The gravitational force of two objects, by definition, is given by:
Where,
G: gravitational constant
m1: mass of object number 1.
m2: mass of object number 2.
d: distance between both objects.
Therefore, according to the given equation, a change that always results in an increase in gravitational force is:
Increase in the mass of the objects and decrease in the distance between them.
Answer:
A change that will always result in an increase in the gravitational force between two objects is:
Increase in the mass of the objects and decrease in the distance between them.
Answer:
(a) α = -0.16 rad/s²
(b) t = 33.2 s
Explanation:
(a)
Applying 3rd equation of motion on the circular motion of the tire:
2αθ = ωf² - ωi²
where,
α = angular acceleration = ?
ωf = final angular velocity = 0 rad/s (tire finally stops)
ωi = initial angular velocity = 5.45 rad/s
θ = Angular Displacement = (14.4 rev)(2π rad/1 rev) = 28.8π rad
Therefore,
2(α)(28.8π rad) = (0 rad/s)² - (5.45 rad/s)²
α = -(29.7 rad²/s²)/(57.6π rad)
<u>α = -0.16 rad/s²</u>
<u>Negative sign shows deceleration</u>
<u></u>
(b)
Now, we apply 1st equation of motion:
ωf = ωi + αt
0 rad/s = 5.45 rad/s + (-0.16 rad/s²)t
t = (5.45 rad/s)/(0.16 rad/s²)
<u>t = 33.2 s</u>
Answer:
d) It will be cut to a fourth of the original force.
Explanation:
The magnitude of the electrostatic force between the charged objects is
where
k is the Coulomb's constant
q1 and q2 are the charges of the two objects
r is the separation between the two objects
In this problem, the initial distance is doubled, so
r' = 2r
Therefore, the new electrostatic force will be
So, the force will be cut to 1/4 of the original value.
