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Mila [183]
4 years ago
13

Monochromatic light shines on the cathode in a photoelectric effect experiment, causing the emission of electrons. If the freque

ncy of the light stays the same but the intensity of the light shining on the cathode is increased:_________
a. The emitted electrons will be moving at a higher speed.
b. There will be more electrons emitted.
c. Neither A nor B are true.
d. Both A and B are true
Physics
1 answer:
professor190 [17]4 years ago
3 0

Answer:

<em>b. There will be more electrons emitted.</em>

Explanation:

In photoelectric emission, the energy of the emitted electron is dependent on the frequency of the wave incident on the plate; but not the intensity. The rate of electron emission per unit time however depends on the intensity of the incident light. So increasing the intensity of the light at constant frequency will only affect the number of electrons emitted per unit time.

You might be interested in
7. What is the acceleration of a 1,300 kg truck with a net force of 5,000 newtons (N)?
Crank

Answer: 1200 kg

Explanation: Taking into account the Newton's second law, the mass of the vehicle is 1200 kg.

In first place, acceleration in a body occurs when a force acts on a body. There are two factors that influence the acceleration of an object: the net force acting on it and the mass of the body.

Newton's second law states that this force will change the speed of an object because the speed and / or direction will change. These changes in velocity are called acceleration.

So, Newton's second law defines the relationship between force and acceleration mathematically. This law says that the acceleration of an object is directly proportional to the sum of all the forces acting on it and inversely proportional to the mass of the object.

Mathematically, Newton's second law is expressed as:

F= m×a

where:

F = Force [N]

m = Mass [kg]

a = Acceleration [m/s²]

In this case, you know:

F= 1800 N

m= ?

a= 1.5 m/s²

Replacing:

1800 N= m× 1.5 m/s²

Solving:

m= 1800 N÷1.5 m/s²

m= 1200 kg

Finally, the mass of the vehicle is 1200 kg.

5 0
3 years ago
Read 2 more answers
The less internal heat a jovian planet emits the more it stirs up its clouds.
g100num [7]
The correct answer would be false. The less internal heat a jovian planet emits the lesser it stirs up its clouds making the atmosphere hotter. All of the four Jovian planets have unique atmospheres. They have more or less the same structures but they differ in their average temperature. As the distance of the planet is closer to the sun the atmosphere would be cooler. These planets are Jupiter, Saturn, Uranus and Neptune. They do not have a solid structure instead they are primarily composed of helium and hydrogen which makes them gas giants or ice planets. They are larger in size than the remaining planets in the solar system. 
7 0
3 years ago
PLEASE HELP. A substance decays according the the equation ae^-0.0025t, where t is in minutes. Find the half life of the substan
mojhsa [17]

First of all, that's not an equation.  An equation needs an 'equals' sign ( = ),
so you need something like 

               Amount present = a e^-0.0025t .

                                   'a' = amount present when t = 0 .

The half-life is the time it takes for the original amount to decay to half of it.

That's just 't' when                      e^-0.0025t  =  1/2

Take the natural log of each side:    -0.0025t  =  ln(0.5)

Divide each side by  -0.0025 :        t = ln(0.5) / (-0.0025) =

                                                     (-0.69315) / (-0.0025) =

                                                   <em>t  =  277.26 minutes</em>

                                                    ( 4hrs 37min 15.5sec)

Rounded to nearest tenth:  t = 277.3 minutes .


3 0
4 years ago
In your own words explain the importance of the cycles to an ecosystem and how the cycles of matter differ from the flow of ener
Rina8888 [55]
The three main cycles of an ecosystem are the water cycle, the carbon cycle and the nitrogen cycle.
8 0
3 years ago
You have a reservoir held at a constant temperature of –30°C. You add 400 J of heat to the reservoir. If you have another reserv
marysya [2.9K]

Answer:

449.38 J

Explanation:

ΔS = ΔQ/T

Where ΔS = entropy change

Q = quantity of heat

T = temperature

First reservoir :

T = –30°C = - 30 + 273 = 243K

Q = 400 J

Second reservoir :

T = 0°C = 273K

Q =?

To have same increase in entropy for both reservoirs :

Q/T of first reservoir = Q/T of second reservoir

400/243 = Q/273

243 * Q = 400 * 273

Q = (400 * 273) / 243

Q = 109,200 / 243

Q = 449.38271

Q = 449.38 J

6 0
4 years ago
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