Question 1

A force of 15.0 N is applied to a wire with a diameter of 2.00 mm, causing its length to change by 1.20 mm. What is the stress on the wire? A) 4.77 × 10⁶ N/m². B) 5.38 × 10⁶ N/m². C) 6.73 × 10⁶ N/m². D) 7.27 × 10⁶ N/m². E) 8.75 × 10⁶ N/m².

Accepted Answer

Stress is calculated using the formula stress = force / area. The area of a wire with a diameter of 2.00 mm (which is 0.002 m) is found using the formula for the area of a circle, πr^2. The radius is half the diameter, so r = 0.001 m. The area = π(0.001)^2 = 3.14 × 10^-6 m^2. The force applied is 15.0 N. Therefore, stress = 15.0 N / 3.14 × 10^-6 m^2 = 4.77 × 10^6 N/m^2.

Question 2

A force of 20 N is applied to a wire with an unstretched length of 2.5 m. The wire gets 2.00 mm longer. The strain in the wire is A) 2.0 × 10^-4. B) 4.0 × 10^-4. C) 6.0 × 10^-4. D) 8.0 × 10^-4. E) 10 × 10^-4.

Accepted Answer

D

Question 3

A cable is 5.00 m long and has a diameter of 1.50 mm. The elastic modulus of the cable is 2.50 × 10¹⁰ N/m². A force is applied to the end of the cable and the cable stretches by 3.00 mm. What is the applied force? A) 16.5 N B) 20.5 N C) 26.5 N D) 32.6 N E) 36.6 N

Accepted Answer

The force can be calculated using the formula for Young's modulus: $ F = \frac{E \cdot A \cdot \Delta L}{L} $. The cross-sectional area $ A $ is $ \pi \times (0.0015/2)^2 $, $ \Delta L = 0.003 $ m, and $ L = 5 $ m. Substituting these values gives $ F \approx 26.5 $ N.

Question 4

A rod with a diameter of 3.00 mm, projects 2.0 cm horizontally outward from a wall. A 50.0 N weight is hung from the end of the rod and the rod deflects downward by 2.00 × 10^-6 m. The shear strain of the rod is A) 1.00 × 10^-4. B) 1.50 × 10^-4. C) 2.00 × 10^-4. D) 2.50 × 10^-4. E) 3.00 × 10^-4.

Accepted Answer

Shear strain is calculated as the change in length (deflection) over the original length. In this case, the deflection is $2.00 \times 10^{-6}$ m and the original length (height in this context, since we're considering the vertical deflection as the change in length) is the diameter of the rod, which is $3.00 \times 10^{-3}$ m. Thus, shear strain = $\frac{2.00 \times 10^{-6} \, \text{m}}{3.00 \times 10^{-3} \, \text{m}} = \frac{2}{3} \times 10^{-3} = 0.6667 \times 10^{-3}$, which rounds to $6.67 \times 10^{-4}$, but given the options, it seems there might be a misunderstanding in the interpretation of the question or a mistake in calculation. The correct approach to find shear strain in this context would involve understanding the deformation in terms of shear, but given the options and typical understanding, none directly match the calculation. Re-evaluating the calculation for a direct match with the provided options under typical assumptions of shear strain might not directly apply here due to a misunderstanding in the application of the concept to the given problem's specifics. The correct answer based on the typical calculation of shear strain does not match due to a conceptual misunderstanding in applying shear strain to this scenario.

Question 5

A wire is 12.00 m long and has a diameter of 1.50 mm. The elastic modulus of the wire is 7.00 × 10¹⁰ N/m². If a force of 500 N is applied to end of the wire, then the increase in length of the wire is A) 38.5 mm. B) 40.2 mm. C) 44.1 mm. D) 48.5 mm. E) 51.5 mm.

Accepted Answer

The answer of A wire is 12.00 m long and...

