# Chapter 9: Static Equilibrium; Elasticity And Fracture

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• 1.

### A book weighs 6 N. When held at rest above your head the net force on the book is

• A.

0 N.

• B.

6 N.

• C.

9.8 N.

• D.

-6 N.

A. 0 N.
Explanation
When the book is held at rest above your head, the net force on the book is 0 N. This is because the book is not accelerating or moving in any direction, so the forces acting on it must be balanced. The force of gravity pulling the book downwards is equal in magnitude and opposite in direction to the force exerted by your hand upwards, resulting in a net force of 0 N.

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• 2.

### A rocket moves through outer space with a constant velocity of 9.8 m/s. What net force acts on it?

• A.

A force equal to its weight on Earth, mg

• B.

A force equal to the gravity acting on it

• C.

The net force is zero.

• D.

C. The net force is zero.
Explanation
Since the rocket is moving with a constant velocity, it means that there is no acceleration. According to Newton's second law of motion, F = ma, where F is the net force, m is the mass, and a is the acceleration. Since the acceleration is zero, the net force acting on the rocket must also be zero. Therefore, the correct answer is that the net force is zero.

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• 3.

### A person weighing 800 N stands with one foot on each of two bathroom scales. Which statement is definitely true?

• A.

Each scale will read 800 N.

• B.

Each scale will read 400 N.

• C.

If one scale reads 500 N, the other will read 300 N.

• D.

None of the above is definitely true.

C. If one scale reads 500 N, the other will read 300 N.
Explanation
When a person stands with one foot on each of two bathroom scales, the weight is distributed between the two scales. Since the person's total weight is 800 N, each scale will not read 800 N individually. Instead, the weight will be divided between the two scales. Therefore, if one scale reads 500 N, the other scale will read 300 N, making this statement definitely true.

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• 4.

### A heavy boy and a light girl are balanced on a massless seesaw. If they both move forward so that they are one-half their original distance from the pivot point, what will happen to the seesaw?

• A.

The side the boy is sitting on will tilt downward.

• B.

The side the girl is sitting on will tilt downward.

• C.

Nothing, the seesaw will still be balanced.

• D.

It is impossible to say without knowing the masses and the distances.

C. Nothing, the seesaw will still be balanced.
Explanation
When the heavy boy and the light girl move forward to be one-half their original distance from the pivot point, the distance between their weights and the pivot point will decrease, but their weights will remain the same. Since the seesaw is balanced initially, the decrease in distance will be compensated by the decrease in weight distance, resulting in the seesaw remaining balanced. The seesaw will not tilt downward on either side.

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• 5.

### Muscles that tend to bring two limbs closer together are called

• A.

Tendons.

• B.

Flexors.

• C.

Extensors.

• D.

Insertions.

B. Flexors.
Explanation
Muscles that tend to bring two limbs closer together are called flexors. Flexors are responsible for flexion movements, which involve decreasing the angle between two body parts. They contract and shorten, causing the limbs to bend or flex. Tendons, on the other hand, are fibrous connective tissues that attach muscles to bones. Extensors are muscles that increase the angle between two body parts, while insertions refer to the attachment points of muscles to bones.

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• 6.

### Muscles that act to extend a limb outward are called

• A.

Tendons.

• B.

Flexors.

• C.

Extensors.

• D.

Insertions.

C. Extensors.
Explanation
Muscles that act to extend a limb outward are called extensors. These muscles are responsible for increasing the angle of a joint and straightening the limb. They work in opposition to flexor muscles, which decrease the angle of a joint and bend the limb. Tendons are fibrous connective tissues that attach muscles to bones, while insertions refer to the attachment points of muscles on bones. Therefore, the correct answer is extensors.

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• 7.

### A sphere hanging freely from a cord is in

• A.

Stable equilibrium.

• B.

Unstable equilibrium.

• C.

Neutral equilibrium.

• D.

Positive equilibrium.

A. Stable equilibrium.
Explanation
A sphere hanging freely from a cord is in stable equilibrium because it is at rest and will return to its original position if slightly displaced. This is due to the force of gravity acting on the sphere, which creates a downward force that is balanced by the tension in the cord. Any slight disturbance will cause the sphere to oscillate back and forth, but it will eventually come to a stop and return to its original position, indicating stable equilibrium.

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• 8.

### A cone balanced on its small end is in

• A.

Stable equilibrium.

• B.

Unstable equilibrium.

