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Sagot :
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'
[tex]=\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[/tex] ......................(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'
[tex]=2\int\limits^9_4 {\frac{1}{u} } \, du\\\\ = 2[(log~u)]_4^{9}\\\\=2[(log~9-log~4)]\\\\=2\times[0.352][/tex]
= 0.704
Thus, we can conclude that the average density of the rod is 0.704 kg/m.
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