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aliya0001 [1]
3 years ago
9

1. In a particular experiment to study the photoelectric effect, the frequency of the incident light and the temperature of the

metal are held constant. Assuming that the light incident on the metal surface causes electrons to be ejected from the metal, what happens if the intensity of the incident light is increased?A. The work function of the metal decreases. B. The number of electrons emitted from the metal per second increases. C. The maximum speed of the emitted electrons increases. D. The stopping potential increases.2. In another experiment, the intensity of the incident light and the temperature of the metal are held constant. Assuming that the initial light incident on the metal surface causes electrons to be ejected from the metal, what happens if the frequency of the incident light is increased? A. The work function of the metal increases. B. The number of electrons emitted from the metal per second increases. C. The maximum speed of the emitted electrons increases. D. The stopping potential increases.
Physics
1 answer:
shutvik [7]3 years ago
4 0

Answer: 1. B. The number of electrons emitted from the metal per second increases.

2. The maximum speed of the emitted electrons increases.

The stopping potential increases

Explanation:

Photoelectric effect is simply referred to as the emission of electrons that occurs when there's an electromagnetic radiation. An example of such electromagnetic radiation is when material is being hit by light.

Assuming that the light incident on the metal surface causes electrons to be ejected from the metal, the number of electrons emitted from the metal per second increases if the intensity of the incident light is increased.

Also, if the initial light incident on the metal surface causes electrons to be ejected from the metal, the maximum speed of the emitted electrons increases and the stopping potential increases.

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In the diagram, q1 = -6.39*10^-9 C and q2 = +3.22*10^-9 C. What is the electric field at point P? pls help
Alexxx [7]

Answer:

Below

Explanation:

First draw the vectors that represent both electric fields.

E1 is the elictric field created by q1, E2 is the one created by q2.

● q1 is negative so E1 will point from P.

● q2 is positive so E2 will point out of P

(Picture below)

■■■■■■■■■■■■■■■■■■■■■■■■■■

The resulting electric field is equal to the sum of the two fields since both vectors are colinear.

Let E be the total field.

● E = E1 + E2

The formula of the electric field intensity is:

● E = K ×(q/d^2)

-K is Coulomb's constant

-d is the distance between the charge and the object ( here P)

-q is the charge

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E1 = K × (q1/d1^2)

The distance between q1 and P is the qum of 0.15 m 0.25 m. (0.4 m)

Coulombs constant is 9×10^9 m^2/C^2

● E1 = 9×10^9 ×[-6.39 × 10^(-9)/ 0.4^2]

● E1 = -359.43 N/C

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E2 = K ×(q2/d^2)

The distance between q2 and P is 0.25 m.

● E2 = 9×10^9×[3.22×10^(-9) /0.25^2]

● E2 = 463.68 N/C

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E = E1 + E2

● E = -359.43+463.68

● E = 105.25 N/C

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