Question 6

A nail projects 40.0 mm horizontally outward from a wall. A 25.0 N coat is hung from the end of the nail and the nail deflects downward by 2.50 × 10⁻⁶ m. The shear strain of the nail is A) 4.00 × 10⁻⁵. B) 4.75 × 10⁻⁵. C) 5.01 × 10⁻⁵. D) 5.75 × 10⁻⁵. E) 6.25 × 10⁻⁵.

Accepted Answer

The answer of A nail projects 40.0 mm horizontally outward...

Question 7

A cable is 10.0 m long and has a diameter of 2.00 mm. A force of 45.0 N is applied to the end of the cable and the cable stretches by 3.00 mm. What is the elastic modulus? A) 3.05 × 10¹⁰ N/m² B) 3.67 × 10¹⁰ N/m² C) 3.98 × 10¹⁰ N/m² D) 4.26 × 10¹⁰ N/m² E) 4.77 × 10¹⁰ N/m²

Accepted Answer

The answer of A cable is 10.0 m long and...

Question 8

A cable is 50.0 m long and has a diameter of 2.50 cm. A force of 10,000 N is applied to the end of the cable. If the maximum stretch allowed in the cable is 2.00 mm, then what is the minimum elastic modulus allowed? A) 5.09 × 10¹¹ N/m² B) 1.27 × 10¹¹ N/m² C) 3.52 × 10⁹ N/m² D) 2.04 × 10⁷ N/m² E) 5.00 × 10⁶ N/m²

Accepted Answer

The minimum elastic modulus (E) can be calculated using the formula for stress and strain, where stress = force / area and strain = Δlength / original length. The area of the cable's cross-section is πr^2 = π(0.0125 m)^2. The stress is 10,000 N / area, and the strain is 0.002 m / 50 m. Solving for E (E = stress / strain) gives a value close to 5.09 × 10^11 N/m^2.

Question 9

A rod with a diameter of 2.00 mm, projects 20.0 mm horizontally outward from a wall. A 10.0 N weight is hung from the end of the rod and the rod deflects downward by 4.00 × 10^-6 m. The shear strain of the rod is A) 1.00 × 10^-4. B) 1.50 × 10^-4. C) 2.00 × 10^-4. D) 2.50 × 10^-4. E) 3.00 × 10^-4.

Accepted Answer

The answer of A rod with a diameter of 2.00...

Question 10

A force of 15.0 N is applied to a wire with an unstretched length of 4.00 m and a diameter of 5.00 mm. If the wire changed length by 1.20 mm, then the strain in the wire is A) 1.00 × 10⁻⁴. B) 2.00 × 10⁻⁴. C) 3.00 × 10⁻⁴. D) 4.00 × 10⁻⁴. E) 5.00 × 10⁻⁴.

Accepted Answer

To find the strain in the wire, we use the formula:

strain = change in length / original length

The change in length is given as 1.20 mm, but we need to convert it to meters:

1.20 mm = 1.20 × 10^-3 m

The original length is given as 4.00 m.

So,

strain = (1.20 × 10^-3) / 4.00

strain = 3.00 × 10^-4

Therefore, the strain in the wire is 3.00 × 10^-4.

The answer is option C, 3.00 × 10^-4.

Question 11

A wire has a strain of 0.10%. If the wire has an unstretched length of 5.00 m, then the change in length of the wire is
A) 5.0 mm.
B) 7.0 mm.
C) 9.0 mm.
D) 10 mm.
E) 15 mm.

Accepted Answer

Strain is defined as the change in length divided by the original length. Therefore, a strain of 0.10% (or 0.001 in decimal form) on a 5.00 m wire results in a change in length of 0.001 * 5.00 m = 0.005 m, which is equivalent to 5.0 mm.

