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2 years ago

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Roku internet service providet

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podryga [215]2 years ago

6 0

Answer:

What?? I do not understand?

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A cyclic tensile load ranging from 0 kN to 55 kN force is applied along the length of a 100 mm long bar with a 15 mm x 15 mm squ

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square cross section. The bar is made of a 7075-T6 aluminum alloy which has a yield strength of 500 MPa, a tensile strength of 575 MPa, and a fracture toughness of 27.5 MPaâm.

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3 years ago

Why did my dad leave me yesterday like whatd i do?

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3 years ago

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3 years ago

An asbestos pad is square in cross section, measuring 5 cm on a side at its small end increasing linearly to 10 cm on a side at

polet [3.4K]

Answer:

q = 1.73 W

Explanation:

given data

small end = 5 cm

large end = 10 cm

high = 15 cm

small end is held = 600 K

large end at = 300 K

thermal conductivity of asbestos = 0.173 W/mK

solution

first we will get here side of cross section that is express as

$S = S1 + \frac{S2-S1}{L} x$ ...............1

here x is distance from small end and S1 is side of square at small end

and S2 is side of square of large end and L is length

put here value and we get

S = 5 + $\frac{10-5}{15} x$

S = $\frac{0.15 + x}{3}$ m

and

now we get here Area of section at distance x is

area A = S² ...............2

area A = $(\frac{0.15 + x}{3})^2$ m²

and

now we take here small length dx and temperature difference is dt

so as per fourier law

heat conduction is express as

heat conduction q = $\frac{-k\times A\ dt}{dx}$ ...............3

put here value and we get

heat conduction q = $-k\times (\frac{0.15 + x}{3})^2 \ \frac{dt}{dx}$

it will be express as

$q \times \frac{dx}{(\frac{0.15 + x}{3})^2} = -k (dt)$

now we intergrate it with limit 0 to 0.15 and take temp 600 to 300 K

$q \int\limits^{0.15}_0 {\frac{dx}{(\frac{0.15 + x}{3})^2 } = -0.173 \int\limits^{300}_{600} {dt}$

solve it and we get

q (30) = (0.173) × (600 - 300)

q = 1.73 W

5 0

3 years ago