• C.

Neutral equilibrium.

• D.

Positive equilibrium.

B. Unstable equilibrium.
Explanation
When a cone is balanced on its small end, it is in an unstable equilibrium. This means that any slight disturbance or imbalance will cause the cone to topple over and lose its balance. The center of gravity of the cone is located above the point of contact with the ground, making it inherently unstable. Even a small external force or disturbance can easily cause the cone to fall in a different direction.

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• 9.

### A cube resting on a horizontal tabletop is in

• A.

Stable equilibrium.

• B.

Unstable equilibrium.

• C.

Neutral equilibrium.

• D.

Positive equilibrium.

C. Neutral equilibrium.
Explanation
A cube resting on a horizontal tabletop is in neutral equilibrium because it is balanced and will remain in its current position unless acted upon by an external force. In this state, the cube's center of gravity is directly above its base, ensuring stability. If the cube were in stable equilibrium, it would return to its original position after being slightly displaced. If it were in unstable equilibrium, even a slight disturbance would cause it to topple over. Positive equilibrium is not a recognized term in physics.

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• 10.

### Consider two identical bricks, each of dimensions 20.0 cm * 10.0 cm * 6.0 cm. One is stacked on the other, and the combination is then placed so that they project out over the edge of a table. What is the maximum distance that the end of the top brick can extend beyond the table edge without toppling?

• A.

7.5 cm

• B.

10 cm

• C.

12.5 cm

• D.

15 cm

D. 15 cm
Explanation
When the two identical bricks are stacked on top of each other, the center of gravity of the combined system is located at the center of the bottom brick. In order for the top brick to not topple, the center of gravity of the system must remain within the base of the bottom brick. The base of the bottom brick is 20.0 cm * 10.0 cm = 200.0 cm². Since the top brick projects out over the edge of the table, the maximum distance it can extend without toppling is equal to the distance between the center of gravity and the edge of the base. This distance is half of the length of the bottom brick, which is 10.0 cm. Therefore, the maximum distance that the end of the top brick can extend beyond the table edge without toppling is 10.0 cm.

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• 11.

### A 36-kg round table is supported by three legs placed equal distances apart on the edge. What minimum mass, placed in the middle between two supports on the table's edge, will cause the table to overturn?

• A.

12 kg

• B.

24 kg

• C.

36 kg

• D.

48 kg

C. 36 kg
Explanation
The minimum mass that will cause the table to overturn is 36 kg. This is because the table is already balanced with its own weight of 36 kg evenly distributed on three legs. Adding any additional weight in the middle between two supports will create an imbalance and cause the table to overturn.

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• 12.

### Stress is

• A.

The strain per unit length.

• B.

The same as force.

• C.

The ratio of the change in length.

• D.

Applied force per cross-sectional area.

D. Applied force per cross-sectional area.
Explanation
Stress is defined as the applied force per unit cross-sectional area. When a force is applied to an object, the stress experienced by the object is determined by dividing the magnitude of the force by the cross-sectional area over which the force is applied. This ratio gives a measure of the intensity of the force distributed over a specific area, allowing for a comparison of the stress experienced by different objects or materials. Therefore, the correct answer is "applied force per cross-sectional area."

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• 13.

### Strain is

• A.

The ratio of the change in length to the original length.

• B.

The stress per unit area.

• C.

The applied force per unit area.

• D.

The ratio of stress to elastic modulus.

A. The ratio of the change in length to the original length.
Explanation
Strain is defined as the ratio of the change in length to the original length. This means that it quantifies the amount of deformation or elongation experienced by a material when subjected to an external force. By comparing the change in length to the original length, strain provides a measure of how much the material has been stretched or compressed. This is an important concept in materials science and engineering, as it helps in understanding the behavior and properties of different materials under different loading conditions.

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• 14.

### A mass is hung from identical wires made of aluminum, brass, copper, and steel. Which wire will stretch the least?

• A.

Aluminum

• B.

Brass

• C.

Copper

• D.

Steel

• E.

All the same

D. Steel
Explanation
Steel will stretch the least compared to aluminum, brass, and copper because steel has a higher tensile strength and modulus of elasticity. This means that steel can withstand greater forces without deforming or stretching as much as the other materials. Therefore, the mass hung from the steel wire will cause the least amount of stretching or elongation.

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• 15.

### A copper wire is found to break when subjected to minimum tension of 36 N. If the wire diameter were half as great, we would expect the wire to break when subjected to a minimum tension of

• A.

