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andrew11 [14]
3 years ago
14

Monochromatic light is incident on a metal surface and electrons are ejected. If the intensity of the light is increased, what w

ill happen to the ejection rate and maximum energy of the electrons
Physics
1 answer:
MAVERICK [17]3 years ago
6 0

Answer:

<u>Increase the rate and the same maximum energy of the electrons</u>

Explanation:

According to the photoelectric effect we can say:

  • The number of electrons, or the electric current, has a linear behaviour with the intensity of the light and a constant behaviour whit the frequency. <u>Therefore, the rate of electrons increases.</u>

  • The kinetic energy of the ejected electrons has a linear dependence on the frequency of the light and has a constant behaviour with the intensity. <u>So, we can say there is the same maximum energy.</u>

<u></u>

I hope it helps you!

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If an object has a mass of 8 kg, what is its approximate weight on Earth?
Fittoniya [83]

Answer:

10^24 kg

Explanation:

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87) Determine the equivalent resistances for the following circuits.
stepladder [879]

Answer:

15    and  11 ohms

Explanation:

First one =   For the parallel resistors 1 / (1/6 + 1/6 + 1/6) =   1/ (3/6 ) = 6/3 = 2 ohms   then add the 3 and the 10  = 15 ohms

second one    for the parallel portion   equiv =    (10+2)*24 / ( (10+2 + 24) = 8

  then add the  3 in series = 11 ohms

8 0
2 years ago
A force of 20 N acts over an area of 2 m2. What is the pressure? ​
ycow [4]

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5 0
2 years ago
Read 2 more answers
22. (I) A car slows down from 28 m/s to rest in a distance of
lidiya [134]

Answer:

a = -8.912 m/s²

Explanation:

Given,

The initial velocity of the car, u = 28 m/s

The final velocity of the car, v = 0

The distance traveled by car, d = 88 m

The velocity displacement relation is given by the formula

                                          v = d/t

∴                                         t = d/v

Substituting in the above values in the given equation

                                           t = 88/28

                                            = 3.142 s

The acceleration is given by the formula

                                         a = (v-u)/t

                                            = (0 - 28)/3.142

                                            = -8.912 m/s²

The negative sign is that the car is decelerating.

Hence, acceleration a = -8.912 m/s²

7 0
4 years ago
An athlete runs at a constant velocity of 5.2 m/s. What is the velocity of the athlete relative to the ground?
dimulka [17.4K]

The relative velocity of the athlete relative to the ground is 5.2 m/s

The given parameters;

constant velocity of the athlete, V = 5.2 m/s

let the velocity of the ground = Vg = 0

The relative velocity concept helps us to determine the velocity of a moving object relative to a stationary observer.

The athlete is the moving object in this question while the ground is stationary.

The relative velocity of the athlete relative to the ground is calculated as follows;

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Thus, the relative velocity of the athlete relative to the ground is 5.2 m/s

Learn more here: brainly.com/question/24430414

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