A 280 km long pipeline connects two pumping stations. It is desired to pump 0.56 m3/s of oil through a 0.62 m diameter line, the
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The question is not complete. We are supposed to find the average value of v_o.
Answer:
v_o,avg = 0.441V
Explanation:
Let t1 and t2 be the start and stop times of the output waveforms. Thus, from the diagram i attached, using similar triangles, we have;
3/(T/4) = 0.7/t1
So, 12/T = 0.7/t1
So, t1 = 0.7T/12
t1 = 0.0583 T
Also, from symmetry of triangles,
t2 = T/2 - t1
So, t2 = T/2 - 0.0583 T
t2 = 0.4417T
Average of voltage output is;
v_o,avg = (1/T) x Area under small triangle
v_o,avg = (1/T) x (3 - 0.7) x (T/4 - t1)
v_o,avg = (1/T) x (2.3) x (T/4 - 0.0583 T)
v_o,avg = (1/T) x 2.3 x 0.1917T
T will cancel out to give;
v_o,avg = 0.441V
Answer:
Vector C = 1.334i + 8.671j + 2k or 1.334x + 8.671y + 2z
Explanation:
The concept applied to solve the question is cross product of vector, AXB since vector C is perpendicular to vector A and B and this is solved by applying the 3X3 determinant method.
A detailed step by step explanation is attached below.
