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3 years ago

The error in the trapezoidal rule is proportional to h^2 True b. False

Engineering

1 answer:

Lynna [10]3 years ago

4 0

Answer:

b) False

Explanation:

in trapezoidal rule the error is proportional to $h^3$ and the order of accuracy is proportional to $h^3$ .

Trapezoidal rule is numerical integration method .Trapezoidal rule is used to find the area of curves.In trapezoidal rule we finds the approximate value of integration.But the real value of integration will not differ to much from the value which finds by using trapezoidal rule.

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The corner store sells candy in ₵20, ₵30 and ₵50 packages. List all the ways in which the Candyman

Novosadov [1.4K]

Answer: I will list them down below!

Explanation:

He can buy 6, 50 cent candies.

He can buy 30, 20 cent candies.

He can buy 6, 30 cent candies and 6, 20 cent candies.

He can buy 15, 20 cent candies and 3, 50 cent candies.

He can by 3, 20 and 30 cent candies and 3, 50 cent candies.

That's it.

Hope this helps!

6 0

1 year ago

Identify the advantages of using 6 tube passes instead of just 2 of the same diameter in shell-and-tube heat exchanger.What are

liq [111]

Answer:

Please check explanation for answer

Explanation:

Here, we are concerned with stating the advantages and disadvantages of using a 6 tube passes instead of a 2 tube passes of the same diameter:

<u>Advantages</u>

* By using a 6 tube passes diameter, we are increasing the surface area of the heat transfer surface

* As a result of increasing the heat transfer surface area, the rate of heat transfer automatically increases too

Thus, from the above, we can conclude that the heat transfer rate of a 6 tube passes is higher than that of a 2 tube passes of the same diameter.

<u>Disadvantages</u>

* They are larger in size and in weight when compared to a 2 tube passes of the same diameter and therefore does not find use in applications where space conservation is quite necessary.

* They are more expensive than the 2 tube passes of the same diameter and thus are primarily undesirable in terms of manufacturing costs

5 0

3 years ago

Thin film deposition is a process where: a)-elemental, alloy, or compound thin films are deposited onto a bulk substrate! b)-Pho

marshall27 [118]

Answer:

(A) elemental, alloy, or compound thin films are deposited on to a bulk substrate

Explanation:

In film deposition there is process of depositing of material in form of thin films whose size varies between the nano meters to micrometers onto a surface. The material can be a single element a alloy or a compound.

This technology is very useful in semiconductor industries, in solar panels in CD drives etc

so from above discussion it is clear that option (a) will be the correct answer

8 0

3 years ago

For laminar flow over a hot flat plate, the local heat transfer coefficient decreases with distance because (select all that are

kotegsom [21]

Answer:

B. The thickness of the heated region near the plate is increasing.

Explanation:

First we know that, a boundary layer is the layer of fluid in the immediate vicinity of a bounding surface where the effects of viscosity are significant. The fluid is often slower due to the effects of viscosity. Advection i.e the transfer of heat by the flow of liquid becomes less since the flow is slower, thereby the local heat transfer coefficient decreases.

From law of conduction, we observe that heat transfer rate will decrease based on a smaller rate of temperature, the thickness therefore increases while the local heat transfer coefficient decreases with distance.

3 0

3 years ago

An automotive fuel has a molar composition of 85% ethanol (C2H5OH) and 15% octane (C8H18). For complete combustion in air, deter

slava [35]

Answer:

a) 1

b) 1813.96 MJ/kmol

c) 32.43 MJ/kg , 1980.39 MJ/Kmol

Explanation:

molar mass of ethanol (C2H5OH) = 46 g/mol

molar mass of octane (C8H18) = 114 g/mol

therefore the moles of ethanol and octane

ethanol = 0.85 / 46

octane = 0.15 / 114

a) determine the molar air-fuel ratio and air-fuel ratio by mass

attached below

mass of air / mass of fuel = 12.17 / 1 = 12.17

b ) Determine the lower heating value

LHV of ( C2H5OH) = 26.8 * 46 = 1232.8 MJ/kmol

LHV of (C8H18). = 44.8 mj/kg * 114 kg/kmol = 5107.2 MJ/Kmol

LHV ( MJ/kmol) for fuel mixture = 0.85 * 1232.8 + 0.15 * 5107.2 = 1813.96 MJ/kmol

c) Determine higher heating value ( HHV )

HHV of (C2H5OH) = 29.7 * 46 = 1366.2 MJ/kmol

HHV of C8H18 = 47.9 MJ/kg * 114 = 5460.6 MJ/kmol

HHV in MJ/kg = 0.85 * 29.7 + 0.15 * 47.9 = 32.43 MJ/kg

HHV in MJ /kmol = 0.85 * 1366.2 + 0.15 * 5460.8 = 1980.39 MJ/Kmol

4 0

3 years ago