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:
Potential energy. Releasing it, the potential energy would convert into motion, kinetic energy.
Potential energy is when an object has some sort of potential eg. for motion such as in this example.
Answer:
Approximately 
.
Explanation:
Since the result needs to be accurate to three significant figures, keep at least four significant figures in the calculations.
Look up the Rydberg constant for hydrogen: 
.
Look up the speed of light in vacuum: 
.
Look up Planck's constant: 
.
Apply the Rydberg formula to find the wavelength 
(in vacuum) of the photon in question:
.
The frequency of that photon would be:
.
Combine this expression with the Rydberg formula to find the frequency of this photon:
.
Apply the Einstein-Planck equation to find the energy of this photon:
.
(Rounded to three significant figures.)
You want v2 = v1 + at
v is measured in m/s, a in m/s2, and t in s.
the dimensions multiply like algebraic quantities.
so because v2 is measured in m/s, then (v1 + at) has to come out in m/s
the units for (v1 + at) are (m/s) + (m/s2)(s)
time "s" cancels out one acceleration "s", so it comes ut to (m/s) + (m/s), which = (m/s).
if you had (v1t + a), then you would have (m/s)(s) + (m/s2) which = (m) + (m/s2), which doesn't work.
