Answer:
Explanation:
This problem relates to interference of light in thin films .
The condition of bright fringe in thin films which is sandwitched by two layers of medium having lesser refractive index is as follows.
2nt = (2n+1) λ / 2 , n is refractive index of thin layer , t is its thickness , λ is wavelength of light .
2 x 1.5 t = λ / 2 , if n = 0 for minimum thickness.
2 x 1.5 t = 600 / 2 nm
t = 100 nm .
Answer:
351 ohm
720 ohm
Explanation:
When c and d are open:
Terminals c and d are open.. If you redraw the circuit as below, you can see that the two resistors in the first column are in parallel as, they are connected together at both pairs of terminals (due to the short).
Hence, we have a pair of parallel resistors:
Req1 = (R1*R2)/ (R1 + R2) = 360*540/(360+540) = 216 ohms
Req2 = (R3*R4)/ (R3 + R4) = 180*540/(180+540) = 135 ohms
Now these two sets are in series with another Hence,
Req = Req1 + Req2 = 216 + 135 = 351 ohms
Answer: 351 ohms
When c and d are shorted:
The current will flow through the least resistant path naturally from resistors R3 and R1 or R4.
Both of these resistor lie in a single path placing the resistors in series to one another, hence
Req = R3 + R1 = 180 + 540 = 720 ohms
Answer:720 ohms
Answer:
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Explanation:
did the studyisland :)
Answer:
0 J
Explanation:
given,
mass of the ball = 5 kg
radius of the horizontal circle = 0.5 m
tension in the string = 10 N
Work done = ?
Work done by the tension in the circular path will be equal to zero.
This is because body moves in circular path, the centripetal force act along the radius of the circle and motion is right angle to the tension on the string.
so, work done = F s cos θ
θ = 90°,
work done = F s cos 90° ∵ cos 90° = 0
Work done = 0 J
