Question

The linear density in a rod 5 m long is 10 x + 4 kg/m, where x is measured in meters from one end of the rod. Find the average density ave (in kg/m) of the rod.

Answer 1

The average density of the rod is  0.704 kg/m.

For given question,

We have been given the linear density in a rod 5 m long is 10 / x + 4 kg/m, where x is measured in meters from one end of the rod.

We need to find the

The length of rod is, L = 5 m.

The linear density of rod is, ρ = 10/( x + 4) kg/m

To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 5,  

The expression for the average density is given as,

⇒ ρ'

$=\int\limits^5_0 {\rho} \, dx\\\\=\int\limits^5_0 {\frac{m}{L} } \, dx\\\\=\int\limits^5_0 {\frac{10}{5(x+4)} }\, dx\\\\=\int\limits^5_0 {\frac{2}{x+4} }\, dx$                        ......................(1)

Using u = x + 4  

du = dx

u₁ = x₁ + 4

u₁ = 0 + 4

u₁ = 4

and

u₂ = x₂ + 4

u₂ = 5 + 4

u₂ = 9

By entering the values above into (1), we have:

⇒ ρ'

$=2\int\limits^9_4 {\frac{1}{u} } \, du\\\\ = 2[(log~u)]_4^{9}\\\\=2[(log~9-log~4)]\\\\=2\times[0.352]$

= 0.704

Thus, we can conclude that the average density of the rod is  0.704 kg/m.

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