Forces in the universe that act over long distance, meaning the distance is greater than the diameter of the nucleus of the atom are:
1. Electrostatic force or Coulomb force: Fc=(k*Q₁*Q₂)/r²,
2. Gravitational force: Fg=(G*m₁*m₂)/r²,
3. Magnetic force: Fm=qvB,
4. London dispersion force, also known as one of the van der Waals forces.
Answer:
0.75
Explanation:
Since the static frictional force is the maximum force applied just before sliding, our frictional force, F is 300 N.
Since F = μN where μ = coefficient of static friction and N = normal force = 400 N (which is the downward force applied against the surface).
So, μ = F/N
= 300 N/400 N
= 3/4
= 0.75
So, the coefficient of static friction μ = 0.75
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Force, newtons 3rd law of motion stated for every action there is an equal and opposite reaction
I'm not sure what "60 degree horizontal" means.
I'm going to assume that it means a direction aimed 60 degrees
above the horizon and 30 degrees below the zenith.
Now, I'll answer the question that I have invented.
When the shot is fired with speed of 'S' in that direction,
the horizontal component of its velocity is S cos(60) = 0.5 S ,
and the vertical component is S sin(60) = S√3/2 = 0.866 S . (rounded)
-- 0.75 of its kinetic energy is due to its vertical velocity.
That much of its KE gets used up by climbing against gravity.
-- 0.25 of its kinetic energy is due to its horizontal velocity.
That doesn't change.
-- So at the top of its trajectory, its KE is 0.25 of what it had originally.
That's E/4 .