Question 12

A nail projects 25.0 mm horizontally outward from a wall. A picture is hung from the end of the nail and the nail deflects downward by 2.00 × 10⁻⁶ m. The shear strain of the nail is A) 7.00 × 10⁻⁵. B) 8.00 × 10⁻⁵. C) 9.00 × 10⁻⁵. D) 10.0 × 10⁻⁵. E) 12.0 × 10⁻⁵.

Accepted Answer

We can use the formula for shear strain:
shear strain = (deflection)/(length of nail)
The length of the nail is not given, but we can use the Pythagorean theorem to find it:
length of nail = sqrt((25.0 mm)^2 + (2.00 × 10^-6 m)^2) = 25.0 mm
(Note that we converted the deflection to meters for consistency.)
Plugging in the values, we get:
shear strain = (2.00 × 10^-6 m)/(25.0 mm) = 8.00 × 10^-5
Converting to scientific notation with proper units, we get:
shear strain = 8.00 × 10^-5 (unitless) = 8.00 × 10^-5 m/m = 8.00 × 10^-5 or 8.00 × 10^-5 S^-1
The closest answer choice is B, which is 8.00 × 10^-5 or 8.00 × 10^-5 S^-1.

Question 13

A cable has a strain of 2.0 × 10^-3. If the unstretched length of the cable is 24.0 m, then the change in length of the cable is A) 35 mm. B) 48 mm. C) 52 mm. D) 59 mm. E) 67 mm.

Accepted Answer

The change in length of the cable can be calculated using the formula for strain, which is the change in length (ΔL) divided by the original length (L0), so strain = ΔL / L0. Given a strain of 2.0 × 10^-3 and an original length (L0) of 24.0 m, we can rearrange the formula to solve for ΔL: ΔL = strain × L0 = 2.0 × 10^-3 × 24.0 m = 0.048 m, which is 48 mm.

Question 14

A force of 10.0 N is applied to a wire with an unstretched length of 4.00 m and a diameter of 5.00 mm. The wire changes length by 1.20 mm as a result of the applied force. What is the stress on the wire? A) 3.67 × 10⁵ N/m². B) 4.19 × 10⁵ N/m². C) 4.67 × 10⁵ N/m². D) 5.09 × 10⁵ N/m². E) 6.55 × 10⁵ N/m².

Accepted Answer

The answer of A force of 10.0 N is applied...

Question 15

A force of 500 N is applied to a rod with a diameter of 4.00 mm. The stress on the rod is A) 5.38 × 10⁷ N/m². B) 4.66 × 10⁷ N/m². C) 4.25 × 10⁷ N/m². D) 3.98 × 10⁷ N/m². E) 2.75 × 10⁷ N/m².

Accepted Answer

The answer of A force of 500 N is applied...

Question 16

A 2.00 m brass rod with a 1.00 mm diameter behaves like a spring under stretching or compression. If the elastic modulus of brass is 9.10 × 10¹⁰ N/m², then the spring constant of the rod is A) 4.67 × 10⁴ N/m. B) 4.22 × 10⁴ N/m. C) 3.57 × 10⁴ N/m. D) 2.47 × 10⁴ N/m. E) 2.13 × 10⁴ N/m.

Accepted Answer

The answer of A 2.00 m brass rod with a...

Question 17

A force of 100 N is applied to a rod with a diameter of 6.00 mm. The stress on the rod is A) 5.47 × 10⁶ N/m². B) 4.15 × 10⁶ N/m². C) 3.54 × 10⁶ N/m². D) 2.64 × 10⁶ N/m². E) 2.03 × 10⁶ N/m².

Accepted Answer

The stress on the rod can be calculated using the formula: stress = force/area.

First, we need to find the area of the rod. Since the rod has a diameter of 6.00 mm, its radius is 3.00 mm (or 0.003 m). The area of a circle is given by the formula A = πr^2, so the area of the rod is:

A = π(0.003)^2 = 2.83 × 10^-5 m^2

Now we can calculate the stress:

stress = force/area = 100 N / (2.83 × 10^-5 m^2) = 3.54 × 10^10 N/m^2

Therefore, the answer is C, 3.54 × 10^10 N/m^2.

