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iogann1982 [59]
4 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]4 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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Answer:

21 coils

Explanation:

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Why are Gas cars Bad?(cons) give me reasons why gasoline cars are bad<br><br>Thx if u help ​
mixas84 [53]

Answer: Gasoline cars are bad because

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3 years ago
Three masses are attached to a uniform meter stick, as shown in Figure 12.9. The mass of the meter
sp2606 [1]

At equilibrium, the sum of clockwise and anticlockwise moments about a point is zero

The mass of that balances the system is \underline {316.\overline 6} kg

The normal reaction force at the fulcrum is <u>5,804.25 N</u>

Rason:

The mass of the stick = 150.0 g

Mass m₁ on the left = 50.0 g, location = 30 cm to the left of m₂

Mass m₂ on the left = 75.0 g, location = 40 cm to the left of the fulcrum

Mass m₃ on the right of the fulcrum. location = 30 cm to the right of the fulcrum

Required:

To find the mass of m₃

Solution:

Taking moment about the fulcrum, we have;

50 × (30 + 40) + 75 × (40) + 150 × 20 = m₃ × 30

9,500 = m₃×30

m_3 = \dfrac{9,500}{30} = 316. \overline 6

The mass of that balances the system when it is attached at the right end of the stick, m₃ = 316.\overline 6 kg

Normal reaction at the fulcrum = (50 + 75 + 150 +  316.\overline 6) × 9.81 = 5804.25

The normal reaction at the fulcrum is 5,804.25 N

Learn more about the moment of a force here:

brainly.com/question/19464450

8 0
2 years ago
Calculate the time taken to completely empty aswimming pool 15
user100 [1]

Answer:

Time needed to empty the pool is 401.35 seconds.

Explanation:

The exit velocity of the water from the orifice is obtained from the Torricelli's law as

V_{exit}=\sqrt{2gh}

where

'h' is the head under which the flow of water occurs

Thus the theoretical discharge through the orifice equals

Q_{th}=A_{orifice}\times \sqrt{2gh}

Now we know that

C_{d}=\frac{Q_{act}}{Q_{th}}

Thus using this relation we obtain

Q_{act}=C_{d}\times A_{orifice}\times \sqrt{2gh}

Now we know by definition of discharge

Q_{act}=\frac{d}{dt}(volume)=\frac{d(lbh)}{dt}=Lb\cdot \frac{dh}{dt}

Using the above relations we obtain

Lb\times \frac{dh}{dt}=AC_{d}\times \sqrt{2gh}\\\\\frac{dh}{\sqrt{h}}=\frac{AC_{d}}{Lb}\times \sqrt{2g}dt\\\\\int_{1.5}^{0}\frac{dh}{\sqrt{h}}=\int_{0}^{t}\frac{0.62\times 0.3}{15\times 9}\times \sqrt{2\times 9.81}\cdot dt\\\\

The limits are put that at time t = 0 height in pool = 1.5 m and at time 't' the height in pool = 0

Solving for 't' we get

\sqrt{6}=6.103\times 10^{-3}\times t\\\\\therefore t=\frac{\sqrt{6}}{6.103\times 10^{3}}=401.35seconds.

4 0
3 years ago
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