The back-and-forth movement of electrons is called alternating current. Electrons go back and forth, the direction of their path alternates from one direction to another.
the movement of electrons in one direction is called direct current. The electrons move in a direct, single path without changing directions.
Newton's 2nd law:
Fnet = ma
Fnet is the net force acting on an object, m is the object's mass, and a is the acceleration.
The electric force on a charged object is given by
Fe = Eq
Fe is the electric force, E is the electric field at the point where the object is, and q is the object's charge.
We can assume, if the only force acting on the proton and electron is the electric force due to the electric field, that for both particles, Fnet = Fe
Fe = Eq
Eq = ma
a = Eq/m
We will also assume that the electric field acting on the proton and electron are the same. The proton and electron also have the same magnitude of charge (1.6×10⁻¹⁹C). What makes the difference in their acceleration is their masses. A quick Google search will provide the following values:
mass of proton = 1.67×10⁻²⁷kg
mass of electron = 9.11×10⁻³¹kg
The acceleration of an object is inversely proportional to its mass, so the electron will experience a greater acceleration than the proton.
They are both in motion because an object is not at rest, but moving so slow it could be at rest. A car going at the same constant velocity is neither speeding up or slowing down, an object "at rest" is also moving at a constant rate, not speeding up or slowing done.
Answer:
radius comes out to be 3 m
height of the cylinder comes out to be 3m
Explanation:
given
volume of cylinder = 27π m³
π r² h = 27π
r² h = 27.............(1)
surface area of cylinder open at the top
S = 2πrh + π r²
for least amount of material requirement.
hence radius comes out to be 3 m
for height put the value in the equation 1
so, height of the cylinder comes out to be 3m
Answer:
W = ½ m v²
Explanation:
In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation
We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved
initial instant. before separation
p₀ = m v
final attempt. after separation
= m /2 0 + m /2 v_{f}
p₀ = p_{f}
m v = m /2 
v_{f}= 2 v
this is the speed of the second part of the ship
now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body
initial energy
K₀ = ½ m v²
final energy
= ½ m/2 0 + ½ m/2 v_{f}²
K_{f} = ¼ m (2v)²
K_{f} = m v²
the expression for work is
W = ΔK = K_{f} - K₀
W = m v² - ½ m v²
W = ½ m v²
