How much power is necessary to do 50 J of work in 5 seconds
2 answers:
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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
Answer:
1.
2.
3.
Explanation:
Given:
- mass of slinky,
- length of slinky,
- amplitude of wave pulse,
- time taken by the wave pulse to travel down the length,
- frequency of wave pulse,
1.
2.
<em>Now, we find the linear mass density of the slinky.</em>
We have the relation involving the tension force as:
3.
We have the relation for wavelength as: