contestada

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)

Respuesta :

Answer:

Extension in the length of the hair is given as

[tex]\Delta L = 2.5 \times 10^{-3}[/tex]

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

[tex]Y = \frac{Stress}{Strain}[/tex]

Now we have

[tex]F = 2 N[/tex]

[tex]Area = 1.0 \times 10^{-7} m^2[/tex]

[tex]L = 0.25 m[/tex]

So we have

[tex]2 \times 10^9 = \frac{2/(1 \times 10^{-7})}{\Delta L/0.25}[/tex]

so we have

[tex]2\times 10^9 = \frac{5 \times 10^6}{\Delta L}[/tex]

So we have

[tex]\Delta L = 2.5 \times 10^{-3}[/tex]

Otras preguntas