When the frequency decreases the wavelength is further apart. When it increases its closer together. Think about a flat line when the frequency is low the wavelengths are wider. When its a high frequency the squiggly lines on the moniter are taller and thinner so the wavelengths are not as wide and not that far from each other depending on how high the frequency is.
By using the Plancks-Einstein equation, we can find the energy;
E = hf
where h is the plancks constant = 6.63 x 10⁻³⁴
f = frequency = 3.55 x 10¹⁷hz
E = (6.63 x 10⁻³⁴) x (3.55 x 10¹⁷)
E = 2.354 x 10⁻¹⁶J
Answer:
5) Displacement = +3.125 m
Displacement is in the same direction as the force vector.
6) Force = -53.89 N
Force is in an opposite direction relative to the displacement.
Explanation:
5) We are given;
Force; F = 160 N.
Workdone; W = +500 J
Now, formula for workdone is;
W = Force × displacement
Thus, displacement = Work/force
Displacement = 500/160
Displacement = +3.125 m
Thus, displacement is in the same direction as the force vector.
6) We are given;
Displacement; d = 18 m.
Workdone; W = -970 J
Like in the first answer above,
Workdone = Force × Displacement
Thus;
Force = Workdone/Displacement
Force = -970/18
Force = -53.89 N
Since force is negative and displacement is positive, it means force is in an opposite direction relative to the displacement.
Answer:

Explanation:
The force on the point charge q exerted by the rod can be found by Coulomb's Law.

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.
In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.
We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.
Applying Coulomb's Law:

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.
Now, we have to write 'dq' in term of the known quantities.

Now, substitute this into 'dF':

Now we can integrate dF over the rod.
