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sweet-ann [11.9K]
3 years ago
15

A 0.39 percent Carbon hypoeutectoid plain-carbon steel is slowly cooled from 950 oC to a temperature just slightly below 723 oC.

Calculate the weight percent proeutectoid ferrite in the steel. Please do not include any numbers after the decimal and give the percentage as "xx" and not "0.xx"
Engineering
1 answer:
belka [17]3 years ago
8 0

Answer:

53%

Explanation:

To find the weight % in proeutectoid ferrite we use:

Wt % in proeutectoid ferrite=

[(0.80 - 0.39) / (0.80 - 0.02)] * 100=

= (0.41 /0.78) * 100 =

52.56%

which is approximately 53%

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Determine the design angle ϕ (0∘≤ϕ≤90 ∘) between struts AB and AC so that the 400 lb horizontal force has a component of 600 lb
Svetllana [295]

Answer:

design angle ∅ = 4.9968 ≈ 5⁰

Explanation:

First calculate the force Fac :

Fac = \sqrt{400^2 + 650^2 - 2(400)(650)cos30}

      = \sqrt{160000 + 422500 - 80210}

      = 708.72 Ib

using the sine law to determine the design angle

\frac{sin}{400}  = \frac{sin 30}{Fac}

hence ∅ = sin^{-1} (\frac{sin 30 *400}{708.72} )

              = sin^{-1} 0.0871 =  4.9968 ≈ 5⁰

4 0
3 years ago
Turn the motor around in the circuit. What happens?
kenny6666 [7]

Answer:

It will create a massive drag and pretty much stop the motor.

Explanation:

7 0
3 years ago
Multiple Choice
mote1985 [20]
I think it’s manufacturing
7 0
3 years ago
Calculate the differential pressure in kPa across the hatch of a submarine 320m below the surface of the sea. Assume the atmosph
kicyunya [14]

Answer:

The pressure difference across hatch of the submarine is 3217.68 kpa.

Explanation:

Gauge pressure is the pressure above the atmospheric pressure. If we consider gauge pressure for finding pressure differential then no need to consider atmospheric pressure as they will cancel out. According to hydrostatic law, pressure varies in the z direction only.  

Given:

Height of the hatch is 320 m

Surface gravity of the sea water is 1.025.

Density of water 1000 kg/m³.

Calculation:

Step1

Density of sea water is calculated as follows:

S.G=\frac{\rho_{sw}}{\rho_{w}}

Here, density of sea water is\rho_{sw}, surface gravity is S.G and density of water is \rho_{w}.

Substitute all the values in the above equation as follows:

S.G=\frac{\rho_{sw}}{\rho_{w}}

1.025=\frac{\rho_{sw}}{1000}

\rho_{sw}=1025 kg/m³.

Step2

Difference in pressure is calculated as follows:

\bigtriangleup p=rho_{sw}gh

\bigtriangleup p=1025\times9.81\times320

\bigtriangleup p=3217680 pa.

Or

\bigtriangleup p=(3217680pa)(\frac{1kpa}{100pa})

\bigtriangleup p=3217.68 kpa.

Thus, the pressure difference across hatch of the submarine is 3217.68 kpa.

6 0
3 years ago
If a signal is transmitted at a power of 250 mWatts (mW) and the noise in the channel is 10 uWatts (uW), if the signal BW is 20M
Bess [88]

Answer:

C = 292 Mbps

Explanation:

Given:

- Signal Transmitted Power P = 250mW

- The noise in channel N = 10 uW

- The signal bandwidth W = 20 MHz

Find:

what is the maximum capacity of the channel?

Solution:

-The capacity of the channel is given by Shannon's Formula:

                            C = W*log_2 ( 1 + P/N)

- Plug the values in:

                            C = (20*10^6)*log_2 ( 1 + 250*10^-3/10)

                            C = (20*10^6)*log_2 (25001)

                            C = (20*10^6)*14.6096

                           C = 292 Mbps

3 0
2 years ago
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