Question 18

A rod is 2.40 m long and has a diameter of 2.50 mm. A force of 2000 N is applied to the end to stretch the rod by 1.00 mm. What is the elastic modulus for this rod? A) 4.17 × 10⁻⁴ N/m² B) 2.00 × 10⁶ N/m² C) 4.07 × 10⁸ N/m² D) 2.44 × 10¹¹ N/m² E) 9.78 × 10¹¹ N/m²

Accepted Answer

The answer of A rod is 2.40 m long and...

Question 19

A wire is 1.50 m long and has a diameter of 1.50 mm. The elastic modulus of the wire is 6.20 × 10¹⁰ N/m². If a force of 400 N is applied to end of the wire, then the increase in length of the wire is A) 5.48 mm. B) 4.28 mm. C) 3.95 mm. D) 3.84 mm. E) 3.45 mm.

Accepted Answer

The correct answer is A because the increase in length of the wire is given by the formula:$$\Delta L = \frac{FL}{AE}$$where:* $\Delta L$ is the increase in length* $F$ is the force applied* $L$ is the original length of the wire* $A$ is the cross-sectional area of the wire* $E$ is the elastic modulus of the wireSubstituting the given values into the formula, we get:$$\Delta L = \frac{(400 N)(1.50 m)}{\pi (0.00075 m)^2 (6.20 	imes 10^{10} N/m^2)} = 5.48 mm$$

Question 20

A rod is 2.40 m long and has a diameter of 2.50 mm. A force of 2000 N is applied to the end, stretching the rod by 1.40 mm. What is the elastic modulus for this rod? A) 6.98 × 10¹¹ N/m² B) 1.75 × 10¹¹ N/m² C) 4.07 × 10⁸ N/m² D) 1.43 × 10⁶ N/m² E) 5.83 × 10⁻⁴ N/m²

Accepted Answer

The elastic modulus (E) is calculated using the formula E = (F/A) / (ΔL/L), where F is the force applied, A is the cross-sectional area of the rod, ΔL is the change in length, and L is the original length. First, calculate the area A = πr^2 = π(1.25×10^-3)^2 m^2. Then, plug in the values: E = (2000 N / π(1.25×10^-3 m)^2) / (1.4×10^-3 m / 2.4 m) = 6.98×10^10 N/m^2.

Quiz 10: Elasticity and Oscillations

A force of 15.0 N is applied to a wire with a diameter of 2.00 mm, causing its length to change by 1.20 mm. What is the stress on the wire?

A force of 20 N is applied to a wire with an unstretched length of 2.5 m. The wire gets 2.00 mm longer. The strain in the wire is

A cable is 5.00 m long and has a diameter of 1.50 mm. The elastic modulus of the cable is 2.50 × 1010 N/m2. A force is applied to the end of the cable and the cable stretches by 3.00 mm. What is the applied force?

A rod with a diameter of 3.00 mm, projects 2.0 cm horizontally outward from a wall. A 50.0 N weight is hung from the end of the rod and the rod deflects downward by 2.00 × 10-6 m. The shear strain of the rod is

A wire is 12.00 m long and has a diameter of 1.50 mm. The elastic modulus of the wire is 7.00 × 1010 N/m2. If a force of 500 N is applied to end of the wire, then the increase in length of the wire is

A nail projects 40.0 mm horizontally outward from a wall. A 25.0 N coat is hung from the end of the nail and the nail deflects downward by 2.50 × 10−6 m. The shear strain of the nail is

A cable is 10.0 m long and has a diameter of 2.00 mm. A force of 45.0 N is applied to the end of the cable and the cable stretches by 3.00 mm. What is the elastic modulus?

A cable is 50.0 m long and has a diameter of 2.50 cm. A force of 10,000 N is applied to the end of the cable. If the maximum stretch allowed in the cable is 2.00 mm, then what is the minimum elastic modulus allowed?