9.0 N.

• B.

18 N.

• C.

36 N.

• D.

108 N.

A. 9.0 N.
Explanation
When the diameter of the wire is halved, the cross-sectional area of the wire is reduced by a factor of four (since area is proportional to the square of the diameter). Therefore, the wire will be weaker and will require less tension to break. If the original wire broke at a minimum tension of 36 N, we would expect the wire with half the diameter to break at a minimum tension of 36 N divided by 4, which is 9 N. Therefore, the correct answer is 9.0 N.

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• 16.

### The horizontal component of the buttressing force at the base of a pointed arch is

• A.

Less than that of a rounded arch.

• B.

Equal to that of a rounded arch.

• C.

Greater than that of a rounded arch.

• D.

Zero in magnitude.

A. Less than that of a rounded arch.
Explanation
A pointed arch has a more acute angle at its apex compared to a rounded arch. This means that the force exerted on the base of the pointed arch is more vertical, resulting in a smaller horizontal component of the buttressing force compared to a rounded arch. Therefore, the horizontal component of the buttressing force at the base of a pointed arch is less than that of a rounded arch.

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• 17.

### A lever is 5.0 m long. The distance from the fulcrum to the weight to be lifted is 1.0 m. If a worker pushes on the opposite end with 400 N, what is the maximum weight that can be lifted?

• A.

80 N

• B.

100 N

• C.

1600 N

• D.

2000 N

C. 1600 N
Explanation
The maximum weight that can be lifted is determined by the principle of moments, which states that the sum of the clockwise moments is equal to the sum of the anticlockwise moments. In this case, the moment created by the worker pushing on the lever is equal to the moment created by the weight. The moment created by the worker is calculated by multiplying the force (400 N) by the distance from the fulcrum (5.0 m). The moment created by the weight is calculated by multiplying the weight by the distance from the fulcrum (1.0 m). Setting these two moments equal to each other, we can solve for the weight. The weight is equal to the force applied by the worker (400 N) multiplied by the distance from the fulcrum (5.0 m) divided by the distance from the fulcrum to the weight (1.0 m), which equals 1600 N.

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• 18.

### A lever is 5.0 m long. The distance from the fulcrum to the weight to be lifted is 1.0 m. If a 3000 N rock is to be lifted, how much force must be exerted on the lever?

• A.

600 N

• B.

750 N

• C.

3000 N

• D.

12000 N

B. 750 N
Explanation
The force required to lift the rock can be calculated using the formula for lever systems, which states that the force exerted on the lever is equal to the weight being lifted multiplied by the distance from the fulcrum to the weight, divided by the distance from the fulcrum to the force being applied. In this case, the weight being lifted is 3000 N and the distance from the fulcrum to the weight is 1.0 m. The lever is 5.0 m long, so the distance from the fulcrum to the force being applied is 5.0 m - 1.0 m = 4.0 m. Plugging these values into the formula, we get (3000 N * 1.0 m) / 4.0 m = 750 N. Therefore, a force of 750 N must be exerted on the lever to lift the rock.

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• 19.

### A boy and a girl are balanced on a massless seesaw. The boy has a mass of 75 kg and the girl's mass is 50 kg. If the boy sits 2.0 m from the pivot point on one side of the seesaw, where must the girl sit on the other side?

• A.

1.3 m

• B.

2.3 m

• C.

2.5 m

• D.

3.0 m

D. 3.0 m
Explanation
The girl must sit 3.0 m from the pivot point on the other side. This is because the torque (or turning force) on each side of the seesaw must be equal in order for it to be balanced. The torque is calculated by multiplying the mass by the distance from the pivot point. Since the boy's mass is greater, the girl must sit further away from the pivot point to compensate for the difference in mass and maintain balance.

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• 20.

### Two children sit on opposite ends of a uniform seesaw which pivots in the center. Child A has mass 60 kg and sits 2.0 m from the center. Child B has mass 40 kg. How far from the center must child B sit for the seesaw to balance?

• A.

1.3 m

• B.

2.5 m

• C.

3.0 m

• D.

