Explanation:
Draw a free body diagram for each disc.
Disc A has three forces acting on it: 86.5 N up, T₁ down, and Wa down.
∑F = ma
86.5 N − T₁ − Wa = 0
Wa = 86.5 N − T₁
ma × 9.8 m/s² = 86.5 N − 55.6 N
ma = 3.2 kg
Disc B has three forces acting on it: T₁ up, T₂ down, and Wb down.
∑F = ma
T₁ − T₂ − Wb = 0
Wb = T₁ − T₂
mb × 9.8 m/s² = 55.6 N − 36.5 N
mb = 1.9 kg
Disc C has three forces acting on it: T₂ up, T₃ down, and Wc down.
∑F = ma
T₂ − T₃ − Wc = 0
Wc = T₂ − T₃
mc × 9.8 m/s² = 36.5 N − 9.6 N
mc = 2.7 kg
Disc D has two forces acting on it: T₃ up and Wd down.
∑F = ma
T₃ − Wd = 0
Wd = T₃
md × 9.8 m/s² = 9.6 N
md = 0.98 kg
A pendulum is not a wave.
-- A pendulum doesn't have a 'wavelength'.
-- There's no way to define how many of its "waves" pass a point
every second.
-- Whatever you say is the speed of the pendulum, that speed
can only be true at one or two points in the pendulum's swing,
and it's different everywhere else in the swing.
-- The frequency of a pendulum depends only on the length
of the string from which it hangs.
If you take the given information and try to apply wave motion to it:
Wave speed = (wavelength) x (frequency)
Frequency = (speed) / (wavelength) ,
you would end up with
Frequency = (30 meter/sec) / (0.35 meter) = 85.7 Hz
Have you ever seen anything that could be described as
a pendulum, swinging or even wiggling back and forth
85 times every second ? ! ? That's pretty absurd.
This math is not applicable to the pendulum.
Answer:
Explanation:
It would actually be A. 30 , as each hour of ascension (i am not sure about the correct terminology) equals 15 .
Answer
given,
angle between two polarizing filters = 45°
filter reduce intensity = ?
a) I = I₀ Cos² θ
here θ = 45⁰
intensity of the light is reduced by 0.500
correct answer from the given option D
b) direction of the polarization
θ = 45°