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Answer:
Explanation:
Given an LC circuit
Frequency of oscillation
f = 299 kHz = 299,000 Hz
AT t = 0 , the plate A has maximum positive charge
A. At t > 0, the plate again positive charge, the required time is
t =
t = 1 / f
t = 1 / 299,000
t = 0.00000334448 seconds
t = 3.34 × 10^-6 seconds
t = 3.34 μs
it will be maximum after integral cycle t' = 3.34•n μs
Where n = 1,2,3,4....
B. After every odd multiples of n, other plate will be maximum positive charge, at time equals
t" = ½(2n—1)•t
t'' = ½(2n—1) 3.34 μs
t" = (2n —1) 1.67 μs
where n = 1,2,3...
C. After every half of t,inductor have maximum magnetic field at time
t'' = ½ × t'
t''' = ½(2n—1) 1.67μs
t"' = (2n —1) 0.836 μs
where n = 1,2,3...
Answer:
The maximum wavelength of light that could liberate electrons from the aluminum metal is 303.7 nm
Explanation:
Given;
wavelength of the UV light, λ = 248 nm = 248 x 10⁻⁹ m
maximum kinetic energy of the ejected electron, K.E = 0.92 eV
let the work function of the aluminum metal = Ф
Apply photoelectric equation:
E = K.E + Ф
Where;
Ф is the minimum energy needed to eject electron the aluminum metal
E is the energy of the incident light
The energy of the incident light is calculated as follows;
The work function of the aluminum metal is calculated as;
Ф = E - K.E
Ф = 8.02 x 10⁻¹⁹ - (0.92 x 1.602 x 10⁻¹⁹)
Ф = 8.02 x 10⁻¹⁹ J - 1.474 x 10⁻¹⁹ J
Ф = 6.546 x 10⁻¹⁹ J
The maximum wavelength of light that could liberate electrons from the aluminum metal is calculated as;
Answer:
in a transverse wave, The directions that energy and matter travel in are perpendicular to one another.
these waves are types of mechanical waves that doesn't require any material medium to transfer.
Answer:
Not 500 points I got 18 but thx
Explanation:
