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How much stronger is the electric force between two electrons than the gravitational force between them?

Is the impulse- momentum relationship related to Newton's second law?

Why is the gravitational potential energy of ideal uniform massive spring $mgx/2$, not $mgx$?
$L=T-<!--\; -\; -->V=\frac{1}{2}\frac{m}{3}{\stackrel{\dot{}<!--\; \dot{}\; -->}{x}}^2+\frac{1}{2}M{\stackrel{\dot{}<!--\; \dot{}\; -->}{x}}^2-<!--\; -\; -->\frac{1}{2}k{x}^2-<!--\; -\; -->\frac{mgx}{2}-<!--\; -\; -->Mgx$
where $mgx/2$ refers to gravitational potential energy of ideal spring. But I do not get why it's $mgx/2$ instead of $mgx$. What is the reason?

Define net force, external force, and system. Apply Newton’s second law to determine the weight of an object

What is the gravitational force between two identical 50,000 kg asteroids whose centers of mass are separated by 1000 m?

A ball was dropped from the window of a building, 33.5 meters above the ground
How fast it is after 2 seconds?
How long will it take if the final velocity is 25 m/s?

Explain Does Newton’s second law hold true for an observer in a van as it speeds up, slows down, or rounds a corner?

A point mass m having charge q is connected with massless spring to a rigid wall on a horizontal surface.A horizontal uniform electric field E is switched on and mass is displaced through a distance x from equilibrium position.
Work done by electric force is $W=qE.x$ as electric force is constant.
But F(external) on spring mas system is equal to kx. So $W=kx.x=k{x}^2$ (equation 1) as external force i.e, electric force is constant.
We know that work done is stored in the form of potential energy and we know that potential energy stored in the spring is $1/2k{x}^2$.
So $W=1/2k{x}^2$ (let it be equation 2)
But the above 2 equations are not equal. From where did the 1/2 come from in the 2nd equation?

Why are non-Newtonian fluids called non-Newtonian when they follow Newton’s third law?
To my understanding, Newton’s third law states that for every action there is an equal and opposite reaction. Therefor if I punch the non-Newtonian fluid harder, there will be a harder reaction force stopping my hand. So why is the fluid called non-Newtonian?

Is there a more fundamental reason for the Classical Lagrangian to be T−V and Electromagnetic Lagrangian to be T−V+qA.v or is it simply because we can derive Newton's Second Law and Lorentz Force Law with these Lagrangians respectively?

Is there a difference between the potential energy of an oscillator and the potential energy of the spring which is associated with? When we compress a spring horizontally without any object associated with it starts to expand? Is the potential energy convert into kinetic energy of the spring?

Each of the pairs of objects listed are the same distance apart. Which pair of objects has the largest gravitational force?
A.car and bowling ball
B. marble and car
C. marble and baseball
D. There is no gravitational force between any of these pairs of objects

Why can't $i\mathrm{\hslash <!--\; \hslash \; -->}\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}t}$ be considered the Hamiltonian operator?

Express Newton’s second law of motion for rotating bodies. What can you say about the angular velocity and angular momentum of a rotating nonrigid body of constant mass if the net torque acting on it is zero?

Let's consider having 2-D Cartesian coordinate xy-plane, a block connected to a spring (placed horizontally on negative part of x-axis and having natural length L) sitting at the origin point with the spring at natural length.It is allowed to oscillate back and forth about the origin. From the equation that relates the work done by conservative force and the potential energy, $U(x_a)-<!--\; -\; -->U(x_b)=-<!--\; -\; -->{\int <!--\; \int \; -->}_{x_b}^{x_a}{F}_x\text{d}x$. If we define the zero point at x=0, we can express its potential energy for $0<<!--\; <\; -->x<<!--\; <\; -->L$, with
$U(x)-<!--\; -\; -->U(x=0)=-<!--\; -\; -->{\int <!--\; \int \; -->}_0^x-<!--\; -\; -->kx\text{d}x\Rightarrow <!--\; \Rightarrow \; -->U(x)=\frac{1}{2}k{x}^2$
while for −L<x<0, we have (the restoring force switch sign as it now points to the positive x- direction):
$U(x)-<!--\; -\; -->U(x=0)=-<!--\; -\; -->{\int <!--\; \int \; -->}_0^{x}kx\text{d}x\Rightarrow <!--\; \Rightarrow \; -->U(x)=-<!--\; -\; -->\frac{1}{2}k{x}^2$
My question: Is my idea and this approach for the energy of the system correct? Because I saw from some reference books, they simply say that it is just $\frac{1}{2}k{x}^2$ for both of the conditions.

The Navier–Stokes equation is also known as:
(a) Newton’s first law
(b) Newton’s second law
(c) Newton’s third law
(d ) Continuity equation
(e) Energy equation

Using all three of Newton's Laws of Motion, describe how a rocket launches into orbit.