Four children are fighting over the same toy. Mike is pulling North with a 50 N force, Justin is pulling East with a 40 N force,
2 answers:
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Answer:
(a) With a short line, the A,B,C,D parameters are:
A = 1pu B = 1.685∠60.8°Ω C = 0 S D = 1 pu
(b) The sending-end voltage for 0.9 lagging power factor is 35.96
(c) The sending-end voltage for 0.9 leading power factor is 33.40
Explanation:
(a)
Considering the short transition line diagram.
Apply kirchoff's voltage law to the short transmission line.
Write the equation showing the relations between the sending end and the receiving end quantities.
Compare the line equations with the A,B,C,D parameter equations.
(b)
Determine the receiving-end current for 0.9 lagging power factor.
Determine the line-to-neutral receiving end voltage.
Determine the sending end voltage of the short transition line.
Determine the line-to-line sending end voltage which is the sending end voltage.
(c)
Determine the receiving-end current for 0.9 leading power factor.
Determine the sending-end voltage of the short transition line.
Determine the line-to-line sending end voltage which is the sending end voltage.
Answer:
The separation between the first two minima on either side is 0.63 degrees.
Explanation:
A diffraction experiment consists on passing monochromatic light trough a small single slit, at some distance a light diffraction pattern is projected on a screen. The diffraction pattern consists on intercalated dark and bright fringes that are symmetric respect the center of the screen, the angular positions of the dark fringes θn can be find using the equation:
with a the width of the slit, n the number of the minimum and λ the wavelength of the incident light. We should find the position of the n=1 and n=2 minima above the central maximum because symmetry the angular positions of n=-1 and n=-2 that are the angular position of the minima below the central maximum, then:
for the first minimum
solving for θ1:
for the second minimum:
So, the angular separation between them is the rest:
Answer:
The answers can be found attached.