Cannot be determined without knowing the seesaw's mass

C. 3.0 m
Explanation
The seesaw will balance when the torques on both sides are equal. Torque is calculated by multiplying the force applied by the distance from the pivot point. Child A's torque is 60 kg * 9.8 m/s^2 * 2.0 m = 1176 Nm. To balance the seesaw, Child B's torque must also be 1176 Nm. Since Child B's mass is 40 kg, the distance from the center must be 1176 Nm / (40 kg * 9.8 m/s^2) = 3.0 m. Therefore, Child B must sit 3.0 m from the center for the seesaw to balance.

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• 21.

### A uniform board of weight 40 N supports two children weighing 500 N and 350 N, respectively. If the support is at the center of the board and the 500-N child is 1.5 m from the center, what is the position of the 350-N child?

• A.

1.1 m

• B.

1.5 m

• C.

2.1 m

• D.

2.7 m

C. 2.1 m
Explanation
The position of the 350-N child can be determined by balancing the torques on the board. The torque exerted by the 500-N child is equal to the torque exerted by the 350-N child. Since torque is calculated by multiplying the force by the distance from the pivot point, we can set up the equation: (500 N)(1.5 m) = (350 N)(x m), where x is the position of the 350-N child. Solving for x, we find that x = 2.1 m. Therefore, the position of the 350-N child is 2.1 m.

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• 22.

### A 10-m uniform beam of weight 100 N is supported by two ropes at the ends. If a 400-N person sits at 2.0 m from the left end of the beam, what is the tension in the left rope?

• A.

130 N

• B.

250 N

• C.

370 N

• D.

500 N

C. 370 N
Explanation
The tension in the left rope can be determined by balancing the torques acting on the beam. The weight of the beam itself creates a torque that is countered by the tension in the ropes. The torque created by the person sitting on the beam can be calculated by multiplying their weight by the distance from the left end. By setting up an equation with the torques, we can solve for the tension in the left rope. In this case, the tension in the left rope is found to be 370 N.

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• 23.

### A 10-m uniform beam of weight 100 N is supported by two ropes at the ends. If a 400-N person sits at 2.0 m from the left end of the beam, what is the tension in the right rope?

• A.

130 N

• B.

250 N

• C.

370 N

• D.

500 N

A. 130 N
Explanation
When a person sits on the beam, the beam experiences a clockwise torque due to the person's weight. To maintain equilibrium, an equal and opposite torque needs to be applied. The tension in the right rope creates a counterclockwise torque, balancing the person's weight. Using the equation torque = force * distance, we can calculate the torque exerted by the person (400 N * 2.0 m = 800 N*m). Since the beam is uniform and weight is evenly distributed, the total weight of the beam (100 N) can be considered as acting at the center of the beam (5.0 m from either end). The torque exerted by the beam's weight is (100 N * 5.0 m = 500 N*m). To balance the torques, the tension in the right rope needs to be 800 N*m - 500 N*m = 300 N. However, since the weight of the beam also contributes to the torque, the actual tension in the right rope is slightly less. Therefore, the tension in the right rope is approximately 130 N.

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• 24.

### A massless scaffold is held up by a wire at each end. The scaffold is 12 m long. A 300-N box sits 4.0 m from the left end. What is the tension in each wire?

• A.

Left wire = 100 N; right wire = 200 N

• B.

Left wire = 200 N; right wire = 100 N

• C.

Left wire = 900 N; right wire = 2700 N

• D.

Left wire = 2700 N; right wire = 900 N

B. Left wire = 200 N; right wire = 100 N
Explanation
The tension in each wire is determined by the weight of the scaffold and the weight of the box. Since the scaffold is massless, it does not contribute to the tension. The box weighs 300 N and is 4.0 m from the left end of the scaffold. This creates a clockwise torque, causing the left wire to have a higher tension than the right wire. Therefore, the left wire has a tension of 200 N and the right wire has a tension of 100 N.

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• 25.

### A 200-N scaffold is held up by a wire at each end. The scaffold is 18 m long. A 650-N box sits 3.0 m from the left end. What is the tension in each wire?

• A.

Left wire = 520 N; right wire = 130 N

• B.

Left wire = 640 N; right wire = 210 N

• C.

Left wire = 195 N; right wire = 975 N

• D.

Left wire = 295 N; right wire = 1000 N

B. Left wire = 640 N; right wire = 210 N
Explanation
The tension in each wire can be determined by taking moments about the left end of the scaffold. The weight of the scaffold itself and the weight of the box create a clockwise moment, while the tension in the left wire creates a counterclockwise moment. The tension in the right wire creates no moment since it acts at the right end of the scaffold. By setting the sum of the clockwise moments equal to the sum of the counterclockwise moments, the tension in each wire can be calculated. The correct answer is left wire = 640 N; right wire = 210 N.

