An electron is accelerated through a potential difference of 3.1 kV and directed into a region between two parallel plates separ
1 answer:
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Explanation:
It is given that,
Spring constant of the spring, k = 15 N/m
Amplitude of the oscillation, A = 7.5 cm = 0.075 m
Number of oscillations, N = 31
Time, t = 15 s
(a) Let m is the mass of the ball. The frequency of oscillation of the spring is given by :
Total number of oscillation per unit time is called frequency of oscillation. Here,
m = 0.0895 kg
or
m = 89 g
(b) The maximum speed of the ball that is given by :
Hence, this is the required solution.
Answer:
λ = a
Explanation:
This is a diffraction exercise that is described by the expression
a sin θ = m λ
sin θ = m λ/ a
the first zero of the diffraction occurs for m = 1
sin θ = λ / a
angles are generally very small and are measured in radians
sin θ = θ = y / x
we substitute
the width of the central maximum is twice the distance to zero
w = 2y
in the exercise indicate that this width is equal to twice the distance to the screen (2x)
W = 2x
2y = 2x
we substitute
1 = λ/ a
λ = a
we see that the width of the slit is equal to the wavelength used.