Answer:
Its D Kr and Xe
Explanation:
Element from the same group have simmilar properties
If you now the atomic number ،its so easy to know how much electrons they have in outermost shell.
We know that d=m/v where d is density, m is mass and v is volume. Lets transform formula to get v in times of d and m
![d= \frac{m}{V} /*V \\ V*d=m /:d \\ V= \frac{m}{d}](https://tex.z-dn.net/?f=d%3D%20%5Cfrac%7Bm%7D%7BV%7D%20%20%2F%2AV%20%5C%5C%20V%2Ad%3Dm%20%2F%3Ad%20%5C%5C%20V%3D%20%5Cfrac%7Bm%7D%7Bd%7D%20)
Now we have to substitute our data to formula
V=
![\frac{3.2}{16.8}=0.19](https://tex.z-dn.net/?f=%20%5Cfrac%7B3.2%7D%7B16.8%7D%3D0.19%20)
ml - its the answeer
Answer:
The magnitude of the electric field is 0.1108 N/C
Explanation:
Given;
number of electrons, e = 8.05 x 10⁶
length of the wire, L = 1.03 m
distance of the field from the center of the wire, r = 0.201 m
Charge of the electron;
Q = (1.602 x 10⁻¹⁹ C/e) x (8.05 x 10⁶ e)
Q = 1.2896 x 10⁻¹² C
Linear charge density;
λ = Q / L
λ = (1.2896 x 10⁻¹² C) / (1.03 m)
λ = 1.252 x 10⁻¹² C/m
The magnitude of electric field at r = 0.201 m;
![E = (\frac{1}{4 \pi \epsilon_o} )\frac{ 2 \lambda}{r} \\\\E = k \frac{ 2 \lambda}{r}\\\\E = (8.89*10^9)*\frac{2*1.252*10^{-12}}{0.201} \\\\E = 0.1108 \ N/C](https://tex.z-dn.net/?f=E%20%3D%20%28%5Cfrac%7B1%7D%7B4%20%5Cpi%20%5Cepsilon_o%7D%20%29%5Cfrac%7B%202%20%5Clambda%7D%7Br%7D%20%5C%5C%5C%5CE%20%3D%20k%20%5Cfrac%7B%202%20%5Clambda%7D%7Br%7D%5C%5C%5C%5CE%20%3D%20%288.89%2A10%5E9%29%2A%5Cfrac%7B2%2A1.252%2A10%5E%7B-12%7D%7D%7B0.201%7D%20%20%5C%5C%5C%5CE%20%3D%200.1108%20%5C%20N%2FC)
Therefore, the magnitude of the electric field is 0.1108 N/C
Answer:
109ohm
Explanation:
Note: when resistors are connected in series
Rt = R1 + R2 + R3 = 23+81+5 = 109ohm