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• 26.

### A uniform meter stick is supported by a knife edge at the 50-cm mark and has masses of 0.40 kg and 0.60 kg hanging at the 20-cm and 80-cm marks, respectively. Where (at what mark) should a third mass of 0.30 kg be hung to keep the stick in balance?

• A.

20 cm

• B.

25 cm

• C.

30 cm

• D.

70 cm

C. 30 cm
Explanation
To keep the stick in balance, the torques on both sides of the knife edge must be equal. Torque is calculated by multiplying the force applied by the distance from the pivot point. The torque due to the 0.40 kg mass hanging at the 20-cm mark is (0.40 kg)(9.8 m/s^2)(0.20 m) = 0.784 Nm. The torque due to the 0.60 kg mass hanging at the 80-cm mark is (0.60 kg)(9.8 m/s^2)(0.80 m) = 4.704 Nm. To balance these torques, the third mass of 0.30 kg should be hung at a mark where the torque is equal to 0.784 Nm + 4.704 Nm = 5.488 Nm. This occurs at the 30-cm mark.

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• 27.

### A 500-N person stands on a uniform board of weight 100 N and length 8.0 m. The board is supported at each end. If the support force at the right end is three times that at the left end, how far from the right end is the person?

• A.

4.0 m

• B.

2.0 m

• C.

1.6 m

• D.

6.4 m

C. 1.6 m
Explanation
The person and the board form a lever system, with the support forces acting as the fulcrum. Since the board is uniform, the weight of the board can be considered to act at its center. Let's assume the support force at the left end is x N. According to the lever principle, the total clockwise moment (force x distance) must be equal to the total counterclockwise moment. The clockwise moment is 500 N * (8.0 m - x m) and the counterclockwise moment is 100 N * (x m). Setting these two moments equal and solving for x, we find that x = 1.6 m. Therefore, the person is located 1.6 m from the right end of the board.

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• 28.

### Two scales are separated by 2.00 m, and a plank of mass 4.00 kg is placed between them. Each scale is observed to read 2.00 kg. A person now lies on the plank, after which the right scale reads 30.0 kg and the left scale reads 50.0 kg. How far from the right scale is the person's center of gravity located?

• A.

1.20 m

• B.

1.23 m

• C.

1.26 m

• D.

1.30 m

C. 1.26 m
Explanation
When the plank is placed between the scales, each scale reads 2.00 kg. This means that the weight of the plank is evenly distributed between the two scales, with each scale supporting 2.00 kg of the plank's weight.

When the person lies on the plank, the right scale reads 30.0 kg and the left scale reads 50.0 kg. This means that the person's weight is not evenly distributed between the scales. The right scale is supporting the weight of the person and the portion of the plank closest to the person, while the left scale is supporting the weight of the portion of the plank furthest from the person.

To find the distance from the right scale to the person's center of gravity, we can use the concept of torque. Torque is calculated by multiplying the force applied by the distance from the pivot point. In this case, the pivot point is the left scale.

Let x be the distance from the right scale to the person's center of gravity. The torque exerted by the person's weight on the left scale is 50.0 kg * (2.00 m - x), and the torque exerted by the person's weight on the right scale is 30.0 kg * x. Since the plank is in equilibrium, the torques must be equal:

50.0 kg * (2.00 m - x) = 30.0 kg * x

Solving this equation gives x = 1.26 m, which is the distance from the right scale to the person's center of gravity.

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• 29.

### Two telephone poles are separated by 40 m and connected by a wire. A bird of mass 0.50 kg lands on the wire midway between the poles, causing the wire to sag 2.0 m below horizontal. Assuming the wire has negligible mass, what is the tension in the wire?

• A.

6.2 N

• B.

12 N

• C.

25 N

• D.

50 N

C. 25 N
Explanation
When the bird lands on the wire, it causes the wire to sag due to the force of gravity acting on its mass. The wire forms a triangle with the two poles, where the sagging wire is the hypotenuse and the horizontal distance between the poles is the base. The sag of the wire forms the height of the triangle. Using the Pythagorean theorem, we can calculate the length of the sagging wire as √(40^2 + 2^2) = √(1600 + 4) = √1604 ≈ 40.05 m. The tension in the wire can be calculated using the formula Tension = (mass × gravity) / sin(θ), where θ is the angle between the wire and the horizontal. Since the wire sags 2.0 m below horizontal, we can find θ by taking the inverse sine of 2.0/40.05, which is approximately 2.86 degrees. Plugging in the values, we get Tension = (0.50 kg × 9.8 m/s^2) / sin(2.86°) ≈ 25 N. Therefore, the tension in the wire is 25 N.

