The formula for the elongation of a wire is as follows: ΔL = (F L) / (A Y) Where: ΔL is the elongation, F is the force, L is the original length of the wire, A is the cross-sectional area, Y is Young's Modulus. For the first wire, the elongation is given by: ΔL1 = (F L) / (A1 Y) For the second wire,Read more
The formula for the elongation of a wire is as follows:
ΔL = (F L) / (A Y)
Where:
ΔL is the elongation,
F is the force,
L is the original length of the wire,
A is the cross-sectional area,
Y is Young’s Modulus.
For the first wire, the elongation is given by:
ΔL1 = (F L) / (A1 Y)
For the second wire, the elongation is given by:
ΔL2 = (F L) / (A2 Y)
Since the area of the second wire is three times that of the first wire, A2 = 3 A1, we can compare the elongations:
ΔL2 / ΔL1 = A1 / A2 = 1 / 3
Thus, the elongation of the second wire is one-third that of the first wire:
We can use the formula to find the bulk modulus as follows: Bulk Modulus (B) = (Pressure × ΔV) / V Where: - Pressure P = 100 atm - Change in volume ΔV = 0.3 c.c. - Initial volume V of the sphere is given by the formula for the volume of a sphere: V = (4/3) π r³ For r = 3 cm: V = (4/3) π (3)³ = 36 πRead more
We can use the formula to find the bulk modulus as follows:
Bulk Modulus (B) = (Pressure × ΔV) / V
Where:
– Pressure P = 100 atm
– Change in volume ΔV = 0.3 c.c.
– Initial volume V of the sphere is given by the formula for the volume of a sphere:
V = (4/3) π r³
For r = 3 cm:
V = (4/3) π (3)³ = 36 π c.c.
Now, let’s calculate the bulk modulus:
B = (100 atm × 0.3 c.c.) / (36 π c.c.)
Simplifying the above expression gives us the answer
We can then use the formula for breaking stress to determine how long the wire will be under its own weight: Breaking Stress = Force/ Area Here, the force due to weight is given by: Force = Weight = Density × Volume × g Here, Density = 3 × 10³ kg/m³ g = 9.8 m/s² - Volume = A × L, where L is the lengRead more
We can then use the formula for breaking stress to determine how long the wire will be under its own weight:
Breaking Stress = Force/ Area
Here, the force due to weight is given by:
Force = Weight = Density × Volume × g
Here,
Density = 3 × 10³ kg/m³
g = 9.8 m/s²
– Volume = A × L, where L is the length of the wire.
The breaking stress is a measure of the stress at which the wire will break so equate the force due to weight to the breaking stress, we get;
Breaking stress = Density × L × g
Rearranging to solve for L:
L = Breaking stress / (Density × g)
Substituting known values
L = (10⁶ N/m²) / (3 × 10³ kg/m³ × 9.8 m/s²)
L = 10⁶ / (3 × 10³ × 9.8)
L = 10⁶ / 2.94 × 10⁴ = 33.3 m
Therefore,
33.3 m
The expression of stress, strain, and Young's modulus is as follows: Stress = Y × Strain where, - Y = 2.0 × 10¹¹ N/m² - Strain = 0.16% = 0.0016 Stress can also be defined as follows: Stress = Force / Area Cross-sectional area of the rod A = πr² = π × (10 × 10⁻³)² = π × 10⁻⁴ m² Put it into the equatiRead more
The expression of stress, strain, and Young’s modulus is as follows:
Stress = Y × Strain
The bulk strain can be described as the fractional alteration in the volume of a body. In this case of uniform compression for the cube, this is computed with: Bulk Strain = 3 × Linear Strain Since the length of the cube has been compressed by 2%, then this gives a value of: Linear Strain = ΔL / L =Read more
The bulk strain can be described as the fractional alteration in the volume of a body. In this case of uniform compression for the cube, this is computed with:
Bulk Strain = 3 × Linear Strain
Since the length of the cube has been compressed by 2%, then this gives a value of:
Linear Strain = ΔL / L = 0.02
This therefore, implies that,
Bulk Strain = 3 × 0.02 = 0.06
A wire whose cross-sectional area is 2 mm² is stretched by 0.1 mm by a certain load, and if a similar wire of triple the area of cross-section is stretched by the same load, then the elongation of the second wire would be
The formula for the elongation of a wire is as follows: ΔL = (F L) / (A Y) Where: ΔL is the elongation, F is the force, L is the original length of the wire, A is the cross-sectional area, Y is Young's Modulus. For the first wire, the elongation is given by: ΔL1 = (F L) / (A1 Y) For the second wire,Read more
The formula for the elongation of a wire is as follows:
ΔL = (F L) / (A Y)
Where:
ΔL is the elongation,
F is the force,
L is the original length of the wire,
A is the cross-sectional area,
Y is Young’s Modulus.
