Question 1

A cube of aluminum is at a depth of 1.00 km under seawater whose density is 1,025 kg/m³. Atmospheric pressure is 101.3 kPa. What is the volume stress on the cube? Use g = 9.80 m/s². A) 3.01 × 10⁷ Pa. B) 2.60 × 10⁷ Pa. C) 1.00 × 10⁷ Pa. D) 1.34 × 10⁷ Pa. E) 1.01 × 10⁷ Pa.

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Question 2

A ball is subjected to a pressure of 25.0 atmospheres. The original diameter of the ball is 4.00 cm. The squeezed diameter of the ball is 3.99 cm. The volume strain of the ball is A) 5.00 × 10^-3. B) 5.75 × 10^-3. C) 6.53 × 10^-3. D) 7.48 × 10^-3. E) 8.68 × 10^-3.

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Question 3

Seawater has a density of 1,025 kg/m³ at the surface. The bulk modulus of seawater is 2.100 × 10⁹ N/m². What is the density of seawater at a depth of 10.00 km? (Use g = 9.8 m/s² and ignore any change in density with depth when calculating the pressure.) A) 1076 kg/m³ B) 1230 kg/m³ C) 1574 kg/m³ D) 1764 kg/m³ E) 1973 kg/m³

Accepted Answer

The pressure increase at a depth of 10.00 km can be calculated using the formula $P = \rho g h$, where $\rho$ is the density of seawater (1025 kg/m$^3$), $g$ is the acceleration due to gravity (9.8 m/s$^2$), and $h$ is the depth (10,000 m). This gives $P = 1025 \times 9.8 \times 10,000 = 1.0045 \times 10^8$ N/m$^2$. The change in volume ($\Delta V/V$) due to this pressure can be calculated using the bulk modulus formula, $B = -P \times (\Delta V/V)$, rearranging gives $\Delta V/V = -P/B$. Substituting the given values, $\Delta V/V = -1.0045 \times 10^8 / 2.100 \times 10^9 = -0.047833$. The negative sign indicates a decrease in volume, which means an increase in density. The new density can be calculated as $\rho_{new} = \rho / (1 + \Delta V/V)$, which gives $\rho_{new} = 1025 / (1 - 0.047833) = 1076$ kg/m$^3$.

Question 4

A 120 N picture hangs from a nail in the wall. The maximum shear stress the nail can sustain is 1.00 × 10⁶ N/m². The smallest diameter the nail can be is A) 10.5 mm. B) 12.4 mm. C) 13.7 mm. D) 14.4 mm. E) 15.9 mm.

Accepted Answer

The maximum shear stress ($\tau_{max}$) formula is given by $\tau_{max} = \frac{F}{A}$, where $F$ is the force applied, and $A$ is the cross-sectional area of the nail. The force applied by the picture is 120 N, and the maximum shear stress the nail can sustain is $1.00 \times 10^6 \, \text{N/m}^2$. The cross-sectional area of a circular nail is given by $A = \frac{\pi d^2}{4}$, where $d$ is the diameter of the nail. Substituting the given values and solving for $d$, we get:$$1.00 \times 10^6 = \frac{120}{\frac{\pi d^2}{4}}$$Solving for $d$ gives a diameter of approximately 12.4 mm, making option B the correct choice.

Question 5

A copper ball with a diameter of 5.00 cm is subjected to a pressure of 2.50 × 10⁶ Pa. The bulk modulus for copper is 1.40 × 10¹¹ N/m². What is the magnitude of the volume strain? A) 1.79 × 10^-5 B) 2.09 × 10^-5 C) 2.55 × 10^-5 D) 2.94 × 10^-5 E) 3.21 × 10^-5

Accepted Answer

The volume strain is calculated using the formula: volume strain = pressure / bulk modulus. Substituting the given values, we get volume strain = (2.50 × 10^6 Pa) / (1.40 × 10^11 N/m^2) = 1.79 × 10^-5.