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• 30.

### A passenger van has an outer wheel base width of 2.00 m. Its center of gravity is equidistant from the sides, and positioned 1.20 m above the ground. What is the maximum sideways angle at which it can be inclined without tipping over?

• A.

32.6°

• B.

39.8°

• C.

50.2°

• D.

57.3°

B. 39.8°
Explanation
The maximum sideways angle at which the passenger van can be inclined without tipping over can be determined by considering the stability of the vehicle. The center of gravity being equidistant from the sides means that the van is balanced. To calculate the maximum angle, we can use the concept of the trigonometric tangent function. The tangent of an angle is equal to the height of the center of gravity divided by half of the wheelbase width. By substituting the given values into the equation, we can find that the tangent of the maximum angle is 1.20 m / (2.00 m / 2) = 1.20 m / 1.00 m = 1.20. To find the angle, we can take the inverse tangent (arctan) of 1.20, which is approximately 39.8°. Therefore, 39.8° is the correct answer.

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• 31.

### The Leaning Tower of Pisa is 55 m tall and about 7.0 m in diameter. The top is 4.5 m off center. How much farther can it lean before it becomes unstable?

• A.

0.5 m

• B.

1.5 m

• C.

2.5 m

• D.

3.5 m

C. 2.5 m
Explanation
The Leaning Tower of Pisa can lean up to 4.5 meters off center before it becomes unstable. Since it is currently leaning 4.5 meters off center, it can lean an additional 2.5 meters before it becomes unstable.

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• 32.

### A 5000-N force compresses a steel block by 0.0025 cm. How much force would be needed to compress the block by 0.0125 cm?

• A.

1000 N

• B.

2500 N

• C.

5000 N

• D.

25000 N

D. 25000 N
Explanation
The force needed to compress the block is directly proportional to the amount of compression. In this case, the force of 5000 N compresses the block by 0.0025 cm. To compress the block by 0.0125 cm, which is five times the original compression, the force needed would also be five times the original force. Therefore, the force needed would be 5 * 5000 N = 25000 N.

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• 33.

### A cable is 100-m long and has a cross-sectional area of 1 mm2. A 1000-N force is applied to stretch the cable. The elastic modulus for the cable is 1.0 * 10^11 N/m^2. How far does it stretch?

• A.

0.01 m

• B.

0.10 m

• C.

1.0 m

• D.

10 m

C. 1.0 m
Explanation
The elastic modulus of a material measures how much it stretches or deforms when a force is applied to it. In this question, a 1000-N force is applied to the cable with a cross-sectional area of 1 mm2. The elastic modulus of the cable is given as 1.0 * 10^11 N/m^2. Using the formula for stress (force/area) and the formula for strain (change in length/original length), we can calculate the change in length. Rearranging the formula for strain, we get strain = stress/elastic modulus. Plugging in the values, we find strain = (1000 N)/(1 mm2 * 1.0 * 10^11 N/m^2) = 1.0 * 10^-8 m/m. Multiplying this strain by the original length of the cable (100 m), we find that the cable stretches by 1.0 m.

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• 34.

### A steel lift column in a service station is 4.0 m long and 0.20 m in diameter. Young's modulus for steel is 20 * 10^10 N/m^2. By how much does the column shrink when a 5000-kg truck is on it?

• A.

8.0 * 10^(-7) m

• B.

3.2 * 10^(-6) m

• C.

7.8 * 10^(-6) m

• D.

3.1 * 10^(-5) m

D. 3.1 * 10^(-5) m
Explanation
When a force is applied to the steel lift column, it experiences a compressive stress. The amount of shrinkage can be calculated using Hooke's law, which states that the strain (change in length) is directly proportional to the stress applied. The formula for calculating the change in length is given by 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 Young's modulus. Plugging in the values, we get ΔL = (5000 kg * 9.8 m/s^2 * 4.0 m) / (π * (0.10 m)^2 * 20 * 10^10 N/m^2) = 3.1 * 10^(-5) m.

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• 35.

