Answer:
, repulsive
Explanation:
The magnitude of the electric force between two charged particles is given by Coulomb's law:
where:
is the Coulomb's constant
are the two charges of the two particles
r is the separation between the two charges
The force is:
- repulsive if the two charges have same sign
- Attractive if the two charges have opposite signs
In this problem, we have two electrons, so:
is the magnitude of the two electrons
is their separation
Substituting into the formula, we find the electric force between them:
And the force is repulsive, since the two electrons have same sign charge.
Answer:
wo = 18.75 rev / s
Explanation:
This is an exercise in endowment kinematics, it indicates that the final angular velocity is w_f = 109 rad / s, the time to reach this velocity is t = 1.87 s and the deceleration a = 4.7 rad / s²
w_f = w₀ - a t
w₀ = w_f + a t
w₀ = 109 + 4.7 1.87
w₀ = 117.8 rad / s
let's reduce to revolutions / s
w₀ = 117.8 rad / s (1 rev / 2pi rad)
w₀ = 18.75 rev / s
Answer:
a) The angular acceleration of the beam is 0.5 rad/s²CW (direction clockwise due the tangential acceleration is positive)
b) The acceleration of point A is 3.25 m/s²
The acceleration of point E is 0.75 m/s²
Explanation:
a) The relative acceleration of B with respect to D is equal:
Where
aB = absolute acceleration of point B = 2.5 j (m/s²)
aD = absolute acceleration of point D = 1.5 j (m/s²)
(aB/D)n = relative acceleration of point B respect to D (normal direction BD) = 0, no angular velocity of the beam
(aB/D)t = relative acceleration of point B respect to D (tangential direction BD)
We have that
(aB/D)t = BDα
Where α = acceleration of the beam
BDα = 1 m/s²
Where
BD = 2
b) The acceleration of point A is:
(aA/D)t = ADαj
The acceleration of point E is:
(aE/D)t = -EDαj