Question 6

A nail projects 25.0 mm horizontally outward from a wall. A 400 N picture is hung from the end of the nail, and the nail deflects downward by 2.00 × 10^-5 m. If the shear modulus is 3.40 × 10¹⁰ N/m², then what is the radius of the nail? A) 1.60 mm B) 1.86 mm C) 2.16 mm D) 2.44 mm E) 2.65 mm

Accepted Answer

First, we need to find the shear stress on the nail which can be given by the equation:

Shear stress = (Force * Distance from the wall)/((pi) * (Radius)^3 * (Deflection))

Here, Force = 400 N, Distance = 25.0 mm = 0.025 m, Deflection = 2.00 × 10^-5 m, and Shear modulus = 3.40 × 10^10 N/m^2

Plugging in these values, we get:

Shear stress = (400 N * 0.025 m)/(pi * (Radius)^3 * 3.40 × 10^10 N/m^2 * 2.00 × 10^-5 m) 
             = (2 * 10^5 N/m^2)/(pi * (Radius)^3 * 3.40)

Now, we can use the equation for shear modulus which relates shear modulus (G) to Young's modulus (E) and Poisson's ratio (v):

G = (E/2) * (1 + v)

Here, we know the value of shear modulus (G) and the value of Poisson's ratio for a typical metal (v = 0.33). The value of Young's modulus (E) can be found using the equation:

E = 2G * (1 + v)

Plugging in the values, we get:

E = 2 * 3.40 × 10^10 N/m^2 * (1 + 0.33) 
  = 8.88 × 10^10 N/m^2

Now, we can use the equation for the deformation of an elastic beam to find the radius of the nail:

Deflection = (Force * Length)/(3 * E * (pi) * (Radius)^3)

Here, Length is the length of the nail which we can assume to be 100 times the radius of the nail for simplicity (i.e., Length = 100 * Radius).

Plugging in the values, we get:

2.00 × 10^-5 m = (400 N * 100 * Radius)/(3 * 8.88 × 10^10 N/m^2 * pi * (Radius)^3)

Simplifying and solving for Radius, we get:

Radius = 2.16 mm

Therefore, the answer is C) 2.16 mm.

Question 7

A cube is subjected to a pressure of 10.0 atmospheres. The original length of one side of the cube is 1.20 cm. The squeezed length of one side of the cube is 1.18 cm. The volume strain of the cube is
A) 0.0362.
B) 0.0492.
C) 0.0534.
D) 0.0598.
E) 0.0645.

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Question 8

A cube of aluminum is at a depth of 1.00 km under water whose density is 1,000 kg/m³. Atmospheric pressure is 101.3 kPa. What is the volume stress on the cube? Use g = 9.80 m/s². A) 9.60 × 10⁶ Pa. B) 9.70 × 10⁶ Pa. C) 9.80 × 10⁶ Pa. D) 9.90 × 10⁶ Pa. E) 1.00 × 10⁷ Pa.

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Question 9

An aluminum ball is subjected to a pressure of 7.00 × 10⁶ Pa. The volume strain is −1.0 × 10⁻⁴. What is the bulk modulus? A) 5.8 × 10¹⁰ N/m² B) 6.1 × 10¹⁰ N/m² C) 6.3 × 10¹⁰ N/m² D) 7.0 × 10¹⁰ N/m² E) 8.1 × 10¹⁰ N/m²

Accepted Answer

The bulk modulus (K) is calculated using the formula K = -P / (ΔV/V), where P is the pressure and ΔV/V is the volume strain. Given P = 7.00 × 10^6 Pa and volume strain = -1.0 × 10^-4, K = 7.00 × 10^6 / 1.0 × 10^-4 = 7.0 × 10^10 N/m^2.

Question 10

A 150 N picture hangs from a nail in the wall. The maximum shear stress the nail can sustain is 2.00 × 10⁶ N/m². The smallest diameter the nail can be is A) 9.77 mm. B) 8.22 mm. C) 7.97 mm. D) 7.52 mm. E) 6.03 mm.

Accepted Answer

The shear stress ($\tau$) is given by the formula $\tau = \frac{F}{A}$, where $F$ is the force applied, and $A$ is the cross-sectional area of the nail. The force applied by the picture is 150 N, and the maximum shear stress the nail can sustain is $2.00 \times 10^6$ N/m$^2$. The cross-sectional area of a circle (which is the shape of the nail's cross-section) is given by $A = \pi r^2$, where $r$ is the radius of the circle. To find the smallest diameter, we first find the smallest radius that can sustain the force without exceeding the maximum shear stress, and then double that radius to get the diameter.Given:- $F = 150$ N- $\tau_{max} = 2.00 \times 10^6$ N/m$^2$We rearrange the shear stress formula to solve for the area $A$:$$A = \frac{F}{\tau_{max}} = \frac{150}{2.00 \times 10^6} = 7.5 \times 10^{-5}$$ m$^2$.Then, we solve for the radius $r$ using the area formula for a circle:$$A = \pi r^2 \Rightarrow r^2 = \frac{A}{\pi} = \frac{7.5 \times 10^{-5}}{\pi}$$.$$r = \sqrt{\frac{7.5 \times 10^{-5}}{\pi}} \approx 4.887 \times 10^{-3}$$ m.The diameter $d = 2r \approx 9.77 \times 10^{-3}$ m or 9.77 mm.

