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A vertical steel wire and a parallel brass wire, each 4 m long and 4.0×10^-7 m^2 cross section, hang from a ceiling and are 0.30 m apart. The lower ends of the wires are attached to a light horizontal bar. Find the mass of the load which must be hung from the bar to cause each wire to extend 1.0×10^-3 m. At what distance from the brass wire must the mass be suspended? Given Young's modulus for steel is 2×10^11 N/m^2 and Young's modulus for brass is 1×10^11 N/m^2.?
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A vertical steel wire and a parallel brass wire, each 4 m long and 4.0...
Given Information:
- Length of the vertical steel wire and parallel brass wire: 4 m
- Cross-sectional area of both wires: 4.0×10^-7 m^2
- Distance between the wires: 0.30 m
- Extension of each wire: 1.0×10^-3 m
- Young's modulus for steel: 2×10^11 N/m^2
- Young's modulus for brass: 1×10^11 N/m^2
Step 1: Calculate the tension in each wire:
The tension in each wire can be calculated using Hooke's Law, which states that the extension of a wire is directly proportional to the applied force.
Tension in steel wire (Ts) = (Young's modulus for steel) × (cross-sectional area of steel wire) × (extension of steel wire) / (length of steel wire)
= (2×10^11 N/m^2) × (4.0×10^-7 m^2) × (1.0×10^-3 m) / (4 m)
Tension in brass wire (Tb) = (Young's modulus for brass) × (cross-sectional area of brass wire) × (extension of brass wire) / (length of brass wire)
= (1×10^11 N/m^2) × (4.0×10^-7 m^2) × (1.0×10^-3 m) / (4 m)
Step 2: Calculate the total load required:
The total load required to cause the wires to extend can be calculated by adding the tension in each wire.
Total load (F) = Ts + Tb
Step 3: Calculate the distance from the brass wire:
The distance from the brass wire can be calculated using the principle of moments. Since the load is symmetrically distributed, the distance from the brass wire to the load is the same as the distance from the load to the steel wire.
Distance from the brass wire (d) = (distance between the wires) / 2 = 0.30 m / 2
Step 4: Calculate the mass of the load:
The mass of the load can be calculated using the equation F = mg, where g is the acceleration due to gravity.
Mass of the load (m) = F / g
Step 5: Calculate the final values:
Substitute the values into the equations and calculate the final values.
- Calculate the tension in each wire using the given values.
- Calculate the total load by adding the tensions.
- Calculate the distance from the brass wire using the given distance.
- Calculate the mass of the load using the total load and acceleration due to gravity.
- Round off the final values to the appropriate significant figures and units.
Final Answer:
The mass of the load that must be hung from the bar to cause each wire to extend is ____ kg. The load must be suspended at a distance of ____ m from the brass wire.
- Length of the vertical steel wire and parallel brass wire: 4 m
- Cross-sectional area of both wires: 4.0×10^-7 m^2
- Distance between the wires: 0.30 m
- Extension of each wire: 1.0×10^-3 m
- Young's modulus for steel: 2×10^11 N/m^2
- Young's modulus for brass: 1×10^11 N/m^2
Step 1: Calculate the tension in each wire:
The tension in each wire can be calculated using Hooke's Law, which states that the extension of a wire is directly proportional to the applied force.
Tension in steel wire (Ts) = (Young's modulus for steel) × (cross-sectional area of steel wire) × (extension of steel wire) / (length of steel wire)
= (2×10^11 N/m^2) × (4.0×10^-7 m^2) × (1.0×10^-3 m) / (4 m)
Tension in brass wire (Tb) = (Young's modulus for brass) × (cross-sectional area of brass wire) × (extension of brass wire) / (length of brass wire)
= (1×10^11 N/m^2) × (4.0×10^-7 m^2) × (1.0×10^-3 m) / (4 m)
Step 2: Calculate the total load required:
The total load required to cause the wires to extend can be calculated by adding the tension in each wire.
Total load (F) = Ts + Tb
Step 3: Calculate the distance from the brass wire:
The distance from the brass wire can be calculated using the principle of moments. Since the load is symmetrically distributed, the distance from the brass wire to the load is the same as the distance from the load to the steel wire.
Distance from the brass wire (d) = (distance between the wires) / 2 = 0.30 m / 2
Step 4: Calculate the mass of the load:
The mass of the load can be calculated using the equation F = mg, where g is the acceleration due to gravity.
Mass of the load (m) = F / g
Step 5: Calculate the final values:
Substitute the values into the equations and calculate the final values.
- Calculate the tension in each wire using the given values.
- Calculate the total load by adding the tensions.
- Calculate the distance from the brass wire using the given distance.
- Calculate the mass of the load using the total load and acceleration due to gravity.
- Round off the final values to the appropriate significant figures and units.
Final Answer:
The mass of the load that must be hung from the bar to cause each wire to extend is ____ kg. The load must be suspended at a distance of ____ m from the brass wire.
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A vertical steel wire and a parallel brass wire, each 4 m long and 4.0×10^-7 m^2 cross section, hang from a ceiling and are 0.30 m apart. The lower ends of the wires are attached to a light horizontal bar. Find the mass of the load which must be hung from the bar to cause each wire to extend 1.0×10^-3 m. At what distance from the brass wire must the mass be suspended? Given Young's modulus for steel is 2×10^11 N/m^2 and Young's modulus for brass is 1×10^11 N/m^2.?