For the first wire, the elongation is given by:
ΔL1 = (F L) / (A1 Y)
For the second wire, the elongation is given by:
ΔL2 = (F L) / (A2 Y)
Since the area of the second wire is three times that of the first wire, A2 = 3 A1, we can compare the elongations:
ΔL2 / ΔL1 = A1 / A2 = 1 / 3
Thus, the elongation of the second wire is one-third that of the first wire:
ΔL2 = ΔL1 / 3 = 0.1 mm / 3 = 0.033 mm
The correct answer is:
0.033 mm
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A sphere of radius 3 cm is subjected to a pressure of 100 atm. Its volume decreases by 0.3 c.c. What will be its bulk modulus?
We can use the formula to find the bulk modulus as follows: Bulk Modulus (B) = (Pressure × ΔV) / V Where: - Pressure P = 100 atm - Change in volume ΔV = 0.3 c.c. - Initial volume V of the sphere is given by the formula for the volume of a sphere: V = (4/3) π r³ For r = 3 cm: V = (4/3) π (3)³ = 36 πRead more
We can use the formula to find the bulk modulus as follows:
Bulk Modulus (B) = (Pressure × ΔV) / V
Where:
– Pressure P = 100 atm
– Change in volume ΔV = 0.3 c.c.
– Initial volume V of the sphere is given by the formula for the volume of a sphere:
V = (4/3) π r³
For r = 3 cm:
V = (4/3) π (3)³ = 36 π c.c.
Now, let’s calculate the bulk modulus:
B = (100 atm × 0.3 c.c.) / (36 π c.c.)
Simplifying the above expression gives us the answer
B = 4 π × 10⁵ atm
Thus the correct answer is
4π x 10⁵ atm
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A substance breaks down by a stress of 10⁶ N/m². If the density of the material of the wire is 3 x 10³ kg/m³, then the length of the wire of the substance which will break under its own weight when suspended vertically will be
We can then use the formula for breaking stress to determine how long the wire will be under its own weight: Breaking Stress = Force/ Area Here, the force due to weight is given by: Force = Weight = Density × Volume × g Here, Density = 3 × 10³ kg/m³ g = 9.8 m/s² - Volume = A × L, where L is the lengRead more
We can then use the formula for breaking stress to determine how long the wire will be under its own weight:
Breaking Stress = Force/ Area
Here, the force due to weight is given by:
Force = Weight = Density × Volume × g
Here,
Density = 3 × 10³ kg/m³
g = 9.8 m/s²
– Volume = A × L, where L is the length of the wire.
The breaking stress is a measure of the stress at which the wire will break so equate the force due to weight to the breaking stress, we get;
Breaking stress = Density × L × g
Rearranging to solve for L:
L = Breaking stress / (Density × g)
Substituting known values
L = (10⁶ N/m²) / (3 × 10³ kg/m³ × 9.8 m/s²)
L = 10⁶ / (3 × 10³ × 9.8)
L = 10⁶ / 2.94 × 10⁴ = 33.3 m
Therefore,
33.3 m
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A steel rod has a radius of 10 mm and a length of 1.0 m. A force stretches it along its length and produce a strain of 0.16%. Young’s modulus of the steel is 2.0 x 10¹¹ Nm⁻². What is the magnitude of the force stretching the rod?
The expression of stress, strain, and Young's modulus is as follows: Stress = Y × Strain where, - Y = 2.0 × 10¹¹ N/m² - Strain = 0.16% = 0.0016 Stress can also be defined as follows: Stress = Force / Area Cross-sectional area of the rod A = πr² = π × (10 × 10⁻³)² = π × 10⁻⁴ m² Put it into the equatiRead more
The expression of stress, strain, and Young’s modulus is as follows:
Stress = Y × Strain
where,
– Y = 2.0 × 10¹¹ N/m²
– Strain = 0.16% = 0.0016
Stress can also be defined as follows:
Stress = Force / Area
Cross-sectional area of the rod
A = πr² = π × (10 × 10⁻³)² = π × 10⁻⁴ m²
Put it into the equation for stress:
Y × Strain = Force / Area
Force =
Force = Y × Strain × A
Put the values:
Force = (2.0 × 10¹¹) × (0.0016) × (π × 10⁻⁴)
Force = 100.5 × 10³ N ≈ 100 kN
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A cube is subjected to a uniform volume compression. If the side of the cube decreases by 2% the bulk strain is
The bulk strain can be described as the fractional alteration in the volume of a body. In this case of uniform compression for the cube, this is computed with: Bulk Strain = 3 × Linear Strain Since the length of the cube has been compressed by 2%, then this gives a value of: Linear Strain = ΔL / L =Read more
The bulk strain can be described as the fractional alteration in the volume of a body. In this case of uniform compression for the cube, this is computed with:
Bulk Strain = 3 × Linear Strain
Since the length of the cube has been compressed by 2%, then this gives a value of:
Linear Strain = ΔL / L = 0.02
This therefore, implies that,
Bulk Strain = 3 × 0.02 = 0.06
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