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## Homework Statement

A 5.0-kg ball hangs from a steel wire 1.00 mm in diameter and 5.00 m long. What would be the speed of a wave in the steel wire?

Hint: Density of steel = þ_steel = 7.8 * 10^3 kg/m^3

Assume that the string is a cylinder to calculate the volume.

m = 5.0 kg

l (length) = 5.00 m

d = 1.00 mm = 0.001 m

r = 0.0005 m

þ_steel = 7.8 * 10^3 kg/m^3

g = 9.8 m/s^2

## Homework Equations

V(volume)_cylinder = πlr^2

m = þV

F = mg

µ = m/l

v(velocity) = √(F/µ)

## The Attempt at a Solution

V_cylinder = πlr^2 = π (5.00 m) (0.0005m)^2 ≈ 4.0 * 10^-6 m^3

m = þV = (7.8 * 10^3 kg/m^3) (4.0 * 10^-6 m^3) ≈ 0.0312 kg

F = mg = (0.0312 kg) (9.8 m/s^2) ≈ 0.306 N

µ = m/l = (5.0 kg)/(5.00 m) = 1 kg/m

v = √(F/µ) = √((0.306 N)/(1 kg/m)) = √(0.306) m/s ≈ 0.553 m/s

So, the speed of the wave in the steel wire would be approximately 0.553 m/s.

^^^ Does all of that look correct?