Question

Before heading out on her big date, Ling stands in front of the bathroom mirror brushing her 0.25-m-long hair with a force of 2.0 N. If the crosssectional area of a piece of hair is 1.0 107 m2, by how much does the hair stretch when it is brushed? (Yhair 2.0 109 N)

Answer 1

Extension in the length of the hair is given as

`\Delta L = 2.5 * 10^(-3)`

Step-by-step explanation:

As we know by the formula of elasticity that the ratio of stress and strain is known as modulus of elasticity

So we will have

`Y = (Stress)/(Strain)`

Now we have

`F = 2 N`

`Area = 1.0 * 10^(-7) m^2`

`L = 0.25 m`

So we have

`2 * 10^9 = (2/(1 * 10^(-7)))/(\Delta L/0.25)`

so we have

`2* 10^9 = (5 * 10^6)/(\Delta L)`

So we have

`\Delta L = 2.5 * 10^(-3)`