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A vertical steel wire and a parallel brass wire, each 4 m long and 4.0×10^-7 m^2 cross section, hang from a ceiling and are 0.30 m apart. The lower ends of the wires are attached to a light horizontal bar. Find the mass of the load which must be hung from the bar to cause each wire to extend 1.0×10^-3 m. At what distance from the brass wire must the mass be suspended? Given Young's modulus for steel is 2×10^11 N/m^2 and Young's modulus for brass is 1×10^11 N/m^2.? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A vertical steel wire and a parallel brass wire, each 4 m long and 4.0×10^-7 m^2 cross section, hang from a ceiling and are 0.30 m apart. The lower ends of the wires are attached to a light horizontal bar. Find the mass of the load which must be hung from the bar to cause each wire to extend 1.0×10^-3 m. At what distance from the brass wire must the mass be suspended? Given Young's modulus for steel is 2×10^11 N/m^2 and Young's modulus for brass is 1×10^11 N/m^2.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A vertical steel wire and a parallel brass wire, each 4 m long and 4.0×10^-7 m^2 cross section, hang from a ceiling and are 0.30 m apart. The lower ends of the wires are attached to a light horizontal bar. Find the mass of the load which must be hung from the bar to cause each wire to extend 1.0×10^-3 m. At what distance from the brass wire must the mass be suspended? Given Young's modulus for steel is 2×10^11 N/m^2 and Young's modulus for brass is 1×10^11 N/m^2.?.
A vertical steel wire and a parallel brass wire, each 4 m long and 4.0×10^-7 m^2 cross section, hang from a ceiling and are 0.30 m apart. The lower ends of the wires are attached to a light horizontal bar. Find the mass of the load which must be hung from the bar to cause each wire to extend 1.0×10^-3 m. At what distance from the brass wire must the mass be suspended? Given Young's modulus for steel is 2×10^11 N/m^2 and Young's modulus for brass is 1×10^11 N/m^2.? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A vertical steel wire and a parallel brass wire, each 4 m long and 4.0×10^-7 m^2 cross section, hang from a ceiling and are 0.30 m apart. The lower ends of the wires are attached to a light horizontal bar. Find the mass of the load which must be hung from the bar to cause each wire to extend 1.0×10^-3 m. At what distance from the brass wire must the mass be suspended? Given Young's modulus for steel is 2×10^11 N/m^2 and Young's modulus for brass is 1×10^11 N/m^2.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A vertical steel wire and a parallel brass wire, each 4 m long and 4.0×10^-7 m^2 cross section, hang from a ceiling and are 0.30 m apart. The lower ends of the wires are attached to a light horizontal bar. Find the mass of the load which must be hung from the bar to cause each wire to extend 1.0×10^-3 m. At what distance from the brass wire must the mass be suspended? Given Young's modulus for steel is 2×10^11 N/m^2 and Young's modulus for brass is 1×10^11 N/m^2.?.
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A vertical steel wire and a parallel brass wire, each 4 m long and 4.0×10^-7 m^2 cross section, hang from a ceiling and are 0.30 m apart. The lower ends of the wires are attached to a light horizontal bar. Find the mass of the load which must be hung from the bar to cause each wire to extend 1.0×10^-3 m. At what distance from the brass wire must the mass be suspended? Given Young's modulus for steel is 2×10^11 N/m^2 and Young's modulus for brass is 1×10^11 N/m^2.?, a detailed solution for A vertical steel wire and a parallel brass wire, each 4 m long and 4.0×10^-7 m^2 cross section, hang from a ceiling and are 0.30 m apart. The lower ends of the wires are attached to a light horizontal bar. Find the mass of the load which must be hung from the bar to cause each wire to extend 1.0×10^-3 m. At what distance from the brass wire must the mass be suspended? Given Young's modulus for steel is 2×10^11 N/m^2 and Young's modulus for brass is 1×10^11 N/m^2.? has been provided alongside types of A vertical steel wire and a parallel brass wire, each 4 m long and 4.0×10^-7 m^2 cross section, hang from a ceiling and are 0.30 m apart. The lower ends of the wires are attached to a light horizontal bar. Find the mass of the load which must be hung from the bar to cause each wire to extend 1.0×10^-3 m. At what distance from the brass wire must the mass be suspended? Given Young's modulus for steel is 2×10^11 N/m^2 and Young's modulus for brass is 1×10^11 N/m^2.? theory, EduRev gives you an
ample number of questions to practice A vertical steel wire and a parallel brass wire, each 4 m long and 4.0×10^-7 m^2 cross section, hang from a ceiling and are 0.30 m apart. The lower ends of the wires are attached to a light horizontal bar. Find the mass of the load which must be hung from the bar to cause each wire to extend 1.0×10^-3 m. At what distance from the brass wire must the mass be suspended? Given Young's modulus for steel is 2×10^11 N/m^2 and Young's modulus for brass is 1×10^11 N/m^2.? tests, examples and also practice Class 11 tests.
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