Question 1

A 1200-kg car is being raised at a constant speed if 25 cm/s by a crane,using a 20-m long steel cable that is 1.5 cm in diameter.Young's modulus for steel is 2.0 × 10¹¹ Pa.What is the change in length of the cable caused by the car?

Accepted Answer

The change in length ($\Delta L$) of the cable can be calculated using the formula derived from Hooke's Law for materials, which is $\Delta L = \frac{FL}{A E}$, where $F$ is the force applied (weight of the car in this case), $L$ is the original length of the cable, $A$ is the cross-sectional area of the cable, and $E$ is Young's modulus for the material of the cable.Given:- Mass of the car ($m$) = 1200 kg- Acceleration due to gravity ($g$) = 9.8 m/s$^2$- Length of the cable ($L$) = 20 m- Diameter of the cable ($d$) = 1.5 cm = 0.015 m (converted to meters)- Young's modulus for steel ($E$) = $2.0 \times 10^{11}$ PaFirst, calculate the force ($F$) due to the weight of the car: $F = mg = 1200 \times 9.8 = 11760$ N.Next, calculate the cross-sectional area ($A$) of the cable: $A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.015}{2}\right)^2 \approx 1.767 \times 10^{-4}$ m$^2$.Now, plug these values into the formula for $\Delta L$:$\Delta L = \frac{11760 \times 20}{1.767 \times 10^{-4} \times 2.0 \times 10^{11}} \approx 0.0067$ m, or 6.7 mm.

Question 2

When a 125-kg piece of equipment is hung from a steel wire of length 75 cm,it stretches the wire by 4.00 mm.If instead the wire had a length of 150 cm,but all else were the same,by what distance would the same equipment stretch the wire?

Accepted Answer

The change in length of the wire, ΔL, is proportional to the original length of the wire, L, and the weight hanging from it, F: ΔL ∝ FL. Therefore, we can write:

ΔL = kFL

where k is a constant of proportionality that depends on the wire's material, cross-sectional area, and other properties.

We can use the given data to find k:

4.00 mm = k(125 kg)(0.75 m)

k = 7.11 × 10^-3 N/m

Now we can use this value of k to find the new elongation when the wire is longer:

ΔL' = kFL' = (7.11 × 10^-3 N/m)(125 kg)(1.50 m) = 1.69 mm

Therefore, the equipment would stretch the longer wire by 1.69 mm (choice B).

Question 3

A solid steel column is 4.0 m long and 0.20 m in diameter.Young's modulus for this steel is 2.0 × 10¹¹ N/m².By what distance does the column shrink when a 5000-kg truck is supported by it?

Accepted Answer

The answer of A solid steel column is 4.0 m...

Question 4

A 50-kg load is suspended from a steel wire of diameter 1.0 mm and length 11.2 m.By what distance will the wire stretch? Young's modulus for steel is 2.0 × 10¹¹ Pa.

Accepted Answer

The stretch of a wire under tension can be found using the formula:

ΔL = (F * L) / (A * E)

where ΔL is the change in length, F is the force applied, L is the original length, A is the cross-sectional area, and E is the Young's modulus.

In this case, the force is the weight of the load, which is 50 kg * 9.8 m/s^2 = 490 N. The cross-sectional area can be found using the formula for the area of a circle: A = π * (d/2)^2 = π * (0.001 m / 2)^2 = 7.85 × 10^-7 m^2.

Plugging in the values, we get:

ΔL = (490 N * 11.2 m) / (7.85 × 10^-7 m^2 * 2.0 × 10^11 Pa) = 0.0035 m = 3.5 cm

Therefore, the wire will stretch by 3.5 cm. The answer is choice C.

Question 5

A 1200-kg car is being raised with a constant acceleration of 2.53 m/s² by a crane,using a 20-m long steel cable that is 1.5 cm in diameter.Young's modulus for steel is 2.0 × 10¹¹ N/m².What is the change in length of the cable caused by lifting the car?

Accepted Answer

The change in length ($\Delta L$) of the cable can be calculated using the formula $\Delta L = \frac{FL}{A E}$, where $F$ is the force applied, $L$ is the original length of the cable, $A$ is the cross-sectional area of the cable, and $E$ is Young's modulus for the material of the cable. The force $F$ includes both the gravitational force on the car and the force due to the acceleration of the car upwards. The gravitational force is $mg$, where $m$ is the mass of the car and $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$). The additional force due to the car's acceleration is $ma$, where $a$ is the acceleration of the car. Thus, $F = mg + ma$. The cross-sectional area $A$ of the cable can be calculated using the formula for the area of a circle, $A = \pi r^2$, where $r$ is the radius of the cable. Given that the diameter of the cable is $1.5 \, \text{cm}$, the radius is $0.75 \, \text{cm}$ or $0.0075 \, \text{m}$. Substituting the given values and calculating gives a change in length of $8.4 \, \text{mm}$.

