<span>Answer:
So it gets to the top of the ramp and stops. The parallel force pushing it down the ramp is mg sin θ, but for it to move, the frictional force must be overcome. This frictional force is μmg cos θ, where μ is the coefficient of static friction. For movement, then,
mg sin θ > μmg cos θ ==> tan θ > μ ==> θ > arctan 0.5 = 26.565° ==> θ = 27°</span>
Given:
The mass of the halfback is m = 107 kg
The speed of the halfback is v = 8 m/s
To find the momentum.
Explanation:
The momentum of the halfback is
Thus, the momentum of the halfback is 856 kg m/s
Answer: a = 1.32 * 10^18m/s² due north
Explanation: The magnitude of the force required to move the electron is given as
F = ma
The force exerted on the charge by the electric field of intensity (E) is given by
F = Eq
Thus
Eq = ma
a = E * q/ m
Where a = acceleration of charge
E = strength of electric field = 7400N/c
q = magnitude of electronic charge = 1.609 * 10^-6c
m = mass of an electronic charge = 9.109 * 10^-31kg
a = 7400 * 1.609 * 10^-16/ 9.109 * 10^-31
a = 11906.6 * 10^-16 / 9.019 * 10^-31
a = 1.19 * 10^-12 / 9.019 * 10^-31
a = 0.132 * 10^19
a = 1.32 * 10^18m/s²
As stated in the question, the direction of the electric field is due north hence, the direction of it force will also be north thus making the electron experience a force due north ( according to Newton second law of motion)
Light from the stars, because the orbits make it difficult to see them.