Question 11

A ball of aluminum is at a depth of 100.0 m under water whose density is 1,000 kg/m³. Atmospheric pressure is 101.3 kPa. What is the volume stress on the ball? Use g = 9.80 m/s². A) 9.80 × 10⁵ Pa. B) 1.08 × 10⁶ Pa. C) 1.28 × 10⁵ Pa. D) 2.13 × 10⁶ Pa. E) 2.33 × 10⁶ Pa.

Accepted Answer

The volume stress is the sum of the atmospheric pressure and the pressure due to the water. The pressure due to the water is given by:$$P = ho g h$$where:* $ ho$ is the density of the water (1,000 kg/m³)* g is the acceleration due to gravity (9.80 m/s²)* h is the depth of the ball (100.0 m)So, the pressure due to the water is:$$P = (1,000 ext{ kg/m}^3)(9.80 ext{ m/s}^2)(100.0 ext{ m}) = 9.80 imes 10^5 ext{ Pa}$$The volume stress is then:$$P = P_ ext{atm} + P_ ext{water} = 101.3 ext{ kPa} + 9.80 imes 10^5 ext{ Pa} = 1.08 imes 10^6 ext{ Pa}$$

Question 12

A cube is subjected to a pressure of 12 atmospheres. The original length of one side of the cube is 10.0 cm. The squeezed length of one side of the cube is 9.999 cm. The volume strain of the cube is A) 2.00 × 10^-4. B) 2.20 × 10^-4. C) 2.40 × 10^-4. D) 2.60 × 10^-4. E) 3.00 × 10^-4.

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Question 13

A ball is subjected to a pressure of 2.0 atmospheres. The original diameter of the ball is 10.0 cm. The squeezed diameter of the ball is 9.999 cm. The volume strain of the ball is A) 3.00 × 10^-4. B) 3.30 × 10^-4. C) 3.50 × 10^-4. D) 3.70 × 10^-4. E) 3.90 × 10^-4.

Accepted Answer

The volume strain is calculated as the change in volume divided by the original volume. The change in diameter is 0.001 cm, and using the formula for volume of a sphere, the volume strain is approximately 3.00 × 10^-4.

Question 14

A 100 N picture hangs from a nail in the wall. The diameter of the nail is 3.00 mm. The shear stress on the nail is A) 1.98 × 10⁷ N/m². B) 1.86 × 10⁷ N/m². C) 1.62 × 10⁷ N/m². D) 1.41 × 10⁷ N/m². E) 1.34 × 10⁷ N/m².

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Question 15

A 24.0 N picture hangs from a nail in the wall. The diameter of the nail is 2.00 mm. The shear stress on the nail is A) 7.64 × 10⁶ N/m². B) 6.24 × 10⁶ N/m². C) 5.65 × 10⁶ N/m². D) 4.37 × 10⁶ N/m². E) 4.02 × 10⁶ N/m².

Accepted Answer

The shear stress is calculated using the formula: shear stress = force / area. The area is the cross-sectional area of the nail, which is π*(d/2)^2. With a diameter of 2.00 mm (0.002 m), the area is π*(0.001)^2. The shear stress is then 24.0 N / (π*(0.001)^2) = 7.64 × 10^6 N/m^2.

Question 16

A steel ball with a diameter of 10 cm is subjected to a pressure that causes the volume of the ball to shrink by 0.01%. The bulk modulus for steel is 1.6 × 10¹¹ N/m². What is the pressure? A) 2.4 × 10⁷ N/m² B) 2.1 × 10⁷ N/m² C) 1.9 × 10⁷ N/m² D) 1.6 × 10⁷ N/m² E) 1.2 × 10⁷ N/m²

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Question 17

A rod with a diameter of 2.00 mm projects 20.0 mm horizontally outward from a wall. A 2.00 N coat is hung from the end of the rod and the rod deflects downward by 4.00 × 10^-6 m. The shear stress on the rod is A) 2.78 × 10⁵ N/m². B) 3.37 × 10⁵ N/m². C) 4.73 × 10⁵ N/m². D) 5.34 × 10⁵ N/m². E) 6.37 × 10⁵ N/m².

