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DedPeter [7]
1 year ago
6

The source term will affect all algebraic equations.

Engineering
1 answer:
Drupady [299]1 year ago
8 0

The statement about the source term affecting all algebraic equations is; False

<h3>What is the source term in Algebraic Equations?</h3>

In the finite-element method of Analysis, we usually go from differential equations to a set of algebraic equations. Now, it is pertinent to note that each algebraic equation will tend to relate a nodal value with all other nodal values.

The assumed polynomial variation that will exist within each element is usually the basis for which we derive the algebraic equations. Thus, to derive the algebraic equations, we will have to assume a polynomial variation for the parameter values within each element.

The above can be done through interpolation of nodal parameter values in the post-processing stage. This is because the assumed polynomial variation within each element that is used for deriving the algebraic equations is also used for post-processing.

Finally, for each algebraic equation, the source term will tend to contribute to the coefficient that is taken to the right hand side of it and then entered into the corresponding row of the {f} vector.

Read more about Source Term at; brainly.com/question/28956378

#SPJ1

Complete Question is;

Select true or false.

The source term will affect all algebraic equations.

A.) True

B.) False

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Inga [223]

Answer:

47.91 sec

Explanation:

it is given that \alpha =\frac{1}{4v^{2}}

at t=0 velocity =0 ( as it is given that it is starting from rest )

we have to find time at which velocity will be 3.3 \frac{m}{sec^{2}}

we know that \alpha =\frac{dv}{dt}=\frac{1}{4v^{2}}

4v^{2}dv=dt

integrating both side

\frac{4v^{3}}{3}=t+c---------------eqn 1

at t=o it is given that v=0 putting these value in eqn 1 c=0

so \frac{4v^{3}}{3}=t

when v=  3.3 \frac{m}{sec^{2}}

t=\frac{4}{3}\times 3.3^{3}

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6 0
4 years ago
An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate (ϵ˙) is found to be
cupoosta [38]

Answer:

Activation energy for creep in this temperature range is Q = 252.2 kJ/mol

Explanation:

To calculate the creep rate at a particular temperature

creep rate, \zeta_{\theta} = C \exp(\frac{-Q}{R \theta} )

Creep rate at 800⁰C, \zeta_{800} = C \exp(\frac{-Q}{R (800+273)} )

\zeta_{800} = C \exp(\frac{-Q}{1073R} )\\\zeta_{800} = 1 \% per hour =0.01\\

0.01 = C \exp(\frac{-Q}{1073R} ).........................(1)

Creep rate at 700⁰C

\zeta_{700} = C \exp(\frac{-Q}{R (700+273)} )

\zeta_{800} = C \exp(\frac{-Q}{973R} )\\\zeta_{800} = 5.5 * 10^{-2}  \% per hour =5.5 * 10^{-4}

5.5 * 10^{-4}  = C \exp(\frac{-Q}{1073R} ).................(2)

Divide equation (1) by equation (2)

\frac{0.01}{5.5 * 10^{-4} } = \exp[\frac{-Q}{1073R} -\frac{-Q}{973R} ]\\18.182= \exp[\frac{-Q}{1073R} +\frac{Q}{973R} ]\\R = 8.314\\18.182= \exp[\frac{-Q}{1073*8.314} +\frac{Q}{973*8.314} ]\\18.182= \exp[0.0000115 Q]\\

Take the natural log of both sides

ln 18.182= 0.0000115Q\\2.9004 = 0.0000115Q\\Q = 2.9004/0.0000115\\Q = 252211.49 J/mol\\Q = 252.2 kJ/mol

3 0
3 years ago
What might cause a shotgun to explode?
Tanzania [10]

Answer:

When you pull the trigger to shoot a shotshell from a shotgun or a cartridge from a rifle or handgun, the firing pin strikes the primer in the base of the cartridge or shotshell. This causes the primer to explode. The spark from the primer ignites the gunpowder, which burns rapidly and converts to a gas.

Explanation:

3 0
3 years ago
Read 2 more answers
Primary Creep: slope (creep rate) decreases with time
Igoryamba

Answer:

true

Explanation:

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Creep is generally seen at high temperature.

Due to creep the length of the structure increases which is not fit for serviceability purpose.

When time passes structure gain strength as the structure strength increases with time so creep tends to decrease.

When we talk about Creep rate for new structure the creep will be more than the old structure i.e. the creep rate decreases with time.

5 0
3 years ago
For welding the most important reason to use jigs and fixtures in a welding shop is to
diamong [38]

Answer:

Reduce manufacturing costs.

Explanation:

Hope This Helps

Have A Great Day

7 0
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