The solution for this problem is:
For 1st minimum, let m be equal to 1.
d = slit width
D = screen distance.
Θ = arcsin (m * lambda/ (d))
= 0.13934 rad, 7.9836 deg
y = D*tan (Θ)
y = 6.50 * tan (7.9836)
= 0.91161 m is the distance from the central maximum to the first-order minimum
Answer:
31.831 Hz.
Explanation:
<u>Given:</u>
The vertical displacement of a wave is given in generalized form as

<em>where</em>,
- A = amplitude of the displacement of the wave.
- k = wave number of the wave =

= wavelength of the wave.- x = horizontal displacement of the wave.
= angular frequency of the wave =
.- f = frequency of the wave.
- t = time at which the displacement is calculated.
On comparing the generalized equation with the given equation of the displacement of the wave, we get,

therefore,

It is the required frequency of the wave.
Answer:
Explained
Explanation:
Newton would resort to the classical mechanics and say that the momentum of the particle that is moving with a constant velocity will be given by: momentum = mass x velocity
this approach will highlight the particle nature and will not be relativistic.
De-Broglie will say that the momentum of the particle is related to its associated matter wave and the relation between them is given by:

where \lambda = wavelength of the matter wave associated to the particle, h = planck's constant
and
thus, this highlights the wave nature of the particle and is also relativistic.
-17.555m/s
first I found the time it took for jacks stone to reach the bottom, using the formula vf = vi + at, vf and vi are final and initial velocities.
then i found the velocity at 6.6m using vf^2 = vi^2 + 2ad
and I found the time it took to get to 6.6m, so that I knew how long Jill waited to throw her stone, I used the formula d = t(vi+vf)/2, then i done total time - the time she waited, to get the time it took for there stones to hit the ground at the same time.
then to find the initial velocity of her throw I used the formula d = vit + (at^2)/2