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2 years ago

Difference between theory and practice?

Engineering

1 answer:

Ronch [10]2 years ago

5 0

Answer:

There is a huge difference between theory vs. practice. Theory assumes an outcome, while practice allows you to test the theory and see if it is accurate.

Theory and Practice Explained

Practice is the observation of disparate concepts (or a phenomenon) that needs explanation. A theory is a proposed explanation of the relationship between two or more concepts, or an explanation for how/why a phenomenon occurs.

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A vertical pole consisting of a circular tube of outer diameter 127 mm and inner diameter 115 mm is loaded by a linearly varying

Anna [14]

Maximum shear stress in the pole is 0.

<u>Explanation:</u>

Given-

Outer diameter = 127 mm

Outer radius, $r_{2}$ = 127/2 = 63.5 mm

Inner diameter = 115 mm

Inner radius, $r_{1}$ = 115/2 = 57.5 mm

Force, q = 0

Maximum shear stress, τmax = ?

τmax $= \frac{4q}{3\pi } (\frac{r2^2 + r2r1 + r1^2}{r2^4 - r1^4} )$

If force, q is 0 then τmax is also equal to 0.

Therefore, maximum shear stress in the pole is 0.

3 0

3 years ago

The question “How do plants convert sunlight to energy?” best represents which of the following?

OverLord2011 [107]

Answer:

Explanation:

I would say this awnser because its the only one that makes sence to me

5 0

3 years ago

(a)Compute the electrical conductivity of a cylindrical silicon specimen 7.0 mm (0.28 in.) diameter and 57 mm (2.25 in.) in leng

igor_vitrenko [27]

Answer:

a) $\sigma = 12.2$ (Ω-m)^{-1}

b) Resistance = 121.4 Ω

Explanation:

given data:

diameter is 7.0 mm

length 57 mm

current I = 0.25 A

voltage v = 24 v

distance between the probes is 45 mm

electrical conductivity is given as

$\sigma = \frac{I l}{V \pi r^2}$

$\sigma = \frac{0.25 \times 45\times 10^{-3}}{24 \pi [\frac{7 \times 10^{-3}}{2}]^2}$

$\sigma = 12.2$ (Ω-m)^{-1}[/tex]

$Resistance = \frac{l}{\sigma A}$

$= \frac{l}{ \sigma \pi r^2}$

$= \frac{57 \times 10^{-3}}{12.2 \times \pi [\frac{7 \times 10^{-3}}{2}]^2}$

Resistance = 121.4 Ω

8 0

3 years ago

Unlike expendable molds, permanent molds do not collapse, so the mold must be opened before appreciable cooling contraction occu

Pani-rosa [81]

Answer: True

Explanation:

Permanent molds do not collapse, unlike expendable molds so the mold must be opened before appreciable cooling contraction occurs in order to prevent cracks from developing in the casting.

The metal casting becomes solid inside the mold after it has been poured. But during the process of manufacture, before the would cools any further, they usually remove the metal cast in order to stop excess contractions of the solid metal casting in the mold. This is done to prevent prevent cracks from developing in the casting since permanent mold do not collapse.

8 0

3 years ago

A 2-m3 rigid tank initially contains air at 100 kPa and 22°C. The tank is connected to a supply line through a valve. Air is flo

Finger [1]

Answer:

9.58 Kg of air has entered the tank.

heat entered=3483.76 Kilo.Joule

Explanation:

(A) R=287 Kilo.J/Kg.K

as per initial conditions P=100 Kilo.Pa ,V=2 cubic meter, T=22 C=295.15 K,

using the relation P*V=m*R*T

m=(100*1000*2)/(287*295.15)=2.36 Kg this is the mass that is already present in tank.

after filling tank at 600 Kilo.Pa.

P=600 Kilo Pa T=77 C=350.15 K

P*V=m*R*T

m=(600*1000*2)/(287*350.15)=11.94 Kg

mass that has entered=11.94-2.36=9.58 Kg

(b) using air psychometric property table

specific heat content initial 100 KILO Pa and 22 C=295.576 Kilo.Joule/Kg

specific heat content final 600 Kilo Pa and 77 C=350.194 Kilo.Joule/Kg

heat at initial stage=295.576*2.36=697.56 Kilo.Joule

heat at final stage=350.194*11.94=4181.32 Kilo.Joule

heat entered=4181.32-697.56=3483.76 Kilo.Joule

3 0

3 years ago

Difference between theory and practice?​

Difference between theory and practice?