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brilliants [131]
2 years ago
15

Calculate the change in heat of the aluminum; show all calculations. Calculate the change in heat of the water; show all calcula

tions. Are the two values the same? Why or why not? See the attached picture for the numbers.
I got -3443.14 J for the aluminum and 3443.595 for the water

Physics
1 answer:
tatyana61 [14]2 years ago
6 0
Calculate the change in heat of the aluminum; show all calculations. Calculate the change in heat of the water; show all calculations. Are the two values the same? Why or why not? See the attached picture for the numbers.

I got -3443.14 J for the aluminum and 3443.595 for the water
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Answer:

\mu=0.98\ Pa.s

Explanation:

Given:

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  • thickness of fluid layer, dy=0.004\ m
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  • velocity of plate, du=1\ m.s^{-1}

<u>Using Newton's law of viscosity:</u>

\tau=\mu.\frac{du}{dy} ..........................................(1)

where:

\tau= shear force on the surface on the fluid

\mu= coefficient of (dynamic) viscosity

Now, shear force:

\tau=\frac{shear\ force}{area}

\tau=\frac{9.8}{0.2\times0.2} \ Pa

Putting respective values in eq.(1)

\frac{9.8}{0.2\times0.2}=\mu\times\frac{1}{0.004}

\mu=0.98\ Pa.s

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Normalize the equations
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Answer:

Solution is in explanation

Explanation:

part a)

For normalization we have

\int_{0}^{\infty }f(x)dx=1\\\\\therefore \int_{0}^{\infty }ae^{-kx}dx=1\\\\\Rightarrow a\int_{0}^{\infty }e^{-kx}dx=1\\\\\frac{a}{-k}[\frac{1}{e^{kx}}]_{0}^{\infty }=1\\\\\frac{a}{-k}[0-1]=1\\\\\therefore a=k

Part b)

\int_{0}^{L }f(x)dx=1\\\\\therefore Re(\int_{0}^{L }ae^{-ikx}dx)=1\\\\\Rightarrow Re(a\int_{0}^{L }e^{-ikx}dx)=1\\\\\therefore Re(\frac{a}{-ik}[\frac{1}{e^{ikx}}]_{0}^{L})=1\\\\\Rightarrow Re(\frac{a}{-ik}(e^{-ikL}-1))=1\\\\\frac{a}{k}Re(\frac{1}{-i}(cos(-kL)+isin(-kL)-1))=1

\frac{a}{k}Re(\frac{1}{-i}(cos(-kL)+isin(-kL)-1))=1\\\\\frac{a}{k}Re(icos(-kL)+sin(kL)+\frac{1}{i})=1\\\\\frac{a}{k}sin(kL)=1\\\\a=\frac{k}{sin(kL)}

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