In Millikan's oil-drop experiment, one looks ata small oil drop held motionless between two plates. Take the voltage betwee the
1 answer:
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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.
Answer:
0.114 kg or 114 g
Explanation:
From the diagram attaches,
Taking the moment about the fulcrum,
sum of clockwise moment = sum of anticlockwise moment.
Wd = W'd'
Where W = weight of the mass, W' = weight of the meter rule, d = distance of the mass from the fulcrum, d' = distance of the meter rule.
make W' the subject of the equation
W' = Wd/d'................ Equation 1
Given: W = mg = 0.0515(9.8) = 0.5047 N, d = (39.2-16) = 23.2 cm, d' = (49.7-39.2) = 10.5 cm
Substitute these values into equation 1
W' = 0.5047(23.2)/10.5
W' = 1.115 N.
But,
m' = W'/g
m' = 1.115/9.8
m' = 0.114 kg
m' = 114 g
