magnetic field due to a finite straight conductor is given by
here since it forms an equilateral triangle so we will have
also the perpendicular distance of the point from the wire is
now from the above equation magnetic field due to one wire is given by
now since in equilateral triangle there are three such wires so net magnetic field will be
Answer:
a) 298.5 nm
, 522.4 nm and b) radiation frequency does not change
Explanation:
When electromagnetic radiation reaches a medium with a different index of refraction, the medium vibrates the molecules, as if it were a resonance process, whereby the medium vibrates at the same frequency as the incident light.
On the other hand, when the light reaches another medium its average speed within the medium changes, it is now less than the speed of light in a vacuum (c) for this to happen as we saw that the frequency is constant there must be a change in the wavelength of the radiation that is characterized by the ratio
λₙ = λ₀ / n
λₙ = 400 nm in the void
λₙ = 400 / 1.34
λₙ= 298.5 nm
λ₀ = 700 nm
λₙ = 700 / 1.34
λₙ = 522.4 nm
The radiation frequency does not change
Answer:
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Answer:
Explanation:
All the displacement will be converted into vector, considering east as x axis and north as y axis.
5.3 km north
D = 5.3 j
8.3 km at 50 degree north of east
D₁= 8.3 cos 50 i + 8.3 sin 50 j.
= 5.33 i + 6.36 j
Let D₂ be the displacement which when added to D₁ gives the required displacement D
D₁ + D₂ = D
5.33 i + 6.36 j + D₂ = 5.3 j
D₂ = 5.3 j - 5.33i - 6.36j
= - 5.33i - 1.06 j
magnitude of D₂
D₂²= 5.33² + 1.06²
D₂ = 5.43 km
Angle θ
Tanθ = 1.06 / 5.33
= 0.1988
θ =11.25 ° south of due west.
Angstrom = 10^-10 m
for nucleus size are used fermi (femtometer 10^-15 m )
