A group of protons and neutrons surrounded by electrons
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.
Answer:
a) 37.70 m/s
b)710.6 m/s²
Explanation:
Given that ;
Mass of object = 2 kg
Radius of the motion = 2m
Frequency of motion = 3 rev/s
The formula to apply is;
v= 2πrf where v is linear speed
v = 2×π×2×3 =12π = 37.70 m/s
Centripetal acceleration is given as;
a= 4×π²×r×f²
a= 4×π²×2×3²
a=710.6 m/s²
Answer:
present
Explanation:
read doesn't change but write is in present tense
Answer:
option A is correct because air friction is greater than gravity
Explanation:
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