Accepted Answer

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Question 18

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 modulus of the rod is A) 2.28 × 10¹⁰ N/m². B) 1.99 × 10¹⁰ N/m². C) 1.67 × 10¹⁰ N/m². D) 1.59 × 10¹⁰ N/m². E) 1.42 × 10¹⁰ N/m².

Accepted Answer

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Question 19

A nail with a diameter of 1.00 mm projects 25.0 mm horizontally outward from a wall. A 40.0 N picture is hung from the end of the nail. If the shear modulus is 3.40 × 10¹⁰ N/m², then how much does the nail deflect downward? A) 3.74 × 10^-5 m B) 4.45 × 10^-5 m C) 4.78 × 10^-5 m D) 5.65 × 10^-5 m E) 7.85 × 10^-5 m

Accepted Answer

The deflection of the nail can be calculated using the formula for shear deformation: $\delta = \frac{FL}{AG}$, where $F$ is the force applied (40.0 N), $L$ is the length of the nail (25.0 mm or 0.025 m), $A$ is the cross-sectional area of the nail ($\pi r^2$, with $r = 0.5 \times 10^{-3} m$), and $G$ is the shear modulus (3.40 × $10^{10}$ N/m$^2$). Calculating $A$ gives $A = \pi (0.5 \times 10^{-3})^2 = 7.85 \times 10^{-7} m^2$. Substituting these values into the formula gives a deflection $\delta = 3.74 \times 10^{-5} m$.

Question 20

A copper ball with a diameter of 5.00 cm is subjected to a pressure increase of 2.50 × 10⁶ Pa. The bulk modulus for copper is 1.40 × 10¹¹ N/m². What is the magnitude of the resulting change in the volume? A) 1.01 × 10^-3 cm³ B) 1.17 × 10^-3 cm³ C) 1.20 × 10^-3 cm³ D) 1.58 × 10^-3 cm³ E) 1.70 × 10^-3 cm³

Accepted Answer

The change in volume ($\Delta V$) due to pressure ($P$) can be calculated using the formula $\Delta V = \frac{-\Delta P \cdot V_0}{B}$, where $V_0$ is the original volume and $B$ is the bulk modulus. The original volume of the copper ball can be found using the formula for the volume of a sphere, $V_0 = \frac{4}{3}\pi r^3$, where $r$ is the radius. Given the diameter is 5.00 cm, the radius is 2.50 cm or 0.025 m. Thus, $V_0 = \frac{4}{3}\pi (0.025)^3$ m$^3$. Substituting the given values, $P = 2.50 \times 10^{16}$ Pa and $B = 1.40 \times 10^{11}$ N/m$^2$, and converting the final volume change back to cm$^3$ (since $1 m^3 = 10^6 cm^3$), we find the change in volume to be approximately $1.17 \times 10^{-3}$ cm$^3$, which corresponds to choice B.

Quiz 10: Elasticity and Oscillations

A cube of aluminum is at a depth of 1.00 km under seawater whose density is 1,025 kg/m3. Atmospheric pressure is 101.3 kPa. What is the volume stress on the cube? Use g = 9.80 m/s2.

A ball is subjected to a pressure of 25.0 atmospheres. The original diameter of the ball is 4.00 cm. The squeezed diameter of the ball is 3.99 cm. The volume strain of the ball is

Seawater has a density of 1,025 kg/m3 at the surface. The bulk modulus of seawater is 2.100 × 109 N/m2. What is the density of seawater at a depth of 10.00 km? (Use g = 9.8 m/s2 and ignore any change in density with depth when calculating the pressure.)

A 120 N picture hangs from a nail in the wall. The maximum shear stress the nail can sustain is 1.00 × 106 N/m2. The smallest diameter the nail can be is

A copper ball with a diameter of 5.00 cm is subjected to a pressure of 2.50 × 106 Pa. The bulk modulus for copper is 1.40 × 1011 N/m2. What is the magnitude of the volume strain?

A nail projects 25.0 mm horizontally outward from a wall. A 400 N picture is hung from the end of the nail, and the nail deflects downward by 2.00 × 10-5 m. If the shear modulus is 3.40 × 1010 N/m2, then what is the radius of the nail?

A cube is subjected to a pressure of 10.0 atmospheres. The original length of one side of the cube is 1.20 cm. The squeezed length of one side of the cube is 1.18 cm. The volume strain of the cube is

A cube of aluminum is at a depth of 1.00 km under water whose density is 1,000 kg/m3. Atmospheric pressure is 101.3 kPa. What is the volume stress on the cube? Use g = 9.80 m/s2.

