The resistance of a wire is given by the formula R = ρL/A, where ρ is the resistivity, L is the length, and A is the cross-sectional area.

When the wire is stretched, its length increases but its volume remains the same. The volume of a cylinder (which is what the wire can be modeled as) is given by V = AL. Since the volume remains constant, if the length increases, the cross-sectional area must decrease.

The diameter of the wire decreases to 94.0% of the original, so the new diameter is 0.94D. The cross-sectional area of the wire is proportional to the square of the diameter, so the new area is (0.94D)^2 = 0.8836A.

Substituting this into the resistance formula gives R' = ρL/0.8836A. Since ρL/A = R, we can write this as R' = R/0.8836.

Therefore, the resistance of the stretched wire is approximately 1.13 times the original resistance, so the answer is B) 1.13R.

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The resistance of a wire is given by the formula R = ρL/A, where ρ is the resistivity, L is the length, and A is the cross-sectional area.

When the wire is stretched, its length increases but its volume remains the same. The volume of a cylinder (which is what the wire can be modeled as) is given by V = AL. Since the volume remains constant, if the length increases, the cross-sectional area must decrease.

The diameter of the wire decreases to 94.0% of the original, so the new diameter is 0.94D. The cross-sectional area of the wire is proportional to the square of the diameter, so the new area is (0.94D)^2 = 0.8836A.

Substituting this into the resistance formula gives R' = ρL/0.8836A. Since ρL/A = R, we can write this as R' = R/0.8836.

Therefore, the resistance of the stretched wire is approximately 1.13 times the original resistance, so the answer is B) 1.13R.

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The resistance of a wire is given by the formula R = ρL/A, where ρ is the resistivity, L is the length, and A is the cross-sectional area.

When the wire is stretched, its length increases but its volume remains the same. The volume of a cylinder (which is what the wire can be modeled as) is given by V = AL. Since the volume remains constant, if the length increases, the cross-sectional area must decrease.

The diameter of the wire decreases to 94.0% of the original, so the new diameter is 0.94D. The cross-sectional area of the wire is proportional to the square of the diameter, so the new area is (0.94D)^2 = 0.8836A.

Substituting this into the resistance formula gives R' = ρL/0.8836A. Since ρL/A = R, we can write this as R' = R/0.8836.

Therefore, the resistance of the stretched wire is approximately 1.13 times the original resistance, so the answer is B) 1.13R.

A metal wire has resistance R.The wire is stretched so that its diameter decreases to 94.0% of the original diameter.The volume of the wire is unchanged.What is the resistance of the stretched wire?A 1.06R B 1.13R C 1.20R D 1.28R

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