Answer:
r = 0.05 m = 5 cm
Explanation:
Applying ampere's law to the wire, we get:
where,
r = distance of point P from wire = ?
μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²
I = current = 2 A
B = Magnetic Field = 8 μT = 8 x 10⁻⁶ T
Therefore,
<u>r = 0.05 m = 5 cm</u>
Since energy cannot be created nor destroyed, the change in energy of the electron must be equal to the energy of the emitted photon.
The energy of the emitted photon is given by:
where
h is the Planck constant
f is the photon frequency
Substituting
, we find
This is the energy given to the emitted photon; it means this is also equal to the energy lost by the electron in the transition, so the variation of energy of the electron will have a negative sign (because the electron is losing energy by decaying from an excited state, with higher energy, to the ground state, with lower energy)
By Newton's second law, the net vertical force acting on the object is 0, so that
<em>n</em> - <em>w</em> = 0
where <em>n</em> = magnitude of the normal force of the surface pushing up on the object, and <em>w</em> = weight of the object. Hence <em>n</em> = <em>w</em> = <em>mg</em> = 196 N, where <em>m</em> = 20 kg and <em>g</em> = 9.80 m/s².
The force of static friction exerts up to 80 N on the object, since that's the minimum required force needed to get it moving, which means the coefficient of <u>static</u> friction <em>µ</em> is such that
80 N = <em>µ</em> (196 N) → <em>µ</em> = (80 N)/(196 N) ≈ 0.408
Moving at constant speed, there is a kinetic friction force of 40 N opposing the object's motion, so that the coefficient of <u>kinetic</u> friction <em>ν</em> is
40 N = <em>ν</em> (196 N) → <em>ν</em> = (40 N)/(196 N) ≈ 0.204
And so the closest answer is C.
(Note: <em>µ</em> and <em>ν</em> are the Greek letters mu and nu)
Answer:
c) Very dangerous and users must not override devices designed to protect from exposures.
Explanation:
X-ray is a form of high energy electromagnetic radiation and are part of the electromagnetic spectrum.
X-ray radiation from diffraction and fluorescence instruments is very dangerous because of their high energy and wavelength.
Hence, users must not override devices designed to protect from exposures. The best shielding device to protect one from exposure is Lead.
Answer:
The field gets weaker
Explanation:
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