Question 6

A wire of diameter 0.20 mm stretches by 0.20% when a 6.28-N force is applied to it.What is Young's modulus for this wire?

Accepted Answer

Young's modulus (Y) is given by the formula Y = (F * L) / (A * ΔL), where F is the force applied, L is the original length, A is the cross-sectional area, and ΔL is the change in length. Given the diameter (d) is 0.20 mm or 0.20 x 10^-3 m, the radius (r) is 0.10 x 10^-3 m. The area (A) = πr^2 = π(0.10 x 10^-3)^2 m^2. The stretch (ΔL/L) is 0.20%, or 0.002 in decimal form. Rearranging the formula for Y gives Y = F / (A * (ΔL/L)). Substituting the given values and solving for Y yields a value close to 1.0 × 10^11 Pa, making option B correct.

Question 7

A steel wire,3.2 m long,has a diameter of 1.2 mm.The wire stretches 1.6 mm when it bears a load.Young's modulus for steel is 2.0 × 10¹¹ Pa.The mass of the load is closest to

Accepted Answer

The correct answer is A.The stress on the wire is:$$\sigma = \frac{F}{A} = \frac{4F}{\pi d^2}$$where F is the load, A is the cross-sectional area of the wire, and d is the diameter of the wire.The strain on the wire is:$$\epsilon = \frac{\Delta L}{L} = \frac{1.6 	ext{ mm}}{3.2 	ext{ m}} = 5.0 	imes 10^{-4}$$Young's modulus is:$$Y = \frac{\sigma}{\epsilon} = \frac{4F}{\pi d^2 \epsilon}$$Substituting the given values into the equation, we get:$$2.0 	imes 10^{11} 	ext{ Pa} = \frac{4F}{\pi (1.2 	imes 10^{-3} 	ext{ m})^2 (5.0 	imes 10^{-4})}$$Solving for F, we get:$$F = 12 	ext{ kg}$$

Question 8

A cable is 100-m long and has a cross-sectional area of 1.0 mm².A 1000-N force is applied to stretch the cable.Young's modulus for the cable is 1.0 × 10¹¹ N/m².How far does the cable stretch?

Accepted Answer

The answer of A cable is 100-m long and has...

Question 9

A vertical 30-cm steel rod,1.0 cm in diameter,supports a 300-kg mass.What is the change in length of the rod caused by this mass? Young's modulus for steel is 2.0 × 10¹¹ N/m².

Accepted Answer

The change in length ($\Delta L$) of the rod can be calculated using the formula $\Delta L = \frac{FL}{AE}$, where $F$ is the force applied, $L$ is the original length of the rod, $A$ is the cross-sectional area, and $E$ is Young's modulus. The force $F$ is the weight of the mass, $F = mg$, where $m = 300$ kg and $g \approx 9.8$ m/s$^2$. The cross-sectional area $A = \pi r^2$, where $r = 0.5$ cm = $0.005$ m. Plugging in the values: $F = 300 \times 9.8$, $A = \pi \times (0.005)^2$, $L = 0.3$ m, and $E = 2.0 \times 10^{11}$ N/m$^2$, we find $\Delta L = 5.6 \times 10^{-5}$ m.

Question 10

A 50-kg air conditioner is placed on top of a concrete pad that is 20 cm thick and has a cross-sectional area of 3.0 m².What is the change in thickness of the pad caused by the air conditioner? Young's modulus for concrete is 2.3 × 10¹⁰ N/m².

Accepted Answer

The change in thickness ($\Delta L$) can be calculated using the formula $\Delta L = \frac{F \cdot L}{A \cdot Y}$, where $F$ is the force applied (weight of the air conditioner), $L$ is the original length (thickness of the pad), $A$ is the cross-sectional area, and $Y$ is Young's modulus. The force $F$ is the weight of the air conditioner, which is $mg = 50 \times 9.8 = 490$ N. The original length $L = 20 \times 10^{-2}$ m, $A = 3.0 \, m^2$, and $Y = 2.3 \times 10^{10}$ N/m$^2$. Plugging these values into the formula gives $\Delta L = \frac{490 \cdot 20 \times 10^{-2}}{3.0 \times 2.3 \times 10^{10}} = 1.4 \times 10^{-9}$ m.

Question 11

A piano wire has a radius of 0.50 mm.One end is fixed and the other is wrapped around a tuning peg,which has a diameter of 3.5 mm and is 80 cm away.After the wire just becomes taut,the peg is given three full turns.What is the tension in the wire? Young's modulus for steel is 2.0 × 10¹¹ N/m².

Accepted Answer

The length of the wire can be calculated as follows:
Length of the straight wire = distance between the fixed end and the tuning peg
= 80 cm
Length of the wire wrapped around the tuning peg = circumference of the peg × number of turns
= π × diameter of the peg × number of turns
= π × 3.5 mm × 3
= 32.985 cm
Total length of the wire = length of the straight wire + length of the wrapped wire
= 80 cm + 32.985 cm
= 112.985 cm
= 1.12985 m

The cross-sectional area of the wire can be calculated as follows:
Cross-sectional area = π × (radius)²
= π × (0.50 mm)²
= 0.7854 × 10⁻⁶ m²

The tension in the wire can be calculated using the formula:
Tension = (4 × load × length) / (π × (radius)² × π × (Young's modulus))
where load is the weight applied to the wire, which in this case is the weight of the tuning peg and any additional tension required to bring the wire to pitch.

