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3 years ago

Select the correct text in the passage.

Engineering

2 answers:

frutty [35]3 years ago

8 0

Answer:

The gauge converts the charge into a measurable electrical signal.

Explanation:

trust me

sineoko [7]3 years ago

5 0

It is habahi Yw with yuuuuuy I am a little more confused about

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Differences between acidic and basic Bessemer process

dybincka [34]

Answer:

In the acid processes, deoxidation can take place in the furnaces, leaving a reasonable time for the inclusions to rise into the sla*g and so be removed before casting. Whereas in the basic furnaces, deoxidation is rarely carried out in the presence of the sla*g, otherwise phosphorus would return to the metal.

5 0

2 years ago

What is the average linear (seepage) velocity of water in an aquifer with a hydraulic conductivity of 6.9 x 10-4 m/s and porosit

jeka94

Answer:

a. 0.28

Explanation:

Given that

porosity =30%

hydraulic gradient = 0.0014

hydraulic conductivity = 6.9 x 10⁻4 m/s

We know that average linear velocity given as

$v=\dfrac{K}{n_e}\dfrac{dh}{dl}$

$v=\dfrac{6.9\times 10^{-4}}{0.3}\times0.0014\ m/s$

$v=3.22\times 10^{-6}\ m/s$

The velocity in m/d ( 1 m/s =86400 m/d)

v= 0.27 m/d

So the nearest answer is 'a'.

a. 0.28

4 0

3 years ago

Liquid flows with a free surface around a bend. The liquid is inviscid and incompressible, and the flow is steady and irrotation

lions [1.4K]

Answer:

9 cm

Explanation:

The liquid on the bend will be affected by two accelerations: gravity and centripetal force.

Gravity will be of 9.81 m/s^2 pointing down at all points.

The centripetal acceleration will be of

ac = v^2/r

Pointing to the center of the bend (perpendicular to gravity).

The velocity will depend on the radius

v = (1 m^2/s) / r

Replacing:

ac = (1/r)^2 / r

ac = (1 m^4/s^2) / r^3

If we set up a cylindrical reference system with origin at the center of the bend, the total acceleration will be

a = (-1/r^3 * i - 9.81 * j)

The surface of the liquid will be an equipotential surface, this means all points on the surface have the same potential energy.

The potential energy of the gravity field is:

pg = g * h

The potential energy of the centripetal force is:

pc = ac * r

Then the potential field is:

p = -1/r^2 * - 9.81*h

Points on the surface at r = 1 m and r = 3 m have the same potential.

-1/1^2 * - 9.81*h1 = -1/3^2 * - 9.81*h2

-1 - 9.81*h1 = -1/9 - 9.81*h2

-1 + 1/9 = 9.81 * (h1 - h2)

h1 - h2 = (-8/9) / 9.81

h2 - h1 = 0.09 m

The outer part will be 9 cm higher than the inner part.

3 0

3 years ago

A/an_ Oscilloscope uses a cathode ray tube and displays all voltages.

MrRa [10]

Answer:

I think it’s B, digital

Explanation:

3 0

3 years ago

Kerosene flows through 3/4 standard type K drawn copper tube. The pressure drop measured at two points 50 m apart is 130 kPa. De

jok3333 [9.3K]

Answer:

$Q=4.98\times 10^{-3}\ m^3/s$ .

Explanation:

Given that

L= 50 m

Pressure drop = 130 KPa

copper tube is 3/4 standard type K drawn tube.

From standard chart ,the dimension of 3/4 standard type K copper tube given as

Outside diameter=22.22 mm

Inside diameter=18.92 mm

Dynamic viscosity for kerosene

$\mu =0.00164\ Pa.s$

We know that

$\Delta P=\dfrac{128\mu QL}{\pi d_i^4}$

Where Q is volume flow rate

L is length of tube

$d_i$ is inner diameter of tube

ΔP is pressure drop

μ is dynamic viscosity

Now by putting the values

$\Delta P=\dfrac{128\mu QL}{\pi d_i^4}$

$130\times 1000=\dfrac{128\times 0.00164\times 50Q}{\pi \times 0.01892^4}$

$Q=4.98\times 10^{-3}\ m^3/s$

So flow rate is $Q=4.98\times 10^{-3}\ m^3/s$ .

6 0

3 years ago