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3 years ago

What is the difference between digital instruments and decimal scaled instruments to measure

1 answer:

8 0

Answer Digital measuring instruments are self-contained devices that automatically present the value of the measured quantity on a digital display. And Decimal Scaled Instruments: Record all digits that you can certainly determine from the scale markings and estimate one more digit. I hope this Helped I´m new to this.

Explanation:

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A controlled process is described by the closed-loop transfer function G(s).

MissTica

Answer:

The answer is "Option B".

Explanation:

Given equation:

$G(s) =\frac{K(s + 1)}{2s^2 + (K-1)s + (K-1)}\\\\$

$\to 2s^2 + (K-1)s + (K-1)=0$

Calculating by the Routh's Hurwitz table:

$\to s^2 \ \ \ \ \ 2 \ \ \ \ \ \ K-1 \\\\\to s^2 \ \ \ \ \ K-1 \ \ \ \ \ \ \\\\\to s^0 \ \ ( \frac{(K-1)(K-1)(-2) (0)}{K-1} \\\\ \ \ \ \ = (K-1) )$

Form the above table:

$\to K-1 > 0 \\\\ \to K > 1$

In the above, the value of k is greater than 1.

3 0

3 years ago

A sandy soil has a total unit weight of 120 pcf, a specific gravity of solids of 2.64, and a water content of 16 percent. Comput

olchik [2.2K]

Answer:

A). Dry unit weight = 1657.08Kg/m3

B). Porosity = 0.37

C). Void ratio = 0.593

D). 0.712

Explanation:

Total unit weight, Y = 120pcf =1922.2 Kg/m3

Specific gravity of solids, Gs = 2.64

Water content, w = 16%

A). Dry unit weight

Yd = Y/(1+w)

= 1922.2/(1+0.16) = 1657.08Kg/m3

B). Porosity

However void ratio, e = Gs×Yw/Yd, where Yw = 1000Kg/m3

Void ratio = 2.64×1000/1657.08 = 0.593

And porosity = e/(1+e) =0.593/(1+0.593) = 0.37

C). void ratio, e = 0.593

D). Degree of saturation, S = m×Gs/e where m =water content

S = 0.16×2.64/0.593 = 0.712

5 0

3 years ago

A certain part of the cast iron piping of a water distribution system involves a parallel section. Both parallel pipes have a di

Bezzdna [24]

Answer :

Explanation:

Given :

Length of pipe A $L_{A} = 1500$ m

Length of pipe B $L_{B} = 2500$ m

Flow rate through pipe A $Q_{A} = 0.4 \frac{m^{3} }{s}$

Diameter of pipe $D = 30 \times 10^{-2}$ m

Velocity from pipe A,

$V _{A} = \frac{Q_{A} }{A}$

$V _{A} = \frac{0.4 \times 4 }{\pi ( 30 \times 10^{-2} )^{2} }$

$V_{A} = 5.66$ $\frac{m}{s}$

Here, head loss is same because height is same.

$h_{a} = h_{b}$

$L_{A} V_{A} ^{2} = L_{B} V_{B} ^{2}$

$V_{B} = \sqrt{\frac{1500}{2500}} (5.66)$

$V_{B} = 4.38$ $\frac{m}{s}$

Now rate of flow from pipe B is,

$Q_{B} = V_{B} A$

$Q_{B} = \frac{\pi }{4} (0.3)^{2} \times 4.38$

$Q_{B} = 0.3094$ $\frac{m^{3} }{s}$

4 0

3 years ago

A tensile specimen with a 12mm initial diameter and 50mm gage length reaches maximum load at 90KN and fractures at 70KN

Aleksandr-060686 [28]

Answer:

i) 796.18 N/mm^2

ii) 1111.11 N/mm^2

Explanation:

Initial diameter ( D ) = 12 mm

Gage Length = 50 mm

maximum load ( P ) = 90 KN

Fractures at = 70 KN

minimum diameter at fracture = 10mm

<u>Calculate the engineering stress at Maximum load and the True fracture stress</u>

<em>i) Engineering stress at maximum load = P/ A </em>

= P / $\pi \frac{D^2}{4}$ = 90 * 10^3 / ( 3.14 * 12^2 ) / 4

= 90,000 / 113.04 = 796.18 N/mm^2

<em>ii) True Fracture stress = P/A </em>

= 90 * 10^3 / ( 3.24 * 10^2) / 4

= 90000 / 81 = 1111.11 N/mm^2

3 0

2 years ago

3. Consider a vortex filament of strength  in the shape of a closed circular loop of radius R. Obtain an expression for the vel

Vsevolod [243]

Answer:

Explanation:

Given that Г is the strength of vortex filament

R is closed circular loop radius

Attached is the expression for the induced velocity in vector form

3 0

3 years ago