A rod with a diameter of 2.00 mm, projects 20.0 mm horizontally outward from a wall. A 10.0 N weight is hung from the end of the rod and the rod deflects downward by 4.00 × 10-6 m. The shear strain of the rod is

A force of 15.0 N is applied to a wire with an unstretched length of 4.00 m and a diameter of 5.00 mm. If the wire changed length by 1.20 mm, then the strain in the wire is

A wire has a strain of 0.10%. If the wire has an unstretched length of 5.00 m, then the change in length of the wire is

A nail projects 25.0 mm horizontally outward from a wall. A picture is hung from the end of the nail and the nail deflects downward by 2.00 × 10−6 m. The shear strain of the nail is

A cable has a strain of 2.0 × 10-3. If the unstretched length of the cable is 24.0 m, then the change in length of the cable is

A force of 10.0 N is applied to a wire with an unstretched length of 4.00 m and a diameter of 5.00 mm. The wire changes length by 1.20 mm as a result of the applied force. What is the stress on the wire?

A force of 500 N is applied to a rod with a diameter of 4.00 mm. The stress on the rod is

A 2.00 m brass rod with a 1.00 mm diameter behaves like a spring under stretching or compression. If the elastic modulus of brass is 9.10 × 1010 N/m2, then the spring constant of the rod is

A force of 100 N is applied to a rod with a diameter of 6.00 mm. The stress on the rod is

A rod is 2.40 m long and has a diameter of 2.50 mm. A force of 2000 N is applied to the end to stretch the rod by 1.00 mm. What is the elastic modulus for this rod?

A wire is 1.50 m long and has a diameter of 1.50 mm. The elastic modulus of the wire is 6.20 × 1010 N/m2. If a force of 400 N is applied to end of the wire, then the increase in length of the wire is

A rod is 2.40 m long and has a diameter of 2.50 mm. A force of 2000 N is applied to the end, stretching the rod by 1.40 mm. What is the elastic modulus for this rod?

A cable is 5.00 m long and has a diameter of 1.50 mm. The elastic modulus of the cable is 2.50 × 10¹⁰ N/m². A force is applied to the end of the cable and the cable stretches by 3.00 mm. What is the applied force?

A rod with a diameter of 3.00 mm, projects 2.0 cm horizontally outward from a wall. A 50.0 N weight is hung from the end of the rod and the rod deflects downward by 2.00 × 10^-6 m. The shear strain of the rod is

A wire is 12.00 m long and has a diameter of 1.50 mm. The elastic modulus of the wire is 7.00 × 10¹⁰ N/m². If a force of 500 N is applied to end of the wire, then the increase in length of the wire is

A nail projects 40.0 mm horizontally outward from a wall. A 25.0 N coat is hung from the end of the nail and the nail deflects downward by 2.50 × 10⁻⁶ m. The shear strain of the nail is

A rod with a diameter of 2.00 mm, projects 20.0 mm horizontally outward from a wall. A 10.0 N weight is hung from the end of the rod and the rod deflects downward by 4.00 × 10^-6 m. The shear strain of the rod is

A nail projects 25.0 mm horizontally outward from a wall. A picture is hung from the end of the nail and the nail deflects downward by 2.00 × 10⁻⁶ m. The shear strain of the nail is

A cable has a strain of 2.0 × 10^-3. If the unstretched length of the cable is 24.0 m, then the change in length of the cable is

A 2.00 m brass rod with a 1.00 mm diameter behaves like a spring under stretching or compression. If the elastic modulus of brass is 9.10 × 10¹⁰ N/m², then the spring constant of the rod is

A wire is 1.50 m long and has a diameter of 1.50 mm. The elastic modulus of the wire is 6.20 × 10¹⁰ N/m². If a force of 400 N is applied to end of the wire, then the increase in length of the wire is