Answer:
a. metallic bond
b. the valence electrons from the s and p orbitals of the interacting metal atoms delocalize. That is to say, instead of orbiting their respective metal atoms, they form a “cloud” of electrons that surrounds the positively charged atomic nuclei of the interacting metal ions.
c. due to the presence of free electrons in its outer energy levels
Answer: A
Explanation: isotopes of the same thing element have the same number of protons in the nucleus but differ in the number of neutrons.
Diagram B .... light shines through at an angle
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Hope it is helpful
Answer:
![\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]](https://tex.z-dn.net/?f=%5Cvec%7BE%7D%20%3D%20%5Cfrac%7B%5Clambda%7D%7B2%5Cpi%5Cepsilon_0%7D%5B%5Cfrac%7B1%7D%7By%7D%28%5C%5Ey%29%20-%20%5Cfrac%7B1%7D%7Bx%7D%28%5C%5Ex%29%5D)
Explanation:
The electric field created by an infinitely long wire can be found by Gauss' Law.
For the electric field at point (x,y), the superposition of electric fields created by both lines should be calculated. The distance 'r' for the first wire is equal to 'y', and equal to 'x' for the second wire.
![\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) + \frac{-\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) - \frac{\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]](https://tex.z-dn.net/?f=%5Cvec%7BE%7D%20%3D%20%5Cvec%7BE%7D_1%20%2B%20%5Cvec%7BE%7D_2%20%3D%20%5Cfrac%7B%5Clambda%7D%7B2%5Cpi%5Cepsilon_0%20y%7D%28%5C%5Ey%29%20%2B%20%5Cfrac%7B-%5Clambda%7D%7B2%5Cpi%5Cepsilon_0%20x%7D%28%5C%5Ex%29%5C%5C%5Cvec%7BE%7D%20%3D%20%5Cfrac%7B%5Clambda%7D%7B2%5Cpi%5Cepsilon_0%20y%7D%28%5C%5Ey%29%20-%20%5Cfrac%7B%5Clambda%7D%7B2%5Cpi%5Cepsilon_0%20x%7D%28%5C%5Ex%29%5C%5C%5Cvec%7BE%7D%20%3D%20%5Cfrac%7B%5Clambda%7D%7B2%5Cpi%5Cepsilon_0%7D%5B%5Cfrac%7B1%7D%7By%7D%28%5C%5Ey%29%20-%20%5Cfrac%7B1%7D%7Bx%7D%28%5C%5Ex%29%5D)
