it would extend it by 3.00 mm under its weight?

Accepted Answer

To find the mass, use the formula for extension: ΔL = (F*L) / (A*Y), where F = mg, L = original length, A = cross-sectional area, Y = Young's modulus. Solving for m gives m = (ΔL * A * Y) / (g * L). Substituting the given values, m = (3.00 mm * π * (0.75 mm)^2 * 9.00 x 10^10 N/m^2) / (9.81 m/s^2 * 3.00 m), which results in approximately 16.2 kg.

Question 7

A wire 2.5-m long has a cross-sectional area of 2.5 mm². It is hung vertically, and a 4.5-kg mass is attached to it. By how much does the wire stretch if Young's modulus for that material is 2.0 x10¹¹ N/m²?

Accepted Answer

The correct answer is A because the wire stretches by 0.22 mm when a 4.5-kg mass is attached to it. This is calculated using the formula:stress = force / areastrain = change in length / original lengthYoung's modulus = stress / strainIn this case, the stress is:stress = (4.5 kg)(9.8 m/s^2) / (2.5 mm^2) = 1.76 x 10^7 N/m^2The strain is:strain = (change in length) / (2.5 m)The Young's modulus is:Young's modulus = 2.0 x 10^11 N/m^2Using the formula Young's modulus = stress / strain, we can solve for the change in length:change in length = (stress)(original length) / Young's moduluschange in length = (1.76 x 10^7 N/m^2)(2.5 m) / (2.0 x 10^11 N/m^2)change in length = 0.22 mm

Question 8

If you double the length of a bungee cord with no change in its cross-sectional area, its spring constant would

Accepted Answer

According to Hooke's law, the spring constant (k) is directly proportional to the length of the cord. If the length is doubled, the spring constant is halved because the cord stretches twice as far for the same applied force. Thus the answer is C - cut in half.

Question 9

Young's modulus depends on the geometric shape of the object:

Accepted Answer

Young's modulus is a material property and is independent of the geometric shape of an object. It only depends on the type of material and its microstructure.

Question 10

An 800-kg mass is hung from a 1.50-m copper wire with a cross-sectional area of 0.350 cm². Young's modulus for copper is 1.10 x 10¹¹ N/m². How much does the wire stretch?

Accepted Answer

The answer of An 800-kg mass is hung from a...

Question 11

An 800-kg mass is hung from a 2.00-m steel wire with a cross-sectional area of 0.250 cm². Young's modulus for steel is 2.0 *10¹¹ N/m². How much does the wire stretch?

Accepted Answer

The answer of An 800-kg mass is hung from a...

Question 12

Fill in the blank. If you double the cross-sectional area of a steel rod, you would __________ the Young's modulus of steel.

Accepted Answer

The Young's modulus of steel is a material property that is independent of the cross-sectional area of the rod. Therefore, doubling the cross-sectional area of a steel rod would not change the Young's modulus of steel.

Question 13

Young's modulus Y has units of

Accepted Answer

Young's modulus has units of pressure, which is force per unit area. It is a measure of the stiffness of a material in response to tensile or compressive stress.

Question 14

A spring is made of a wire of radius r and length L. If the radius of the wire and the length of the wire were doubled, the spring constant of the spring would

Accepted Answer

The answer of A spring is made of a wire...

Question 15

When an object is under stress,

Accepted Answer

When an object is under stress, it can deform, undergo strain, or change its shape or size, so all of the above choices are correct.

Question 16

A wire has a diameter of 0.75 mm. It is stretched 0.20% of its original length when a mass of 5.0 kg is hanging from it. What is the Young's modulus of the wire?

Accepted Answer

Young's modulus (Y) is calculated using the formula Y = (F/A) / (ΔL/L), where F is the force, A is the cross-sectional area, ΔL is the change in length, and L is the original length. The force F is the weight of the mass (mg), and the area A is π(d/2)^2. Substituting the given values and solving gives approximately 5.6 x 10^10 N/m^2.

Question 17

If you measure the magnitude of the force required to stretch a typical laboratory spring as a function of the MAGNITUDE of stretch, which of the graphs shown would best represent your data?

Accepted Answer

Graph 2 shows a linear relationship between the force required and the magnitude of stretch, which is typical for a laboratory spring. The other graphs either show no change in force with stretch (graphs 1 and 5) or a non-linear relationship (graphs 3 and 4).

Question 18

Which of the following statements are correct?

Accepted Answer

All the statements are correct: tensile stress is applied when stretching an object, volume stress is applied when squeezing an object on all sides, and shear stress is applied when offset forces deform an object.

Question 19

If you grab each end of a pencil and try to pull the ends apart, you are putting the pencil under tension or tensile stress; if you push the ends toward each other, you are putting the pencil under

Accepted Answer

pushing the ends of the pencil toward each other creates compressive stress.

Question 20

A design project requires you to construct a crane to lift a maximum of QUOTE 5.00×103 kg and with a maximum upward acceleration of 2.00 m/s², and with a safety factor of 5. If the cable is 20.0 m long, by how much does the cable stretch when the crane is operating at the design limits?

Accepted Answer

The answer of A design project requires you to construct...

Quiz 9: Elastic Properties of Matter: Stress and Strain