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.
Meters per second squared:
If you think about it, acceleration is about how fast speed changes. Speed is measured in meters per second:
So if you take that and just measure it over time, you get meters per second squared.
Answer:
D . Sound energy
Explanation:
When the strings of a violin vibrate it produces sound which is sound energy. Due to the vibration of the strings the air present near the strings also vibrate in resonance with the strings. This compreesion and decompression's produced in the air is nothing but the sound. So the form of energy given off by the vibrating strings of the violin is Sound energy.
Answer:
100 cm³
Explanation:
Use ideal gas law:
PV = nRT
where P is absolute pressure, V is volume, n is number of moles, R is ideal gas constant, and T is absolute temperature.
n and R are constant, so:
P₁V₁/T₁ = P₂V₂/T₂
If we say point 1 is at 40m depth and point 2 is at the surface:
P₂ = 1.013×10⁵ Pa
T₂ = 20°C + 273.15 = 293.15 K
P₁ = ρgh + P₂
P₁ = (1000 kg/m³ × 9.8 m/s² × 40 m) + 1.013×10⁵ Pa
P₁ = 4.933×10⁵ Pa
T₁ = 4.0°C + 273.15 = 277.15 K
V₁ = 20 cm³
Plugging in:
(4.933×10⁵ Pa) (20 cm³) / (277.15 K) = (1.013×10⁵ Pa) V₂ / (293.15 K)
V₂ = 103 cm³
Rounding to 1 sig-fig, the bubble's volume at the surface is 100 cm³.
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