The zinc plate is coated with mercury
1 answer:
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To solve this problem it is necessary to apply the concepts related to the principle of superposition and constructive interference, that is to say everything that refers to an overlap of two or more equal frequency waves, which when interfering create a new pattern of waves of greater intensity (amplitude) whose cusp is the antinode.
Mathematically its definition can be given as:
Where
d = Width of the slit
Angle between the beam and the source
m = Order (any integer) which represent the number of repetition of the spectrum, at this case 1 (maximum respect the wavelength)
Since the point of the theta angle for which the diffraction becomes maximum will be when it is worth one then we have to:
Applying the given relation of frequency, speed and wavelength then we will have that the frequency would be:
Here the velocity is equal to the speed of light and the wavelength to the value previously found.
Therefore the smallest microwave frequency for which only the central maximum occurs is 1.5Ghz
Answer:
a) k = Mg / d
, b) v = √2gh
, c) v_{f} = , d) x² + 6d x -
dh = 0
e)the spring must compress a greater distance.
Explanation:
a) when the block of mass M is placed on the spring, we have an equilibrium condition,
∑ F = 0
- W = 0
k d = Mg
k = Mg / d
b) let's use the concepts of energy to find the velocity of the block just before the collision
starting point. Position when released
Em₀ = U = m g h
lowest point. Right at the point of shock
= K = ½ m v²2
as there is no friction, energy is conserved
Em₀ = Em_{f}
mg h = ½ m v²
v = √2gh
c) The velocity of the two blocks after the collision, we define a system formed by the two blocks, in such a way that the forces during the collision are internal and the moment is conserved
initial instant. Just before the crash
p₀ = 2M v + M 0
final instant. Just after the shock, before the spring compression begins
p_{f} = (2M + M) v_{f}
the moment is preserved
p₀ = p_{f}
2M v = 3M v_{f}
v_{f} = ⅔ v
v_{f} =
d) now we work with the joined system after the collision, let's use the concepts of energy
starting point. After shock, before beginning spring compression
Em₀ = K = ½ (3M)
Em₀ = 3/2 M (\frac{2}{3} \ \sqrt{2gh})²
Em₀ = 4/3 M gh
final point. With the spring fully compressed
Em_f = K_e + U = ½ k x² + (3M) g x
in this case we have taken the zero of gravitational potential energy at the point where the blocks collide, as there is no friction, the energy is conserved
Em₀ = Em_f
4/3 M g h = ½ k x² + 3M g x
½ k x² + 3Mg x - 4/3 Mgh = 0
we substitute the expression for k
(
) x² + 3Mg x -
Mgh = 0
+ 3 x -
h = 0
to find the value of the spring compression, the second degree equation must be solved
x² + 6d x - dh = 0
x = [-6d ± ] / 2
x = [-6d ± 6d ]/2
x = 3d ( -1± )
e) If the collision elastic force would not lose any part of the kinetic energy during the collision, therefore the speed of the block of mass M would be much higher and therefore the spring must compress a greater distance.