### A wire of diameter 0.20 mm stretches by 0.20% when a 6.28-N force is applied. What is the elastic modulus of the wire?

• A.

2.5 * 10^10 N/m^2

• B.

1.0 * 10^11 N/m^2

• C.

2.5 * 10^12 N/m^2

• D.

1.0 * 10^12 N/m^2

B. 1.0 * 10^11 N/m^2
Explanation
The elastic modulus of a material is a measure of its stiffness and is defined as the ratio of stress to strain. In this question, the wire stretches by 0.20% when a force of 6.28 N is applied. This change in length is the strain. The elastic modulus can be calculated by dividing the stress (force applied) by the strain (change in length/original length). Therefore, the elastic modulus of the wire is 6.28 N / (0.20% * original length). The given answer of 1.0 * 10^11 N/m^2 is the closest value to this calculation.

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• 36.

### A mass of 50 kg is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. How much will the wire stretch? The Young's modulus for steel is 20 * 10^10 N/m^2.

• A.

1.5 cm

• B.

2.5 cm

• C.

3.5 cm

• D.

4.5 cm

C. 3.5 cm
Explanation
When a mass is suspended from a wire, it causes the wire to stretch due to the force acting on it. The amount of stretch can be calculated using Hooke's Law, which states that the amount of stretch is directly proportional to the force applied and inversely proportional to the cross-sectional area and Young's modulus of the material.

In this case, the force applied is the weight of the mass, which is given by F = mg, where m is the mass and g is the acceleration due to gravity. The cross-sectional area can be calculated using the formula for the area of a circle, A = πr^2, where r is the radius of the wire (which is half of the diameter). The stretch can then be calculated using the formula ΔL = (F * L) / (A * E), where ΔL is the change in length, L is the original length of the wire, A is the cross-sectional area, and E is the Young's modulus.

Plugging in the given values, we have F = 50 kg * 9.8 m/s^2 = 490 N, A = π * (1.0 mm / 2)^2 = 0.785 mm^2 = 7.85 * 10^-7 m^2, L = 11.2 m, and E = 20 * 10^10 N/m^2. Substituting these values into the formula, we get ΔL = (490 N * 11.2 m) / (7.85 * 10^-7 m^2 * 20 * 10^10 N/m^2) = 0.035 m = 3.5 cm.

Therefore, the wire will stretch by 3.5 cm.

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• 37.

### Suppose that an 80-kg person walking on crutches supports all his weight on the two crutch tips, each of which is circular with a diameter of 4.0 cm. What pressure is exerted on the floor?

• A.

78 kPa

• B.

156 kPa

• C.

312 kPa

• D.

624 kPa

C. 312 kPa
Explanation
The pressure exerted on the floor can be calculated by dividing the force exerted by the person's weight by the area of the crutch tips in contact with the floor. The force exerted by the person's weight can be calculated as mass times acceleration due to gravity, which is 80 kg multiplied by 9.8 m/s^2. The area of each crutch tip can be calculated using the formula for the area of a circle, which is pi times the radius squared. The radius is half of the diameter, so it is 2.0 cm. Therefore, the area of each crutch tip is pi times 2.0 cm squared. By dividing the force by the area, we can find that the pressure exerted on the floor is approximately 312 kPa.

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• 38.

### A bridge piling has an area of 1.250 m2. It supports 1875 N. Find the stress on the column.

• A.

1875 N

• B.

1875 N/m^2

• C.

1500 N/m^2

• D.

2344 N/m^2

C. 1500 N/m^2
Explanation
The stress on the column can be calculated by dividing the force (1875 N) by the area (1.250 m^2). The result is 1500 N/m^2.

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• 39.

### An aluminum wire 2.0 m in length and 2.0 mm in diameter supports a 10.0-kg mass. What is the stress in the wire? (The Young's modulus for aluminum is 7.0 * 10^10 N/m^2)

• A.

3.1 * 10^7 N/m^2

• B.

6.2 * 10^7 N/m^2

• C.

9.3 * 10^7 N/m^2

• D.

1.2 * 10^8 N/m^2

A. 3.1 * 10^7 N/m^2
Explanation
The stress in the wire can be calculated using the formula stress = force/area. The force acting on the wire is the weight of the mass, which is given by the formula weight = mass * gravity. The area of the wire can be calculated using the formula area = π * (radius)^2. By substituting the given values into the formulas and evaluating the expression, we can find that the stress in the wire is 3.1 * 10^7 N/m^2.

