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Liula [17]
3 years ago
11

Different Gauss quadrature formulae predict different values for the same integral a. True b. False

Engineering
1 answer:
scoray [572]3 years ago
5 0

Answer: False

Explanation:

The given statement are false as, the gauss quadrature predicted same value only when their is a minute error at one point and two point rule. It basically given the more precise answer and the answer are only changed by the decimal place. The gauss quadrature method are basically used for calculating the certain integral.

You might be interested in
The period of a pendulum T is assumed to depend only on the mass m, the length of the pendulum `, the acceleration due to gravit
zzz [600]

Answer:

The expression is shown in the explanation below:

Explanation:

Thinking process:

Let the time period of a simple pendulum be given by the expression:

T = \pi \sqrt{\frac{l}{g} }

Let the fundamental units be mass= M, time = t, length = L

Then the equation will be in the form

T = M^{a}l^{b}g^{c}

T = KM^{a}l^{b}g^{c}

where k is the constant of proportionality.

Now putting the dimensional formula:

T = KM^{a}L^{b}  [LT^{-} ^{2}]^{c}

M^{0}L^{0}T^{1} = KM^{a}L^{b+c}

Equating the powers gives:

a = 0

b + c = 0

2c = 1, c = -1/2

b = 1/2

so;

a = 0 , b = 1/2 , c = -1/2

Therefore:

T = KM^{0}l^{\frac{1}{2} } g^{\frac{1}{2} }

T = 2\pi \sqrt{\frac{l}{g} }

where k = 2\pi

8 0
3 years ago
Those in this career install, maintain, and repair electrical wiring,
Ad libitum [116K]
I’m pretty sure it’s C and G
8 0
3 years ago
Read 2 more answers
What process is used to remove collodal and dissolved organic matter in waste water ​
Juli2301 [7.4K]

Answer:

Aerobic biological treatment process

Explanation:

Aerobic biological treatment process in which micro-organisms, in the presence of oxygen, metabolize organic waste matter in the water, thereby producing more micro-organisms and inorganic waste matter like CO₂, NH₃ and H₂O.

3 0
2 years ago
Explain the difference between thermoplastics and thermosets giving structure property correlation.
Misha Larkins [42]

Answer:

Explanation:

Thermosetting polymers are infusible and insoluble polymers. The reason for such behavior is that the chains of these materials form a three-dimensional spatial network, intertwining with strong equivalent bonds. The structure thus formed is a conglomerate of interwoven chains giving the appearance and functioning as a macromolecule, which as the temperature rises, simply the chains are more compacted, making the polymer more resistant to the point where it degrades.

Macromolecules are molecules that have a high molecular mass, formed by a large number of atoms. Generally they can be described as the repetition of one or a few minimum units or monomers, forming the polymers. In contrast, a thermoplastic is a material that at relatively high temperatures, becomes deformable or flexible, melts when heated and hardens in a glass transition state when it cools sufficiently. Most thermoplastics are high molecular weight polymers, which have associated chains through weak Van der Waals forces (polyethylene); strong dipole-dipole and hydrogen bond interactions, or even stacked aromatic rings (polystyrene). Thermoplastic polymers differ from thermosetting polymers or thermofixes in that after heating and molding they can overheat and form other objects.

Thermosetting plastics have some advantageous properties over thermoplastics. For example, better resistance to impact, solvents, gas permeation and extreme temperatures. Among the disadvantages are, generally, the difficulty of processing, the need for curing, the brittle nature of the material (fragile) and the lack of reinforcement when subjected to tension. But even so in many ways it surpasses the thermoplastic.

The physical properties of thermoplastics gradually change if they are melted and molded several times (thermal history), these properties are generally diminished by weakening the bonds. The most commonly used are polyethylene (PE), polypropylene (PP), polybutylene (PB), polystyrene (PS), polymethylmethacrylate (PMMA), polyvinylchloride (PVC), ethylene polyterephthalate (PET), Teflon (or polytetrafluoroethylene, PTFE) and nylon (a type of polyamide).

They differ from thermosets or thermofixes (bakelite, vulcanized rubber) in that the latter do not melt when raised at high temperatures, but burn, making it impossible to reshape them.

Many of the known thermoplastics can be the result of the sum of several polymers, such as vinyl, which is a mixture of polyethylene and polypropylene.

When they are cooled, starting from the liquid state and depending on the temperatures to which they are exposed during the solidification process (increase or decrease), solid crystalline or non-crystalline structures may be formed.

This type of polymer is characterized by its structure. It is formed by hydrocarbon chains, like most polymers, and specifically we find linear or branched chains

4 0
3 years ago
For a fluid with a Prandtl Number of 1000.0, the hydrodynamic layer is thinner than the thermal boundary layers. a) True b) Fals
kvv77 [185]

Answer:

(b)False

Explanation:

Given:

 Prandtl number(Pr) =1000.

We know that   Pr=\dfrac{\nu }{\alpha }

  Where \nu is the molecular diffusivity of momentum

             \alpha is the molecular diffusivity of heat.

 Prandtl number(Pr) can also be defined as

    Pr=\left (\dfrac{\delta }{\delta _t}\right )^3

Where \delta is the hydrodynamic boundary layer thickness and \delta_t is the thermal boundary layer thickness.

So if Pr>1 then hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.

In given question Pr>1 so  hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.

So hydrodynamic layer will be thicker than the thermal boundary layer.

8 0
3 years ago
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