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kolbaska11 [484]
3 years ago
10

PLS HURRY!!!

Engineering
2 answers:
DanielleElmas [232]3 years ago
4 0

Answer:

A) composition of space near the earth

Explanation:

I hope I answered your question.

Hitman42 [59]3 years ago
3 0

Answer:

Composition of space near the earth

Explanation:

on edg

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That’s too hard for me lol oof
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Which of the following approaches to lean engineering consists of teams of people from different departments sharing
AURORKA [14]

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PDSA, where you work as a group

Explanation:

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3 years ago
If you were to plot the voltage versus the current for a given circuit, what would you expect the slope of the line to be? If no
Brut [27]

Answer:

Part 1: It would be a straight line, current will be directly proportional to the voltage.

Part 2: The current would taper off and will have negligible increase after the voltage  reaches a certain  value. Graph attached.

Explanation:

For the first part, voltage and current have a linear relationship as dictated by the Ohm's law.

V=I*R

where V is the voltage, I is the current, and R is the resistance. As the Voltage increase, current is bound to increase too, given that the resistance remains constant.

In the second part, resistance is not constant. As an element heats up, it consumes more current because the free sea of electrons inside are moving more rapidly, disrupting the flow of charge. So, as the voltage increase, the current does increase, but so does the resistance. Leaving less room for the current to increase. This rise in temperature is shown in the graph attached, as current tapers.

7 0
4 years ago
The voltage valve at which a zirconia O2S switches from rich to lean and lean to rich is
frez [133]
I think the answer is C) 0.25v I’m not sure tho
6 0
3 years ago
2. Given that a quality-control inspection can ensure that a structural ceramic part will have no flaws greater than 100 µm (100
MA_775_DIABLO [31]

Answer:

SiC=169.26 Mpa

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Explanation:

<u>SiC </u>

Stress intensity, K is given by

K=\sigma Y\sqrt{\pi a} hence making \sigma the subject where \sigma is applied stress, Y is shape factor, a is crack length

\sigma=\frac {K}{Y\sqrt{\pi a}} substituting the given figures and assuming shape factor, Y of 1

\sigma=\frac {3}{1\sqrt{\pi 100*10^{-6}}}= 169.2569\approx 169.26 Mpa

<u>Stabilized zirconia </u>

Stress intensity, K is given by

K=\sigma Y\sqrt{\pi a} hence making \sigma the subject where \sigma is applied stress, Y is shape factor, a is crack length

\sigma=\frac {K}{Y\sqrt{\pi a}} substituting the given figures and assuming shape factor, Y of 1

\sigma=\frac {9}{1\sqrt{\pi 100*10^{-6}}}= 507.7706\approx 507.77 Mpa

7 0
3 years ago
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