Answer: A symbolic expression for the net force on a third point charge +Q located along the y axis
![F_N=k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}](https://tex.z-dn.net/?f=F_N%3Dk_e%5Cfrac%7BQ%5E2%7D%7Bd%5E2%7D%5Ctimes%20%5Csqrt%7B%5B4%2B%5Cfrac%7B1%7D%7B4%7D-%5Csqrt%7B2%7D%5D%7D)
Explanation:
Let the force on +Q charge y-axis due to +2Q charge be
and force on +Q charge y axis due to -Q charge on x-axis be
.
Distance between the +2Q charge and +Q charge = d units
Distance between the -Q charge and +Q charge =
units
= Coulomb constant


Net force on +Q charge on y-axis is:




![|F_N|=|k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}|](https://tex.z-dn.net/?f=%7CF_N%7C%3D%7Ck_e%5Cfrac%7BQ%5E2%7D%7Bd%5E2%7D%5Ctimes%20%5Csqrt%7B%5B4%2B%5Cfrac%7B1%7D%7B4%7D-%5Csqrt%7B2%7D%5D%7D%7C)
The net froce on the +Q charge on y-axis is
![F_N=k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}](https://tex.z-dn.net/?f=F_N%3Dk_e%5Cfrac%7BQ%5E2%7D%7Bd%5E2%7D%5Ctimes%20%5Csqrt%7B%5B4%2B%5Cfrac%7B1%7D%7B4%7D-%5Csqrt%7B2%7D%5D%7D)
Answer:Actually the wavelength is proportional to its speed and frequency. therefore the wavelength has no affect on the speed or frequency its the other way around.
Explanation:
Answer:
The answer is the shirt would disappear because I would eat him
To solve this problem we will apply the concepts related to Frequency (Reverse to Period) and the description of the wavelength as a function of the speed of light at the rate of frequency.
Our Laser frequency is given as

Therefore the laser wavelength would be

Where,
c = Speed of light
f = Frequency


The laser pulse is emitted at a period (T) of 
Therefore the pulse wavelength would be





Finally the number of wavelengths is the ratio between the two wavelengths, then



The number of wavelengths in the beam length is closer to
Answer:
(a) The motion is uniform
(b) 11.11 m/s
Explanation:
(a)
From the table below, the motion of the bus is uniform.
(b)
Speed(s) = Δd/Δt
s = Δd/Δt............. Equation 1
From the table,
Given: Δd = 10 km = 10000 m, Δt = 15 minutes = (15×60) = 900 seconds
Substitute these values into equation 1
s = 10000/900
s = 11.11 m/s