Answer:
It attracts ferrous materials
Explanation:
A magnet attracts ferrous materials A ferrous materials are metallic substances or conductors that can conduct heat and electricity. Example of this ferrous materials includes iron, metal etc. Since magnets only can attracts metallic substance to itself, then we can also conclude that they attract ferrous materials since ferrous materials. possesses properties of a metal.
Magnets possesses both north and south poles.
The same of the bar magnets are known to repel each other while unlike poles attract each other.
Answer:
E(final)/E(initial)=2
Explanation:
Applying the law of gauss to two parallel plates with charge density equal σ:
![E=\sigma/\epsilon_{o}=Q/(L^{2}*\epsilon_{o})\\](https://tex.z-dn.net/?f=E%3D%5Csigma%2F%5Cepsilon_%7Bo%7D%3DQ%2F%28L%5E%7B2%7D%2A%5Cepsilon_%7Bo%7D%29%5C%5C)
So, if the charge is doubled the Electric field is doubled too
E(final)/E(initial)=2
'A' is correct. B, C, and D are false statements.
The representation of this problem is shown in Figure 1. So our goal is to find the vector
![\overrightarrow{R}](https://tex.z-dn.net/?f=%5Coverrightarrow%7BR%7D)
. From the figure we know that:
![\left | \overrightarrow{A} \right |=12m \\ \\ \left | \overrightarrow{B} \right |=20m \\ \\ \theta_{A}=20^{\circ} \\ \\ \theta_{B}=40^{\circ}](https://tex.z-dn.net/?f=%5Cleft%20%7C%20%5Coverrightarrow%7BA%7D%20%5Cright%20%7C%3D12m%20%5C%5C%20%5C%5C%20%5Cleft%20%7C%20%5Coverrightarrow%7BB%7D%20%5Cright%20%7C%3D20m%20%5C%5C%20%5C%5C%20%5Ctheta_%7BA%7D%3D20%5E%7B%5Ccirc%7D%20%5C%5C%20%5C%5C%20%5Ctheta_%7BB%7D%3D40%5E%7B%5Ccirc%7D)
From geometry, we know that:
![\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}](https://tex.z-dn.net/?f=%5Coverrightarrow%7BR%7D%3D%5Coverrightarrow%7BA%7D%2B%5Coverrightarrow%7BB%7D)
Then using
vector decomposition into components:
![For \ A: \\ \\ A_x=-\left | \overrightarrow{A} \right |sin\theta_A=-12sin(20^{\circ})=-4.10 \\ \\ A_y=\left | \overrightarrow{A} \right |cos\theta_A=12cos(20^{\circ})=11.27 \\ \\ \\ For \ B: \\ \\ B_x=-\left | \overrightarrow{B} \right |cos\theta_B=-20cos(40^{\circ})=-15.32 \\ \\ B_y=-\left | \overrightarrow{B} \right |sin\theta_B=-20sin(40^{\circ})=-12.85](https://tex.z-dn.net/?f=For%20%5C%20A%3A%20%5C%5C%20%5C%5C%20A_x%3D-%5Cleft%20%7C%20%5Coverrightarrow%7BA%7D%20%5Cright%20%7Csin%5Ctheta_A%3D-12sin%2820%5E%7B%5Ccirc%7D%29%3D-4.10%20%5C%5C%20%5C%5C%20A_y%3D%5Cleft%20%7C%20%5Coverrightarrow%7BA%7D%20%5Cright%20%7Ccos%5Ctheta_A%3D12cos%2820%5E%7B%5Ccirc%7D%29%3D11.27%20%5C%5C%20%5C%5C%20%5C%5C%20For%20%5C%20B%3A%20%5C%5C%20%5C%5C%20B_x%3D-%5Cleft%20%7C%20%5Coverrightarrow%7BB%7D%20%5Cright%20%7Ccos%5Ctheta_B%3D-20cos%2840%5E%7B%5Ccirc%7D%29%3D-15.32%20%5C%5C%20%5C%5C%20B_y%3D-%5Cleft%20%7C%20%5Coverrightarrow%7BB%7D%20%5Cright%20%7Csin%5Ctheta_B%3D-20sin%2840%5E%7B%5Ccirc%7D%29%3D-12.85)
Therefore:
![R_x=A_x+B_x=-4.10-15.32=-19.42m \\ \\ R_y=A_y+B_y=11.27-12.85=-1.58m](https://tex.z-dn.net/?f=R_x%3DA_x%2BB_x%3D-4.10-15.32%3D-19.42m%20%5C%5C%20%5C%5C%20R_y%3DA_y%2BB_y%3D11.27-12.85%3D-1.58m)
So if you want to find out <span>
how far are you from your starting point you need to know the magnitude of the vector
![\overrightarrow{R}](https://tex.z-dn.net/?f=%5Coverrightarrow%7BR%7D)
, that is:
</span>
![\left | \overrightarrow{R} \right |= \sqrt{R_x^2+R_y^2}=\sqrt{(-19.42)^2+(-1.58)^2}=\boxed{19.48m}](https://tex.z-dn.net/?f=%5Cleft%20%7C%20%5Coverrightarrow%7BR%7D%20%5Cright%20%7C%3D%0A%5Csqrt%7BR_x%5E2%2BR_y%5E2%7D%3D%5Csqrt%7B%28-19.42%29%5E2%2B%28-1.58%29%5E2%7D%3D%5Cboxed%7B19.48m%7D)
Finally, let's find the <span>
compass direction of a line connecting your starting point to your final position. What we are looking for here is an angle that is shown in Figure 2 which is an angle defined with respect to the positive x-axis. Therefore:
</span>