Answer:
7.2 cm
Explanation:
magnetic field, B = 0.301 T
speed, v = 7.92 x 10^5 m/s
mass, m = 4.39 x 10^-27 kg
q = 1.6 x 10^-19 C
The radius of singly changed ion is given by

where, m is the mass of ion, v be the speed of ion, B is the magnetic field and q be the charge

r = 0.072 m
r = 7.2 cm
Answer:
1. The period is 1.74 s.
2. The frequency is 0.57 Hz
Explanation:
1. Determination of the the period.
Spring constant (K) = 30 N/m
Mass (m) = 2.3 Kg
Pi (π) = 3.14
Period (T) =?
The period of the vibration can be obtained as follow:
T = 2π√(m/K)
T = 2 × 3.14 × √(2.3 / 30)
T = 6.28 × √(2.3 / 30)
T = 1.74 s
Thus, the period of the vibration is 1.74 s.
2. Determination of the frequency.
Period (T) = 1.74 s
Frequency (f) =?
The frequency of the vibration can be obtained as follow:
f = 1/T
f = 1/1.74
f = 0.57 Hz
Thus, the frequency of the vibration is 0.57 Hz
Answer:
True

Explanation:
As we know that total charge is always quantized and in the terms of multiples of charge of single electron
so we have
Q = Ne
here we know that

also we know that

so we have


Since this charge is in multiple of micro coulomb units so yes it is possible to measure it
and the magnitude of this charge is given as
