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
Answer:
C. Alpha
Explanation:
Gamma radiation, unlike alpha or beta, does not consist of any particles, instead consisting of a photon of energy being emitted from an unstable nucleus. Having no mass or charge, gamma radiation can travel much farther through air than alpha or beta, losing (on average) half its energy for every 500 feet.
The solution you should use is Hooke's law: F=-kx
It should have the same signs because they repel due to the stretch of the spring.
a. Since there is a constant energy within the spring, then Hooke's law will determine the possible algebraic signs. The solution should be
<span>F = kx
270 N/m x 0.38 m = 102.6 N
</span>
b. Then use Coulomb's law; F=kq1q2/r^2 to find the charges produced in the force.
Adults. 18years olds and older were affected because back then only 21 years olds and older could vote but the 26th amendment changed that
