Answer:
E = 1,873 10³ N / C
Explanation:
For this exercise we can use Gauss's law
Ф = E. dA = 
/ ε₀
Where q_{int} is the charge inside an artificial surface that surrounds the charged body, in this case with the body it has a spherical shape, the Gaussian surface is a wait with radius r = 1.35 m that is greater than the radius of the sphere.
The field lines of the sphere are parallel to the radii of the Gaussian surface so the scald product is reduced to the algebraic product.
The surface of a sphere is
A = 4π r²
E 4π r² = q_{int} /ε₀
The net charge within the Gauussian surface is the charge in the sphere of q1 = + 530 10⁻⁹ C and the point charge in the center q2 = -200 10⁻⁹ C, since all the charge can be considered in the center the net charge is
q_{int} = q₁ + q₂
q_{int} = (530 - 200) 10⁻⁹
q_{int} = 330 10⁻⁹ C
The electric field is
E = 1 / 4πε₀ q_{int} / r²
k = 1 / 4πε₀
E = k q_{int}/ r²
Let's calculate
E = 8.99 10⁹ 330 10⁻⁹/ 1.32²
E = 1,873 10³ N / C
Answer:
Explanation:
This is a simple Law of Momentum Conservation problem of the inelastic type. The equation for this is
![[m_1v_1+m_2v_2]_b=[(m_1+m_2)v]_a](https://tex.z-dn.net/?f=%5Bm_1v_1%2Bm_2v_2%5D_b%3D%5B%28m_1%2Bm_2%29v%5D_a)
Filling in:
![[1200(4.5)+2100(0)]=[(1200+2100)v]](https://tex.z-dn.net/?f=%5B1200%284.5%29%2B2100%280%29%5D%3D%5B%281200%2B2100%29v%5D)
which simplifies to
5400 + 0 = 3300v
so v = 1.6 m/s to the east, choice B
Answer:
oil floats in water due to the fact that it is less dense. since the water is more dense, it is basically heavier, and heavier items always sink, while lighter items float, the same can be said about liquids. denser liquids will sink while less dense items will float.
Answer:
Kinetic - a box moving
Thermal - sand that feels warm
Electrical - lightning
Radiant - radio waves
Gravitational Potential - fruit hanging from a tree
An electron shell can hold 2(n^2) electrons (technically) where n is the shell number, i.e. shell 1 can hold 2, shell 2 can hold 8, 3 holds 18 and so on.
The atomic number of Nitrogen is 7, i.e. it has 7 electrons (to match its 7 protons, assuming it isn't an ion).
With the atomic number, you simply start from shell 1 and work out. So we put 2 electrons in shell 1, leaving us with 5 left. Shell 2 can hold 6 so we can fit all 5 in.
In other words, you should have 2 electron shells on the atom, shell 1 with 2 e- and shell 2 with 5 e-.