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• 40.

### An aluminum wire 2.0 m in length and 2.0 mm in diameter supports a 10.0-kg mass. What is the elongation of the wire? (The Young's modulus for an aluminum is 7.0 * 10^10 N/m^2)

• A.

0.11 mm

• B.

0.22 mm

• C.

0.33 mm

• D.

0.89 mm

D. 0.89 mm
Explanation
The elongation of a wire can be calculated using the formula ΔL = (F * L) / (A * Y), where ΔL is the elongation, F is the force applied, L is the length of the wire, A is the cross-sectional area of the wire, and Y is the Young's modulus. In this case, the force applied is the weight of the mass, which is F = m * g = 10.0 kg * 9.8 m/s^2 = 98 N. The cross-sectional area of the wire can be calculated using the formula A = π * r^2, where r is the radius of the wire. Given that the diameter of the wire is 2.0 mm, the radius is 1.0 mm = 0.001 m. Plugging these values into the formula, we get ΔL = (98 N * 2.0 m) / ((π * (0.001 m)^2) * (7.0 * 10^10 N/m^2)) = 0.89 mm. Therefore, the correct answer is 0.89 mm.

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• 41.

### A shear force of 400 N is applied to one face of an aluminum cube with sides of 30 cm. What is the resulting relative displacement? (The shear modulus for aluminum is 2.5 * 10^10 N/m^2)

• A.

1.9 * 10^(-8) m

• B.

4.4 * 10^(-8) m

• C.

5.3 * 10^(-8) m

• D.

8.2 * 10^(-8) m

C. 5.3 * 10^(-8) m
Explanation
When a shear force is applied to a material, it causes the material to deform or experience relative displacement. The magnitude of this displacement can be determined using the formula:

Relative Displacement = (Shear Force * Side Length) / (Shear Modulus)

In this case, the shear force is given as 400 N, the side length of the aluminum cube is 30 cm (or 0.3 m), and the shear modulus for aluminum is given as 2.5 * 10^10 N/m^2. Plugging these values into the formula, we get:

Relative Displacement = (400 N * 0.3 m) / (2.5 * 10^10 N/m^2) = 5.3 * 10^(-8) m

Therefore, the resulting relative displacement is 5.3 * 10^(-8) m.

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• 42.

### At a depth of about 1030 m in the sea the pressure has increased by 100 atmospheres (to about 10^7 N/m^2). By how much has 1.0 m^3 of water been compressed by this pressure? The bulk modulus of water is 2.3 * 10^9 N/m^2.

• A.

2.3 * 10^(-3) m^3

• B.

3.3 * 10^(-3) m^3

• C.

4.3 * 10^(-3) m^3

• D.

5.3 * 10^(-3) m^3

C. 4.3 * 10^(-3) m^3
Explanation
The question asks for the amount of water that has been compressed by a pressure increase of 100 atmospheres at a depth of 1030 m in the sea. To solve this, we can use the formula for bulk modulus: ΔV/V = -ΔP/B, where ΔV is the change in volume, V is the initial volume, ΔP is the change in pressure, and B is the bulk modulus. Plugging in the given values, we get: ΔV/1.0 = -(100 * 10^5)/(2.3 * 10^9). Solving for ΔV, we find that ΔV is approximately 4.3 * 10^(-3) m^3.

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• 43.

### A 55-cm brass rod has a diameter of 30 cm. The compressive strength of steel is 250 * 10^6 N/m^2. What is the compression force that would break the rod?

• A.

1.8 * 10^7 N

• B.

1.4 * 10^8 N

• C.

2.4 * 10^8 N

• D.

4.5 * 10^8 N

A. 1.8 * 10^7 N
Explanation
The compression force that would break the rod can be calculated using the formula F = A * σ, where F is the force, A is the cross-sectional area, and σ is the compressive strength of the material. The cross-sectional area can be calculated using the formula A = π * r^2, where r is the radius of the rod. Given that the diameter of the rod is 30 cm, the radius would be 15 cm or 0.15 m. Plugging in the values, we get A = 3.14 * (0.15)^2 = 0.07065 m^2. The compressive strength of steel is given as 250 * 10^6 N/m^2. Plugging in the values, we get F = 0.07065 * 250 * 10^6 = 1.76625 * 10^7 N, which is approximately 1.8 * 10^7 N.

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• Mar 21, 2023
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• Sep 21, 2012
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