Answer:
a. The very first liquid process, when heated from 1250 degree Celsius, is expected to form at the temperature by which the vertical line crosses the phase boundary (a -(a + L)) which is about <em>1310 degree Celsius. </em>
b. The structure of that first liquid is identified by the intersection with ((a+ L)-L) phase boundary; <em>47wt %of Ni</em> is of a tie line formed across the (a+ L) phase area <em>at 1310 degrees.</em>
c. To find the alloy's full melting, it is determined that the intersection of the same vertical line at 60 wt percent Ni with (a -(a+L)) phase boundary is around <em>1350 degrees.</em>
c. The structure of the last remaining solid before full melting correlates to the intersection with the phase boundary (a -(a + L), of the tie line built at 1350 degrees across the (a + L) phase area, <em>being 72wt % of Ni.</em>
Answer: This spark energy trigger ignition and combustion in the compressed air-fuel mixture. This discharge is of extremely brief duration (about 1/1000 of a second) and is extraordinarily complex!
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Answer:
From the main bearings, the oil passes through feed-holes into drilled passages in the crankshaft and on to the big-end bearings of the connecting rod.
Different lever designs can be engineered and developed to alter the brake pedal effort required of the driver by using different levels of <u>mechanical advantage</u>.
<h3>What is
mechanical advantage?</h3>
Mechanical advantage can be defined as a ratio of the output force of a lever to the force acting on it (input force or effort), assuming no losses due to wear, flexibility, tear or friction.
This ultimately implies that, different lever designs can be suitably engineered and developed to alter the brake pedal effort (input force) that is required of the driver, especially by using different levels of <u>mechanical advantage</u>.
Read more on mechanical advantage here: brainly.com/question/18345299
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Answer:
a, c
Explanation:
As the problem statement tells you, the phasor technique cannot be used when the frequencies are different. The frequencies are different when the coefficients of t are different. The different ones are highlighted.
a. 45 sin(2500t – 50°) + 20 cos(1500t +20°)
b. 25 cos(50t + 160°) + 15 cos(50t +70°)
c. 100 cos(500t +40°) + 50 sin(500t – 120°) – 120 cos(500t + 60°) -100 sin(10,000t +90°) + 40 sin(10, 100t – 80°) + 80 cos(10,000t)
d. 75 cos(8t+40°) + 75 sin(8t+10°) – 75 cos(8t + 160°)
