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iogann1982 [59]
3 years ago
12

Are spheroidized steels considered as composite? If so, what is the dispersed phase a)- No b)- Yes, Chromium Carbides c)- Yes, I

ron Carbides d)- Yes, Intermetallic Compounds
Engineering
1 answer:
Kruka [31]3 years ago
3 0

Answer: c)-Yes, spheroidized steel are considered as composite.the dispersed phase is  iron carbide.

Explanation: Spheroidized steel are the alloy that have iron as the basic part that have been heat treated to increase their ductility and malleability property .They are considered as composite because they are made up of  iron alloys.  The heat treatment is usually for the carbon steel and so the dispersed phase that is obtained is iron carbide.

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A single-cylinder pump feeds a boiler through a delivery
Studentka2010 [4]

Answer:

Net discharge per hour will be 3.5325 m^3/hr

Explanation:

We have given internal diameter d = 25 mm

Time = 1 hour = 3600 sec

So radius r=\frac{d}{2}=\frac{25}{2}=12.5mm=12.5\times 10^{-3}m

We know that area is given by

A=\pi r^2=3.14\times (12.5\times 10^{-3})^2=490.625\times 10^{-6}m^2

We know that discharge is given by Q=AV, here A is area and V is velocity

So Q=AV=490.625\times 10^{-6}\times 2=981.25\times 10^{-6}m^3/sec

So net discharge in 1 hour = 981.25\times 10^{-6}m^3/sec\times 3600=3.5325m^3/hour

8 0
3 years ago
A disk of radius 2.1 cm has a surface charge density of 5.6 µC/m2 on its upper face. What is the magnitude of the electric field
Assoli18 [71]

Answer:

=6.3*10^3 N/C

Explanation:

solution:

from this below equation (1)

E=σ/2εo(1-\frac{z}{\sqrt{z^2-R^2} } )...........(1)

we obtain:

=5.6*10^-6 \frac{c}{m^2} /2(8.85*10^-12\frac{c^2}{N.m^2} ).(1-\frac{9.5 cm}{\sqrt{9.5^2-2.1^2} } )

=6.3*10^3 N/C

8 0
3 years ago
I wish to have a computer whose machine-level instructions are all 32 bits each. If I want to have all instructions of the form
melisa1 [442]

Answer:

Maximum number that can be represented by 13 bits  is 8192 Instructions

Explanation:

number of instructions = 1000

number of bits = log(1000) x number of register

                          = 6 bits

Since the complete instruction must have 32 bits, then

remaining number of bits = 32 - 6 = 236

number of registers in instruction = 2

number of bits per register = 26/2 = 13

Maximum number that can be represented by 13 bits = 2^{n}

                       = 2¹³ = 8192

4 0
3 years ago
How s the weather like​
scoray [572]

weather is like corono everywhere #stay safe

3 0
2 years ago
It is desired to produce a continuous and oriented carbon fiber-reinforced epoxy having a modulus of elasticity of at least 83 G
lukranit [14]

Answer:

0.3129

Explanation:

Calculate The minimum fiber volume fraction using the relation below

E = x * E1 + ( 1 - x ) E2  ------ ( 1 )

given that :

E = 83 GPa , x = ?

E1 = 260 GPa ,  E2 = 2.4 GPA

Insert values back into equation 1

Then ; 80.6 = 257.6 x

∴ x ( volume fraction ) = 80.6 / 257.6  = 0.3129

3 0
2 years ago
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