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WINSTONCH [101]
3 years ago
14

Problema sobre programacion orientada a objetos!!

Engineering
1 answer:
Alecsey [184]3 years ago
8 0

Answer:If they are used to top-down programming or functional programming, which treats elements of code as precise mathematical functions, it takes .

Explanation:

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2 years ago
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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

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3 years ago
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3 years ago
Do a summary what happen to titanic in the aspect of material(body) and the ductile brittle temperature (DBT) of the material.
vodka [1.7K]

Explanation:

A ductile material can convert into brittle material due to following reasons

1.At very low temperature

2.Due to presence of notch

In titanic ,the base of ship strike to the large ice cube and lower part of titanic ship material was made of steel .We know that steel is ductile material and when steel came with very low temperature of ice due to this ductile material converted in to brittle material and titanic ship failed.Brittle material does not show any indication before failure.

7 0
3 years ago
Water flows in a constant diameter pipe with the following conditions measured:
Burka [1]

Answer:

a) h_L=-3.331ft

b) The flow would be going from section (b) to section (a)

Explanation:

1) Notation

p_a =31.1psi=4478.4\frac{lb}{ft^2}

p_b =27.3psi=3931.2\frac{lb}{ft^2}

For above conversions we use the conversion factor 1psi=144\frac{lb}{ft^2}

z_a =56.7ft

z_a =68.8ft

h_L =? head loss from section

2) Formulas and definitions

For this case we can apply the Bernoulli equation between the sections given (a) and (b). Is important to remember that this equation allows en energy balance since represent the sum of all the energies in a fluid, and this sum need to be constant at any point selected.

The formula is given by:

\frac{p_a}{\gamma}+\frac{V_a^2}{2g}+z_a =\frac{p_b}{\gamma}+\frac{V_b^2}{2g}+z_b +h_L

Since we have a constant section on the piple we have the same area and flow, then the velocities at point (a) and (b) would be the same, and we have just this expression:

\frac{p_a}{\gamma}+z_a =\frac{p_b}{\gamma}+z_b +h_L

3)Part a

And on this case we have all the values in order to replace and solve for h_L

\frac{4478.4\frac{lb}{ft^2}}{62.4\frac{lb}{ft^3}}+56.7ft=\frac{3931.2\frac{lb}{ft^2}}{62.4\frac{lb}{ft^3}}+68.8ft +h_L

h_L=(71.769+56.7-63-68.8)ft=-3.331ft

4)Part b

Analyzing the value obtained for \h_L is a negative value, so on this case this means that the flow would be going from section (b) to section (a).

5 0
3 years ago
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