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3 years ago

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My Notes The linear density rho in a rod 5 m long is 11/ x + 4 kg/m, where x is measured in meters from one end of the rod. Find

the average density rhoave of the rod.

2 answers:

Lostsunrise [7]3 years ago

5 0

<h2>Corrected Question:</h2>

The linear density ρ in a rod 5 m long is (11x+4) kg/m, where x is measured in meters from one end of the rod. Find the average density ρ(ave)

<h2>Answer:</h2>

157.5kg/m

<h2>Explanation:</h2>

The linear density, ρ, of a rod is given by;

ρ = $\frac{m}{l}$

Where;

m = mass of the rod

l = length of the rod

The average density ρ(ave) is then, the average of the linear density taken over the entire length of the rod. i.e

ρ(ave) = $\int\limits^a_b {p} \, dx$ -----------------(ii)

Where;

a and b are the limits of the length of the rod.

b = 0

a = 5m

p = (11x + 4) kg/m [as given in the question]

Substitute these values into equation (ii) as follows;

ρ(ave) = $\int\limits^5_0 {(11x + 4)} \, dx$

ρ(ave) = [ $\frac{11x^2}{2}$ + 4x ]₀⁵

Substitute the integral limits into the equation;

ρ(ave) = [ $\frac{11(5)^2}{2}$ + 4(5) ] - [ $\frac{11(0)^2}{2}$ + 4(0)]

ρ(ave) = [ $\frac{11(5)^2}{2}$ + 4(5)]

ρ(ave) = [ 137.5 + 20)]

ρ(ave) = 157.5

Therefore, the average density of the rod is 157.5kg/m

dedylja [7]3 years ago

3 0

Answer:

ρ(ave) = 29.1 kg/m

Explanation:

The average value of a function over an interval = definite integral computed with the boundaries of the interval divided by the extent of the integral

ρ(ave) = [∫⁵₀ (11x + 4) dx]/(5-0)

ρ(ave) = [(11x²/2) + 4x]⁵₀ ÷ (5)

ρ(ave) = [(11(5²)/2) + (4×2)]/5

ρ(ave) = (137.5 + 8)/5

ρ(ave) = 29.1 kg/m

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1.

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

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Part b)

Now at higher temperature we have

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now we know that

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$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

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2.

Answer:

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Explanation:

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As we know that current density is defined as

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$v_d = 1.4 \times 10^{-4} m/s$

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3 years ago