Answer:
The lever is a movable bar that pivots on a fulcrum attached to a fixed point. The lever operates by applying forces at different distances from the fulcrum, or a pivot. As the lever rotates around the fulcrum, points farther from this pivot move faster than points closer to the pivot.
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Answer:
b
Explanation:
they both have a neutral charge so they couldn't be positive or negative since that wouldn't come from anywhere
Answer:
what r the questions i can’t see them
Explanation:
It is required an infinite work. The additional electron will never reach the origin.
In fact, assuming the additional electron is coming from the positive direction, as it approaches x=+1.00 m it will become closer and closer to the electron located at x=+1.00 m. However, the electrostatic force between the two electrons (which is repulsive) will become infinite when the second electron reaches x=+1.00 m, because the distance d between the two electrons is zero:

So, in order for the additional electron to cross this point, it is required an infinite amount of work, which is impossible.
Answer:
Given:
Thermal Kinetic Energy of an electron, 
= Boltzmann's constant
Temperature, T = 1800 K
Solution:
Now, to calculate the de-Broglie wavelength of the electron,
:

(1)
where
h = Planck's constant = 
= momentum of an electron
= velocity of an electron
= mass of electon
Now,
Kinetic energy of an electron = thermal kinetic energy



(2)
Using eqn (2) in (1):

Now, to calculate the de-Broglie wavelength of proton,
:

(3)
where
= mass of proton
= velocity of an proton
Now,
Kinetic energy of a proton = thermal kinetic energy



(4)
Using eqn (4) in (3):
