2.37eV stopping potential would be required to arrest the current of photoelectrons.
<h3 /><h3>What is stopping potential ?</h3>
The minimal negative voltage that must be provided to the anode to halt the photocurrent is known as stopping potential. When expressed in electron volts, the maximal kinetic energy of the electrons is equal to the stopping voltage.
Kmax = eV₀
2.37eV = eV₀
V₀ = 2.37eV
to learn more about stopping potential go to - brainly.com/question/4655588
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Impulse is also known as the change in momentum of an object. Momentum has an equation of,
momentum = m x v
where m is mass and v is velocity. Therefore, impulse has a formula which is,
impulse = m x (v₂ - v₁)
Substituting the known values in the equation,
impulse = 975 kg x (3 m/s - 0.5 m/s)
impulse = 2437.5 kg m/s
Answer:
These energy exchanges are not changes in kinetic energy. They are changes in bonding energy between the molecules. If heat is coming into a substance during a phase change, then this energy is used to break the bonds between the molecules of the substance. The example we will use here is ice melting into water.
Explanation:
Aha! This is a fun question, likely AP Physics 1 if I'm not mistaken. This is how you solve it, you need the following equation.
![F=G\frac{m_{1} *m_{2}}{r^2}](https://tex.z-dn.net/?f=F%3DG%5Cfrac%7Bm_%7B1%7D%20%2Am_%7B2%7D%7D%7Br%5E2%7D%20)
Now on this question you don't need to actually use the mass and diameter of the earth as we are only
looking at the
factor of how much your weight would change, so you can just plug in some random numbers for the mass and radius. In my case I will use the original radius as being
2 meters and the original mass of the earth as being
20kg. You can actually just omit the mass of your body and the gravitational constant isn't necessary to be multiplied by because they will remain the same in both scenarios.
Original weight:
![F= \frac{20kg}{2^2m}](https://tex.z-dn.net/?f=F%3D%20%5Cfrac%7B20kg%7D%7B2%5E2m%7D%20)
![F=5N](https://tex.z-dn.net/?f=F%3D5N)
So the original weight in this circumstance is 5N. Now then if we double the mass to
40kg and the radius to
4m you will have the following.
![F= \frac{40kg}{(4)^2m}](https://tex.z-dn.net/?f=F%3D%20%5Cfrac%7B40kg%7D%7B%284%29%5E2m%7D%20)
![F=2.5N](https://tex.z-dn.net/?f=F%3D2.5N)
So the original weight was 5N and after doubling the mass and radius the weight reduced to 2.5N. This means the factor which your weight would change by is .5 or decrease by 1/2.
I hope this helps you :) You can ask me if you have any additional physics questions.