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'
......................(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]](https://tex.z-dn.net/?f=%3D2%5Cint%5Climits%5E9_4%20%7B%5Cfrac%7B1%7D%7Bu%7D%20%7D%20%5C%2C%20du%5C%5C%5C%5C%20%3D%202%5B%28log~u%29%5D_4%5E%7B9%7D%5C%5C%5C%5C%3D2%5B%28log~9-log~4%29%5D%5C%5C%5C%5C%3D2%5Ctimes%5B0.352%5D)
= 0.704
Thus, we can conclude that the average density of the rod is 0.704 kg/m.
Learn more about the average density here:
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An equivalent ratio would be 4/10, and the number of cattle would be 136,000.Explanation:The question asks for an equivalent ratio. This means a fraction that is equivalent to the fraction between 700-799, which is 2/5.To create an equivalent fraction, we multiply the numerator and denominator by the same number. Multiplying both by 2, we get (2*2)/(5*2) = 4/10.Tenths are useful when dealing with thousands, because taking a tenth involves getting rid of a zero. This means that 1/10 would be 34000; 4/10 would be 4 times that: 4(34000) = 136000.
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The answer would be the top one, since 2.25 inches of water are lost over 5 days.
Ah sorry I can’t , so sorry
Answer:
2 7/12
Step-by-step explanation:
