(a)
The relationship between frequency and wavelength of an electromagnetic wave is given by
where
is the speed of light
is the frequency
is the wavelength
In this problem, we are considering light with wavelength of
Substituting into the equation and re-arranging it, we can find the corresponding frequency:
(b)
The period of a wave is equal to the reciprocal of the frequency:
And using as we found in the previous part, we can find the period of this wave:
Answer: Yes.
Explanation:
Oxygen has a bond order of two. The bond order of Oxygen molecule is calculated, where the [<em>eight valence electrons in bonding molecular orbitals</em> minus (-) <em>four valence electrons in antibonding molecular orbitals</em>]/2 in the electron configuration.
Atoms/molecules where electrons are paired are diamagnetic (repelled by both poles of a magnetic); while atoms/molecules that have one or more unpaired electrons are paramagnetic (attracted to magnetic field).
The two unpaired electrons of dioxygen molecules has made it <u>paramagnetic</u>. By pouring liquid oxygen between the poles of a strong magnet, the liquid stream will be contained by the filed and fills up the space between the poles.
Answer:
a) 0 b) 29.9 N c) 21.7 N d) -17.4 N e) -13.0 N f) -17.4 N g) 16.9 N
h) 24.3 N θ = 44.2º
Explanation:
a) As the electric force is exerted along the line that joins the charges, due to any of the charges A or C has non-zero x-coordinates, the force has no x components either.
So, Fcax = 0
b) Similarly, as Fx = 0, the entire force is directed along the y-axis, and is going upward, due both charges repel each other.
Fyca = k*qa*qc / rac² = (9.10⁹ N*m²/C²*(2.79)*(1.07)*10⁻⁸ C²) / 9.00 m²
Fyca = 29. 9 N
c) In order to get the magnitude of the force exerted by B on C, we need to know first the distance between both charges:
rbc² = (3.00 m)² + (4.00m)² = 25.0 m²
⇒ Fbc = k*qb*qc / rbc² = (9.10⁹ N*m²/C²*(5.64)*(1.07)*10⁻⁸ C²) / 25.0 m²
⇒ Fbc = 21.7 N
d) In order to get the x component of Fbc, we need to get the projection of Fcb over the x axis, taking into account that the force on particle C is attractive, as follows:
Fbcₓ = Fbc * cos (-θ) where θ, is the angle that makes the line of action of the force, with the x-axis, so we can write:
cos θ = x/r = 4.00 / 5.00 m =
Fcbx = 21.7*(-0.8) = -17.4 N
e) The y component can be calculated in the same way, projecting the force over the y-axis, as follows:
Fcby = Fcb* sin (-θ) = 21.7* (-3.00/5.00) = -13.0 N
f) The sum of both x components gives :
Fcx = 0 + (-17.4 N) = -17.4 N
g) The sum of both y components gives :
Fcy = 29.9 N + (-13.0 N) = 16.9 N
h) The magnitude of the resultant electric force acting on C, can be found just applying Pythagorean Theorem, as follows:
Fc = √(Fcx)²+(Fcy)² = (17.4)² + (16.9)² = 24.3 N
The angle from the horizontal can be found as follows:
Ф = arc tg (16.9 / 17.4) = 44.2º