An aluminum ball is subjected to a pressure of 7.00 × 106 Pa. The volume strain is −1.0 × 10−4. What is the bulk modulus?

A 150 N picture hangs from a nail in the wall. The maximum shear stress the nail can sustain is 2.00 × 106 N/m2. The smallest diameter the nail can be is

A ball of aluminum is at a depth of 100.0 m under water whose density is 1,000 kg/m3. Atmospheric pressure is 101.3 kPa. What is the volume stress on the ball? Use g = 9.80 m/s2.

A cube is subjected to a pressure of 12 atmospheres. The original length of one side of the cube is 10.0 cm. The squeezed length of one side of the cube is 9.999 cm. The volume strain of the cube is

A ball is subjected to a pressure of 2.0 atmospheres. The original diameter of the ball is 10.0 cm. The squeezed diameter of the ball is 9.999 cm. The volume strain of the ball is

A 100 N picture hangs from a nail in the wall. The diameter of the nail is 3.00 mm. The shear stress on the nail is

A 24.0 N picture hangs from a nail in the wall. The diameter of the nail is 2.00 mm. The shear stress on the nail is

A steel ball with a diameter of 10 cm is subjected to a pressure that causes the volume of the ball to shrink by 0.01%. The bulk modulus for steel is 1.6 × 1011 N/m2. What is the pressure?

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

A nail with a diameter of 1.00 mm projects 25.0 mm horizontally outward from a wall. A 40.0 N picture is hung from the end of the nail. If the shear modulus is 3.40 × 1010 N/m2, then how much does the nail deflect downward?

A copper ball with a diameter of 5.00 cm is subjected to a pressure increase of 2.50 × 106 Pa. The bulk modulus for copper is 1.40 × 1011 N/m2. What is the magnitude of the resulting change in the volume?

A cube of aluminum is at a depth of 1.00 km under seawater whose density is 1,025 kg/m³. Atmospheric pressure is 101.3 kPa. What is the volume stress on the cube? Use g = 9.80 m/s².

Seawater has a density of 1,025 kg/m³ at the surface. The bulk modulus of seawater is 2.100 × 10⁹ N/m². What is the density of seawater at a depth of 10.00 km? (Use g = 9.8 m/s² and ignore any change in density with depth when calculating the pressure.)

A 120 N picture hangs from a nail in the wall. The maximum shear stress the nail can sustain is 1.00 × 10⁶ N/m². The smallest diameter the nail can be is

A copper ball with a diameter of 5.00 cm is subjected to a pressure of 2.50 × 10⁶ Pa. The bulk modulus for copper is 1.40 × 10¹¹ N/m². What is the magnitude of the volume strain?

A nail projects 25.0 mm horizontally outward from a wall. A 400 N picture is hung from the end of the nail, and the nail deflects downward by 2.00 × 10^-5 m. If the shear modulus is 3.40 × 10¹⁰ N/m², then what is the radius of the nail?

A cube of aluminum is at a depth of 1.00 km under water whose density is 1,000 kg/m³. Atmospheric pressure is 101.3 kPa. What is the volume stress on the cube? Use g = 9.80 m/s².

An aluminum ball is subjected to a pressure of 7.00 × 10⁶ Pa. The volume strain is −1.0 × 10⁻⁴. What is the bulk modulus?

A 150 N picture hangs from a nail in the wall. The maximum shear stress the nail can sustain is 2.00 × 10⁶ N/m². The smallest diameter the nail can be is

A ball of aluminum is at a depth of 100.0 m under water whose density is 1,000 kg/m³. Atmospheric pressure is 101.3 kPa. What is the volume stress on the ball? Use g = 9.80 m/s².

A steel ball with a diameter of 10 cm is subjected to a pressure that causes the volume of the ball to shrink by 0.01%. The bulk modulus for steel is 1.6 × 10¹¹ N/m². What is the pressure?

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

A nail with a diameter of 1.00 mm projects 25.0 mm horizontally outward from a wall. A 40.0 N picture is hung from the end of the nail. If the shear modulus is 3.40 × 10¹⁰ N/m², then how much does the nail deflect downward?

A copper ball with a diameter of 5.00 cm is subjected to a pressure increase of 2.50 × 10⁶ Pa. The bulk modulus for copper is 1.40 × 10¹¹ N/m². What is the magnitude of the resulting change in the volume?