Answer:
the reflected wave is inverted and the transmitted wave is up
Explanation:
To answer this question we must analyze the physical phenomenon, with an wave reaching a discontinuity, we can analyze it as a shock.
Let's start when the discontinuity is with a fixed, very heavy and rigid obstacle, in this case the reflected wave is inverted, since the contact point cannot move
In the event that it collides with an object that can move, the reflected wave is not inverted, this is because the point can rise, they form a maximum at this point.
In the proposed case the shock is when the thickness changes, in this case we have the above phenomena, a part of the wave is reflected by being inverted and a part of the wave is transmitted without inverting.
The amplitude sum of the amplitudes of the two waves is proportional to the lanería that is distributed between them.
When checking the answers the correct one is the reflected wave is inverted and the transmitted wave is up
Answer:
A first-class lever: fulcrum is between input and output force; second-class lever: output force is between input force and fulcrum; third-class lever: input force is between fulcrum and output force
In addition to acceleration of gravity we experience centrifugal acceleration away from the axis of rotation of the earth. this additional acceleration has value ac = r w^2 where w = angular velocity and r is distance from your spot on earth to the earth's axis of rotation so r = R cos(l) where l = 60 deg is the lattitude and R the earth's radius and w = 1 / (24hr x 3600sec/hr)
<span>now you look up R and calculate ac then you combine the centrifugal acc. vector ac with the gravitational acceleration vector ag = G Me/R^2 to get effective ag' = ag -</span>
well it would be A because 55 degrees is going strait well 75 is going literally straight up
Answer:
A.2.95 m
B.7
Explanation:
We are given that
Diffraction grating=600 lines/mm
d=
Wavelength of light,
l=4.6 m
A.We have to find the distance between the two m=1 bright fringes
For first bright fringe, =1
The distance between two m=1 fringes
Hence, the distance between two m=1 fringes=2.95 m
B.For maximum number of fringes,
Substitute the values
Maximum number of bright fringes on the scree=
