If a 1.0-kg-mass block is on the left cap, how much total mass must be placed on the right cap so that the caps equilibrate at e
1 answer:
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Answer:
λ = 5.656 x 10⁻⁷ m = 565.6 nm
Explanation:
Using the formula of fringe spacing from the Young's Double Slit experiment, which is given as follows:
where,
λ = wavelength = ?
Δx = fringe spacing = 1.6 cm = 0.016 m
L = Distance between slits and screen = 4.95 m
d = slit separation = 0.175 mm = 0.000175 m
Therefore,
<u>λ = 5.656 x 10⁻⁷ m = 565.6 nm</u>
Answer:
Explanation:
m = Mass of object =
mg = Weight of object = 20 N
g = Acceleration due to gravity =
v = Final velocity = 15 m/s
u = Initial velocity = 0
d = Distance moved by the object = 150 m
= Angle of slope =
f = Force of friction
fd = Work done against friction
The force balance of the system is
The work done against friction is .