Assuming the weight of the tuning peg is negligible compared to the additional tension required to bring the wire to pitch, the load can be approximated as follows:
Load = mass per unit length × length × g
where g is the acceleration due to gravity, which is 9.81 m/s² for this problem.

The mass per unit length can be calculated using the density of steel (7.8 g/cm³) and the cross-sectional area of the wire as follows:
Mass per unit length = density × cross-sectional area
= 7.8 g/cm³ × 0.7854 × 10⁻⁶ m²
= 6.11332 × 10⁻³ kg/m

Substituting all the values into the tension formula, we get:
Tension = (4 × load × length) / (π × (radius)² × π × (Young's modulus))
= (4 × 6.11332 × 10⁻³ kg/m × 1.12985 m × 9.81 m/s²) / (π × (0.50 × 10⁻³ m)² × π × (2.0 × 10¹¹ N/m²))
= 6.472 kN
= 6.5 kN (rounded to one decimal place)

Therefore, the best choice is (C) 6.5 kN.

Question 12

A brass wire 2.0 m long and 2.0 mm in diameter supports a 10.0-kg fixture.By what distance did this fixture stretch the wire? Young's modulus for brass is 10 × 10¹⁰ N/m².

Accepted Answer

The answer of A brass wire 2.0 m long and...

Question 13

When a 125-kg piece of equipment is hung from a steel wire of diameter 6.00 mm,it stretches the wire by 4.00 mm.If instead the wire had a diameter of 3.00 mm,but all else were the same,by what distance would the same equipment stretch the wire?

Accepted Answer

The stretching of the wire is inversely proportional to the cross-sectional area, which is proportional to the square of the diameter. Halving the diameter (from 6.00 mm to 3.00 mm) reduces the area by a factor of 4, thus increasing the stretch by the same factor. Therefore, the stretch increases from 4.00 mm to 16.00 mm.

Quiz 17: Phases and Phase Changes

A 1200-kg car is being raised at a constant speed if 25 cm/s by a crane,using a 20-m long steel cable that is 1.5 cm in diameter.Young's modulus for steel is 2.0 × 1011 Pa.What is the change in length of the cable caused by the car?

When a 125-kg piece of equipment is hung from a steel wire of length 75 cm,it stretches the wire by 4.00 mm.If instead the wire had a length of 150 cm,but all else were the same,by what distance would the same equipment stretch the wire?

A solid steel column is 4.0 m long and 0.20 m in diameter.Young's modulus for this steel is 2.0 × 1011 N/m2.By what distance does the column shrink when a 5000-kg truck is supported by it?

A 50-kg load is suspended from a steel wire of diameter 1.0 mm and length 11.2 m.By what distance will the wire stretch? Young's modulus for steel is 2.0 × 1011 Pa.

A 1200-kg car is being raised with a constant acceleration of 2.53 m/s2 by a crane,using a 20-m long steel cable that is 1.5 cm in diameter.Young's modulus for steel is 2.0 × 1011 N/m2.What is the change in length of the cable caused by lifting the car?

A wire of diameter 0.20 mm stretches by 0.20% when a 6.28-N force is applied to it.What is Young's modulus for this wire?

A steel wire,3.2 m long,has a diameter of 1.2 mm.The wire stretches 1.6 mm when it bears a load.Young's modulus for steel is 2.0 × 1011 Pa.The mass of the load is closest to

A cable is 100-m long and has a cross-sectional area of 1.0 mm2.A 1000-N force is applied to stretch the cable.Young's modulus for the cable is 1.0 × 1011 N/m2.How far does the cable stretch?

A vertical 30-cm steel rod,1.0 cm in diameter,supports a 300-kg mass.What is the change in length of the rod caused by this mass? Young's modulus for steel is 2.0 × 1011 N/m2.

A 50-kg air conditioner is placed on top of a concrete pad that is 20 cm thick and has a cross-sectional area of 3.0 m2.What is the change in thickness of the pad caused by the air conditioner? Young's modulus for concrete is 2.3 × 1010 N/m2.

A piano wire has a radius of 0.50 mm.One end is fixed and the other is wrapped around a tuning peg,which has a diameter of 3.5 mm and is 80 cm away.After the wire just becomes taut,the peg is given three full turns.What is the tension in the wire? Young's modulus for steel is 2.0 × 1011 N/m2.

A brass wire 2.0 m long and 2.0 mm in diameter supports a 10.0-kg fixture.By what distance did this fixture stretch the wire? Young's modulus for brass is 10 × 1010 N/m2.