The greater the amplitude, the greater the wave's:
1 answer:
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Answer:
f ’= 97.0 Hz
Explanation:
This is an exercise of the doppler effect use the frequency change due to the relative movement of the fort and the observer
in this case the source is the police cases that go to vs = 160 km / h
and the observer is vo = 120 km / h
the relationship of the doppler effect is
f ’= f₀ (v + v₀ / v- )
let's reduce the magnitude to the SI system
v_{s} = 160 km / h (1000 m / 1km) (1h / 3600s) = 44.44 m / s
v₀ = 120 km / h (1000m / 1km) (1h / 3600s) = 33.33 m / s
we substitute in the equation of the Doppler effect
f ‘= 100 (330+ 33.33 / 330-44.44)
f ’= 97.0 Hz
Answer:
i. The radius 'r' of the electron's path is 4.23 × m.
ii. The frequency 'f' of the motion is 455.44 KHz.
Explanation:
The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.
r =
Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.
From the question, B = 1.63 × T, v = 121 m/s, Θ =
(since it enters perpendicularly to the field), q = e = 1.6 ×
C and m = 9.11 ×
Kg.
Thus,
r = ÷ sinΘ
But, sinΘ = sin = 1.
So that;
r =
= (9.11 × × 121) ÷ (1.6 ×
× 1.63 ×
)
= 1.10231 × ÷ 2.608 ×
= 4.2266 ×
= 4.23 × m
The radius 'r' of the electron's path is 4.23 × m.
B. The frequency 'f' of the motion is called cyclotron frequency;
f =
= (1.6 × × 1.63 ×
) ÷ (2 ×
× 9.11 ×
)
= 2.608 × ÷ 5.7263 ×
= 455442.4323
f = 455.44 KHz
The frequency 'f' of the motion is 455.44 KHz.