Answer:
μs = 0.75
μk = 0.58
Explanation:
From a force diagram:
(1)
(2)
When it starts slipping, friction force is the maximum and acceleration is 0. Replacing these conditions on (1):
Solving for μs:

μs = tan 36.7° = 0.75
When it moves at constant speed, friction force is kinetic friction and acceleration is 0. With these conditions the coefficient is:
μk = tan 30.1° = 0.58
Answer:
3.39724 seconds
23.0824792352 m, 101.917520765 m
13.58896 m/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
The equation of motion will be


The time at which the cars collide is 3.39724 seconds

Car B traveled 23.0824792352 m and Car A traveled 125-23.0824792352 = 101.917520765 m

The speed of car B is 13.58896 m/s
Explanation:
The magnitude of centripetal acceleration is:
a = v² / r
The velocity is:
v = d/t
v = 2 × 2π (5 m) / 60 s
v = 1.05 m/s
a = (1.05 m/s)² / (5 m)
a = 0.219 m/s²
The direction is towards the center. So at point A, the centripetal acceleration is 0.219 m/s² downward, and at point B, the centripetal acceleration is 0.219 m/s² upward.
<span>The electron is a subatomic particle, symbol e− or β−, with a negative elementary electric charge. Electrons belong to the first generation of the lepton particle family, and are generally thought to be elementary particles because they have no known components or substructure. The electron has a mass that is approximately 1/1836 that of the proton. Quantum mechanical properties of the electron include an intrinsic angular momentum of a half-integer value in units of ħ, which means that it is a fermion. Being fermions, no two electrons can occupy the same quantum state, in accordance with the Pauli exclusion principle. Like all matter, electrons have properties of both particles and waves, and so can collide with other particles and can be diffracted like light. The wave properties of electrons are easier to observe with experiments than those of other particles like neutrons and protons because electrons have a lower mass and hence a higher De Broglie wavelength for typical energies. Hope this HELPS :D</span>