Answer:
the frequency of this mode of vibration is 138.87 Hz
Explanation:
Given;
length of the copper wire, L = 1 m
mass per unit length of the copper wire, μ = 0.0014 kg/m
tension on the wire, T = 27 N
number of segments, n = 2
The frequency of this mode of vibration is calculated as;

Therefore, the frequency of this mode of vibration is 138.87 Hz
Answer:
Velocity of Pauli relative to Daniel = (-1.50ï + 3.90ĵ) m/s
x-component = -1.50 m/s
y-component = 3.90 m/s
Explanation:
Relative velocity of a body A relative to another body B, Vab, is given as
Vab = Va - Vb
where
Va = Relative velocity of body A with respect to another third body or frame of reference C
Vb = Relative velocity of body B with respect to that same third body or frame of reference C.
So, relative velocity can be given further as
Vab = Vac - Vbc
Velocity of Newton relative to Daniel = Vnd = 3.90 m/s due north = (3.90ĵ) m/s in vector form.
Velocity of Newton relative to Pauli = Vnp = 1.50 m/s due East = (1.50î) m/s in vector form
What is Pauli's velocity relative to Daniel?
Vpd = Vp - Vd
(Pauli's velocity relative to Daniel) = (Pauli's velocity relative to Newton) - (Daniel's velocity relative to Newton)
Vpd = Vpn - Vdn
Vpn = -Vnp = -(1.50î) m/s
Vdn = -Vnd = -(3.90ĵ) m/s
Vpd = -1.50î - (-3.90ĵ)
Velocity of Pauli relative to Daniel = (-1.50ï + 3.90ĵ) m/s
Hope this Helps!!!!
Answer:

Explanation:
Given that
Wavelength λ=192 nm
So energy of photon,E

Now by putting the values




We know that
Kinetic energy given as




As per the question the wavelength of spectral line of the neutral hydrogen atom is 21 cm.
Spectral line of hydrogen atom will be observed when an electron will jump from higher excited to the ground state. If E is the energy of the ground state and E' is the energy of the higher excited state,the energy emitted when electron will jump from higher excited to ground state is E'-E.
As per Planck's quantum theory -

Where f is the frequency of emitted radiation and h is the Planck's constant.Actually this energy emitted is also electromagnetic in nature. We know that all the electromagnetic radiation will move with the same velocity which is equal to the velocity of light.
The velocity of light c= 3×10^8 m/s.Hence the velocity of wave corresponding to this spectral line is 3×10^8 m/s.
We know that velocity of a wave is the product of frequency and wavelength of that corresponding wave. Mathematically we can write it as-

[
is the wavelength of the wave ]
It is given that wavelength =21 cm=0.21 m
The frequency is calculated as -




Hence the antenna will be tuned to a frequency of 14.2857×10^8 Hz
Here Hertz[ Hz] is the unit of frequency.