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.
Answer:
ω = 12.023 rad/s
α = 222.61 rad/s²
Explanation:
We are given;
ω0 = 2.37 rad/s, t = 0 sec
ω =?, t = 0.22 sec
α =?
θ = 57°
From formulas,
Tangential acceleration; a_t = rα
Normal acceleration; a_n = rω²
tan θ = a_t/a_n
Thus; tan θ = rα/rω² = α/ω²
tan θ = α/ω²
α = ω²tan θ
Now, α = dω/dt
So; dω/dt = ω²tan θ
Rearranging, we have;
dω/ω² = dt × tan θ
Integrating both sides, we have;
(ω, ω0)∫dω/ω² = (t, 0)∫dt × tan θ
This gives;
-1[(1/ω_o) - (1/ω)] = t(tan θ)
Thus;
ω = ω_o/(1 - (ω_o × t × tan θ))
While;
α = dω/dt = ((ω_o)²×tan θ)/(1 - (ω_o × t × tan θ))²
Thus, plugging in the relevant values;
ω = 2.37/(1 - (2.37 × 0.22 × tan 57))
ω = 12.023 rad/s
Also;
α = (2.37² × tan 57)/(1 - (2.37 × 0.22 × tan 57))²
α = 8.64926751525/0.03885408979 = 222.61 rad/s²
Answer:
Every action has an equal and opposite reaction. If the student doesn't push, nothing moves, is one student pushes, both move which is an example of newtons third law.
Explanation:
