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Bumek [7]
3 years ago
10

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

l happen to the ejection rate of the electrons? Monochromatic light is incident on a metal surface, and electrons are ejected. If the intensity of the light increases, what will happen to the ejection rate of the electrons? The ejection rate will remain constant. Changes in the ejection rate cannot be determined without additional information. The ejection rate will decrease. The ejection rate will increase.
Physics
1 answer:
Nadusha1986 [10]3 years ago
6 0
<h2>Answer: The ejection rate will increase.</h2>

The photoelectric effect is a phenomenon that consists of the emission of electrons by certain metals when a beam of light impacts on its surface.

For this phenomenon to occur, certain conditions must be met, such as when the photon collides with the electron, in order to "pull it" from the metal, the photon must have a minimum energy equal to the ionization energy of the atom, so that the electron can leave the influence of the nucleus.  

This is achieved with the adequate intensity of the incident radiation, which is related to the number of photons that impact the metal.

This means:

<h2>The greater the intensity, the greater the number of photons, hence the greater number of electrons emitted.</h2>
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Johannes Kepler was a main stargazer of the Scientific Revolution known for detailing the Laws of Planetary Motion. A stargazer, obviously, is a man who contemplates the sun, stars, planets and different parts of room. Kepler was German and lived in the vicinity of 1571 and 1630. 
Despite the fact that Kepler is best known for characterizing laws in regards to planetary movement, he made a few other striking commitments to science. He was the first to discover that refraction drives vision in the eye and that utilizing two eyes empowers profundity recognition.
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3 years ago
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This phenomenon is known as the bends. If a scuba diver rises quickly from a depth of 25.0 m in Lake Michigan (which is fresh wa
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Explanation:

Attachments of the solution are below.

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3 years ago
g An astronaut must journey to a distant planet, which is 189 light-years from Earth. What speed will be necessary if the astron
Butoxors [25]

Answer:

The value is v  =  2.999 *10^{8} \  m/s

Explanation:

From the question we are told that

   The time taken to travel to the planet from earth is t = 189 \ light-years

    The  time to be spent on the ship is  t_{s} =  12 \  years

Generally speed can be obtained using the mathematical relation represented below

       t_s  =  2 * t *  \sqrt{1 -  \frac{v^2}{c^2 } }

The 2 in the equation show that the trip is a round trip i.e going and coming back

=>    12 =  2 * 189 *  \sqrt{1 -  \frac{v^2}{(3.0*10^{8})^2 } }

=>     v  =  2.999 *10^{8} \  m/s

5 0
3 years ago
A body falls from the top of the tower and during the last second of its fall it fall through 23mvfind height of tower.
DanielleElmas [232]

Answer:

39.7 m

Explanation:

First, we conside only the last second of fall of the body. We can apply the following suvat equation:

s=ut+\frac{1}{2}at^2

where, taking downward as positive direction:

s = 23 m is the displacement of the body

t = 1 s is the time interval considered

a=g=9.8 m/s^2 is the acceleration

u is the velocity of the body at the beginning of that second

Solving for u, we find:

ut=s-\frac{1}{2}at^2\\u=\frac{s}{t}-\frac{1}{2}at=\frac{23}{1}-\frac{1}{2}(9.8)(1)=18.1 m/s

Now we can call this velocity that we found v,

v = 18 m/s

And we can now consider the first part of the fall, where we can apply the following suvat equation:

v^2-u^2 = 2as'

where

v = 18 m/s

u = 0 (the body falls from rest)

s' is the displacement of the body before the last second

Solving for s',

s'=\frac{v^2-u^2}{2a}=\frac{18.1^2-0}{2(9.8)}=16.7 m

Therefore, the total heigth of the building is the sum of s and s':

h = s + s' = 23 m + 16.7 m = 39.7 m

7 0
3 years ago
you push a sled of mass 15 kg across the snow with a force of 180 N for a distance of 2.5 m. There is no friction. if the sled s
Oxana [17]
<span>7.7 m/s First, determine the acceleration you subject the sled to. You have a mass of 15 kg being subjected to a force of 180 N, so 180 N / 15 kg = 180 (kg m)/s^2 / 15 kg = 12 m/s^2 Now determine how long you pushed it. For constant acceleration the equation is d = 0.5 A T^2 Substitute the known values getting, 2.5 m = 0.5 12 m/s^2 T^2 2.5 m = 6 m/s^2 T^2 Solve for T 2.5 m = 6 m/s^2 T^2 0.41667 s^2 = T^2 0.645497224 s = T Now to get the velocity, multiply the time by the acceleration, giving 0.645497224 s * 12 m/s^2 = 7.745966692 m/s After rounding to 2 significant figures, you get 7.7 m/s</span>
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