Alfisols can be found in humid areas and semiarid areas. They are formed in forest and savanna vegetation. They are very fertile and productive soils. These soils contains high concentrations of nutrient that are cations like calcium, magnesium, potassium and sodium.
Answer:
Mono-chromatic, Poly-chromatic. When a wave moves from one medium into another at an angle other than 90 degrees the wave will change direction and continue to follow a new straight-line path
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To solve this problem we will apply the concept related to the electric field defined from the laws of Coulomb. For this purpose we will remember that the electric field is equivalent to the product of the Coulomb constant due to the change of the charge over the squared distance, mathematically this is
![E = \frac{kq}{r^2}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7Bkq%7D%7Br%5E2%7D)
Here,
k = Coulomb's constant
r = Distance from center of terminal to point where electric field is to found
q = Excess charge placed on the center of terminal of Van de Graff's generator
Replacing we have that,
![E = \frac{(9*10^{9})(8*10^{-3})}{3^2}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B%289%2A10%5E%7B9%7D%29%288%2A10%5E%7B-3%7D%29%7D%7B3%5E2%7D)
![E = 8*10^6N/C](https://tex.z-dn.net/?f=E%20%3D%208%2A10%5E6N%2FC)
Therefore the electric field is ![8*10^6N/C](https://tex.z-dn.net/?f=8%2A10%5E6N%2FC)
Answer:
The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area
Explanation:
Answer:
the speed of the electron at the given position is 106.2 m/s
Explanation:
Given;
initial position of the electron, r = 9 cm = 0.09 m
final position of the electron, r₂ = 3 cm = 0.03 m
let the speed of the electron at the given position = v
The initial potential energy of the electron is calculated as;
![U_i = Fr = \frac{kq^2}{r^2} \times r = \frac{kq^2}{r} \\\\U_i = \frac{(9\times 10^9)(1.602\times 10^{-19})^2}{0.09} \\\\U_i = 2.566 \times 10^{-27} \ J](https://tex.z-dn.net/?f=U_i%20%3D%20Fr%20%3D%20%5Cfrac%7Bkq%5E2%7D%7Br%5E2%7D%20%5Ctimes%20r%20%3D%20%5Cfrac%7Bkq%5E2%7D%7Br%7D%20%5C%5C%5C%5CU_i%20%3D%20%5Cfrac%7B%289%5Ctimes%2010%5E9%29%281.602%5Ctimes%2010%5E%7B-19%7D%29%5E2%7D%7B0.09%7D%20%5C%5C%5C%5CU_i%20%3D%202.566%20%5Ctimes%2010%5E%7B-27%7D%20%5C%20J)
When the electron is 3 cm from the proton, the final potential energy of the electron is calculated as;
![U_f = \frac{kq^2}{r_2} \\\\U_f = [\frac{(9\times 10^9)\times (1.602 \times 10^{-19})^2}{0.03} ]\\\\U_f = 7.669 \times 10^{-27} \ J \\\\\Delta U = U_f -U_i\\\\\Delta U = (7.699\times 10^{-27} \ J ) - (2.566 \times 10^{-27} \ J)\\\\\Delta U = 5.133 \times 10^{-27} \ J](https://tex.z-dn.net/?f=U_f%20%3D%20%5Cfrac%7Bkq%5E2%7D%7Br_2%7D%20%5C%5C%5C%5CU_f%20%3D%20%20%5B%5Cfrac%7B%289%5Ctimes%2010%5E9%29%5Ctimes%20%281.602%20%5Ctimes%2010%5E%7B-19%7D%29%5E2%7D%7B0.03%7D%20%5D%5C%5C%5C%5CU_f%20%3D%207.669%20%5Ctimes%2010%5E%7B-27%7D%20%5C%20J%20%5C%5C%5C%5C%5CDelta%20U%20%3D%20U_f%20-U_i%5C%5C%5C%5C%5CDelta%20U%20%3D%20%287.699%5Ctimes%2010%5E%7B-27%7D%20%5C%20J%20%29%20-%20%282.566%20%5Ctimes%2010%5E%7B-27%7D%20%5C%20J%29%5C%5C%5C%5C%5CDelta%20U%20%3D%205.133%20%5Ctimes%2010%5E%7B-27%7D%20%5C%20J)
Apply the principle of conservation of energy;
ΔK.E = ΔU
![K.E_f -K.E_i = \Delta U\\\\initial \ velocity \ of \ the \ electron = 0\\\\K.E_f - 0 = \Delta U\\\\K.E_f = \Delta U\\\\\frac{1}{2} mv^2 = \Delta U\\\\where;\\\\m \ is \ the \ mass \ of\ the \ electron = 9.1 1 \times 10^{-31} \ kg\\\\v^2 = \frac{ 2 \Delta U}{m} \\\\v = \sqrt{\frac{ 2 \Delta U}{m}} \\\\v = \sqrt{\frac{ 2 (5.133\times 10^{-27})}{9.11\times 10^{-31}}}\\\\v = \sqrt{11268.935} \\\\v = 106.2 \ m/s](https://tex.z-dn.net/?f=K.E_f%20-K.E_i%20%3D%20%5CDelta%20U%5C%5C%5C%5Cinitial%20%5C%20velocity%20%5C%20of%20%20%5C%20the%20%5C%20electron%20%3D%200%5C%5C%5C%5CK.E_f%20-%200%20%3D%20%5CDelta%20U%5C%5C%5C%5CK.E_f%20%3D%20%5CDelta%20U%5C%5C%5C%5C%5Cfrac%7B1%7D%7B2%7D%20mv%5E2%20%3D%20%5CDelta%20U%5C%5C%5C%5Cwhere%3B%5C%5C%5C%5Cm%20%5C%20is%20%5C%20the%20%5C%20mass%20%5C%20of%5C%20the%20%5C%20electron%20%3D%209.1%201%20%5Ctimes%2010%5E%7B-31%7D%20%5C%20kg%5C%5C%5C%5Cv%5E2%20%3D%20%5Cfrac%7B%202%20%5CDelta%20U%7D%7Bm%7D%20%5C%5C%5C%5Cv%20%3D%20%5Csqrt%7B%5Cfrac%7B%202%20%5CDelta%20U%7D%7Bm%7D%7D%20%5C%5C%5C%5Cv%20%3D%20%5Csqrt%7B%5Cfrac%7B%202%20%285.133%5Ctimes%2010%5E%7B-27%7D%29%7D%7B9.11%5Ctimes%2010%5E%7B-31%7D%7D%7D%5C%5C%5C%5Cv%20%3D%20%5Csqrt%7B11268.935%7D%20%5C%5C%5C%5Cv%20%3D%20106.2%20%5C%20m%2Fs)
Therefore, the speed of the electron at the given position is 106.2 m/s