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Please explain part B of the question.

Please explain part B of the question.-example-1
User Alphadogg
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The resistance (R2) of the wire with a diameter of 1.0 mm is calculated to be 21Ω based on the given proportional relationship.

The calculation is based on the relationship between resistance (R), current (I), and the cross-sectional area (A) of a wire.

1. Cross-sectional Area (A):

The cross-sectional area (A) of a wire with radius (r) is A = πr².

Using the diameter (d)

A = (π/4)d²

2. Comparison of Cross-sectional Areas (A1/A2):

The wire has two different diameters: d1= 0.5mm and d2 = 1mm

The ratio of their cross-sectional areas is A1 / A2 = d1² / d2² = 0.5²mm / 1²mm = 0.25

3. Resistance Proportionality (R ∝ 1 / A):

Resistance (R) of a wire is inversely proportional to its cross-sectional area.

R ∝ 1 / A

4. Equating Resistance Ratios (R2 / R1 = A1 / A2):

If R ∝ 1 / A

The ratio of resistances R2 / R1 = A1 / A2 is equal to the inverse ratio of cross-sectional areas A 1 / A2.

5. Calculating (R2):

Rearrang the equation, (R2 = R1 / (A1 / A2).

Substituting the values,

R2 = 84Ω × 0.25 = 21 Ω

Therefore, the resistance (R2) of the wire with a diameter of 1.0 mm is calculated to be 21Ω based on the given proportional relationship.

User Simas Joneliunas
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