You own a high speed digital camera that can take a picture every 0.5 seconds. You decide to take a picture every 0.5 seconds of
2 answers:
Answer:
here distance covered by drop in each picture is increasing
Explanation:
here as we know that drop is falling from certain height and we started to took the picture after every 0.5 seconds of time
now from the beginning it will fall by distance
here we have
now distance covered in next picture will be given as
then in next 5 second interval the distance will be
so here the distance is continuously increasing with time in each interval of 0.5 s
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car starts from rest
final speed attained by the car is
acceleration of the car will be
now the time to reach this final speed will be
so it required 1.39 s to reach this final speed
Answer:
λ = 102.78 nm
This radiation is in the UV range,
Explanation:
Bohr's atomic model for the hydrogen atom states that the energy is
E = - 13.606 / n²
where 13.606 eV is the ground state energy and n is an integer
an atom transition is the jump of an electron from an initial state to a final state of lesser emergy
ΔE = 13.606 (1 / - 1 / n_{i}^{2})
the so-called Lyman series occurs when the final state nf = 1, so the second line occurs when ni = 3, let's calculate the energy of the emitted photon
DE = 13.606 (1/1 - 1/3²)
DE = 12.094 eV
let's reduce the energy to the SI system
DE = 12.094 eV (1.6 10⁻¹⁹ J / 1 ev) = 10.35 10⁻¹⁹ J
let's find the wavelength is this energy, let's use Planck's equation to find the frequency
E = h f
f = E / h
f = 19.35 10⁻¹⁹ / 6.63 10⁻³⁴
f = 2.9186 10¹⁵ Hz
now we can look up the wavelength
c = λ f
λ = c / f
λ = 3 10⁸ / 2.9186 10¹⁵
λ = 1.0278 10⁻⁷ m
let's reduce to nm
λ = 102.78 nm
This radiation is in the UV range, which occurs for wavelengths less than 400 nm.