A book that weighs 16N is on top of a shelf that is 3m high. What is the work required to
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Answer:
spacing between the slits is 405.32043 × m
Explanation:
Given data
wavelength = 610 nm
angle = 2.95°
central bright fringe = 85%
to find out
spacing between the slits
solution
we know that spacing between slit is
I = 4 × cos²∅/2
so
I/4 = cos²∅/2
here I/4 is 85 % = 0.85
so
0.85 = cos²∅/2
cos∅/2 = √0.85
∅ = 2 × 0.921954
∅ = 45.56°
∅ = 45.56° ×π/180 = 0.7949 rad
and we know that here
∅ = 2π d sinθ / wavelength
so
d = ∅× wavelength / ( 2π sinθ )
put all value
d = 0.795 × 610× / ( 2π sin2.95 )
d = 405.32043 × m
spacing between the slits is 405.32043 × m
This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so:
<span>The current I is any motion of charge from one region to another, so this is given by:
</span>
The magnitude of the current density is:
Being:
<span>
Finally, for the drift velocity magnitude vd, we find:
</span>
Notice: The current I is very high for this wire. The given values of the variables are a little bit odd
<span>The current I is any motion of charge from one region to another, so this is given by:
</span>
The magnitude of the current density is:
Being:
<span>
Finally, for the drift velocity magnitude vd, we find:
</span>
Notice: The current I is very high for this wire. The given values of the variables are a little bit odd