Answer:
<em>D. The total force on the particle with charge q is perpendicular to the bottom of the triangle.</em>
Explanation:
The image is shown below.
The force on the particle with charge q due to each charge Q = 
we designate this force as N
Since the charges form an equilateral triangle, then, the forces due to each particle with charge Q on the particle with charge q act at an angle of 60° below the horizontal x-axis.
Resolving the forces on the particle, we have
for the x-component
= N cosine 60° + (-N cosine 60°) = 0
for the y-component
= -f sine 60° + (-f sine 60) = -2N sine 60° = -2N(0.866) = -1.732N
The above indicates that there is no resultant force in the x-axis, since it is equal to zero (
= 0).
The total force is seen to act only in the y-axis, since it only has a y-component equivalent to 1.732 times the force due to each of the Q particles on q.
<em>The total force on the particle with charge q is therefore perpendicular to the bottom of the triangle.</em>
It typically take longer for a heavier object to slow down therefor, a train will take more time. <span />
Answer:
Mass number - ⦁ The number of protons and neutrons in the nucleus of an atom.
Isotopes - ⦁ Atoms with the same number of protons, but different number of neutrons.
Nitrogen - ⦁ The name of the element with atomic number 7.
Atomic number - ⦁ The number of protons in the nucleus of an atom.
Answer:
FB = 0.187 N
Explanation:
To find the magnetic force FB in the wire you use the following formula:

the angle between B and L is given by:

Due to B depends on "y" you take into account the contribution of each element dy of the wire to the magnitude of the magnetic force. Thus, you have to integrate the following expression:
![|\vec{F_B}|=Isin\theta\int_0^{0.25}B(y)dy=Isin\theta\int_0^{0.25}(0.5y)dy\\\\|\vec{F_B}|=(2.0*10^{-3}A)(sin36.86\°)(0.5T)[\frac{0.25^2}{2}m]=0.187\ N](https://tex.z-dn.net/?f=%7C%5Cvec%7BF_B%7D%7C%3DIsin%5Ctheta%5Cint_0%5E%7B0.25%7DB%28y%29dy%3DIsin%5Ctheta%5Cint_0%5E%7B0.25%7D%280.5y%29dy%5C%5C%5C%5C%7C%5Cvec%7BF_B%7D%7C%3D%282.0%2A10%5E%7B-3%7DA%29%28sin36.86%5C%C2%B0%29%280.5T%29%5B%5Cfrac%7B0.25%5E2%7D%7B2%7Dm%5D%3D0.187%5C%20N)
hence, the magnitude of the magnetic force is 0.187N