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.
2000÷330=6.06 repatant so the answer would be about 6.06 seconds
It will land in your lap because there's different frames of motion relative to yourself. For example, if you're running at a speed of 6 mph, it doesn't mean you'll run as fast as the Earth spins. Also, since you're on the interior of the plane, any kind of wind or weather on the outside will not affect the coin. A law to back up this claim is Einsteins Special Law of Relativity.
Answer:

Explanation:
<u>Motion in The Plane</u>
When an object is launched in free air with some angle respect to the horizontal, it describes a known parabolic path, comes to a maximum height and finally drops back to the ground level at a certain distance from the launching place.
The movement is split into two components: the horizontal component with constant speed and the vertical component with variable speed, modified by the acceleration of gravity. If we are given the values of
and
as the initial speed and angle, then we have




If we want to know the maximum height reached by the object, we find the value of t when
becomes zero, because the object stops going up and starts going down

Solving for t

Then we replace that value into y, to find the maximum height

Operating and simplifying

We have

The maximum height is


<span>The answer to this problem is magnesium. I hoped I helped someone with this</span>