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2 years ago

18. Bela is doing a continuity test. What's he checking?

Engineering

1 answer:

murzikaleks [220]2 years ago

8 0

Answer:

the bela is doing a continuity test he checking current

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What are four engineering degrees that can help lead you to becoming an aerospace engineer

mamaluj [8]

Answer:

Mechanical Engineering, Aerospace Engineering, Electrical Engineering, Aeronautical Engineering

Explanation:

I'm a Mechanical Engineer, and I considered Aerospace.

7 0

3 years ago

In a wheatstone bridge three out of four resistors have of 1K ohm each ,and the fourth resistor equals 1010 ohm. If the battery

Dima020 [189]

Answer:

248.756 mV

49.7265 µA

Explanation:

The Thevenin equivalent source at one terminal of the bridge is ...

voltage: (100 V)(1000/(1000 +1000) = 50 V

impedance: 1000 || 1000 = (1000)(1000)/(1000 +1000) = 500 Ω

The Thevenin equivalent source at the other terminal of the bridge is ...

voltage = (100 V)(1010/(1000 +1010) = 100(101/201) ≈ 50 50/201 V

impedance: 1000 || 1010 = (1000)(1010)/(1000 +1010) = 502 98/201 Ω

The open-circuit voltage is the difference between these terminal voltages:

(50 50/201) -(50) = 50/201 V ≈ 0.248756 V . . . . open-circuit voltage

The current that would flow is given by the open-circuit voltage divided by the sum of the source resistance and the load resistance:

(50/201 V)/(500 +502 98/201 +4000) = 1/20110 A ≈ 49.7265 µA

8 0

3 years ago

Compute the volume percent of graphite, VGr, in a 3.2 wt% C cast iron, assuming that all the carbon exists as the graphite phase

Yanka [14]

Answer:

The volume percentage of graphite is 10.197 per cent.

Explanation:

The volume percent of graphite is the ratio of the volume occupied by the graphite phase to the volume occupied by the graphite and ferrite phases. The weight percent in the cast iron is 3.2 wt% (graphite) and 96.8 wt% (ferrite). The volume percentage of graphite is:

$\%V_{gr} = \frac{V_{gr}}{V_{gr}+V_{fe}} \times 100\,\%$

Where:

$V_{gr}$ - Volume occupied by the graphite phase, measured in cubic centimeters.

$V_{fe}$ - Volume occupied by the graphite phase, measured in cubic centimeters.

The expression is expanded by using the definition of density and subsequently simplified:

$\%V_{gr} = \frac{\frac{m_{gr}}{\rho_{gr}} }{\frac{m_{gr}}{\rho_{gr}}+\frac{m_{fe}}{\rho_{fe}}}\times 100\,\%$

Where:

$m_{fe}$ , $m_{gr}$ - Masses of the ferrite and graphite phases, measured in grams.

$\rho_{fe}, \rho_{gr}$ - Densities of the ferrite and graphite phases, measured in grams per cubic centimeter.

$\%V_{gr} = \frac{1}{1+\frac{\frac{m_{fe}}{\rho_{fe}} }{\frac{m_{gr}}{\rho_{gr}} } }\times 100\,\%$

$\%V_{gr} = \frac{1}{1 + \left(\frac{\rho_{gr}}{\rho_{fe}} \right)\cdot\left(\frac{m_{fe}}{m_{gr}} \right)} \times 100\,\%$

If $\rho_{gr} = 2.3\,\frac{g}{cm^{3}}$ , $\rho_{fe} = 7.9\,\frac{g}{cm^{3}}$ , $m_{gr} = 3.2\,g$ and $m_{fe} = 96.8\,g$ , the volume percentage of graphite is: