For this problem, we use the Coulomb's law written in equation as:
F = kQ₁Q₂/d²
where
F is the electrical force
k is a constant equal to 9×10⁹
Q₁ and Q₂ are the charge of the two objects
d is the distance between the two objects
Substituting the values:
F = (9×10⁹)(-22×10⁻⁹ C)(-22×10⁻⁹ C)/(0.10 m)²
F = 0.0004356 N
Answer:
14.2 m
Explanation:
Using conservation of energy:
PE at top = KE at bottom
mgh = ½ mv²
h = v² / (2g)
h = (16.7 m/s)² / (2 × 9.8 m/s²)
h = 14.2 m
Using kinematics:
Given:
v₀ = 16.7 m/s
v = 0 m/s
a = -9.8 m/s²
Find: Δy
v² = v₀² + 2aΔy
(0 m/s)² = (16.7 m/s)² + 2 (-9.8 m/s²) Δy
Δy = 14.2 m
Answer:
The girl has greater tangential acceleration
Explanation:
The angular acceleration (
) of the merry go round is equal to the rate of the change of the angular velocity,
:

Since all the points of the merry go round complete 1 circle in the same time, the angular velocity of each point of the merry go round is the same, and so all the points also have the same angular acceleration.
The tangential acceleration instead is given by

where
is the angular acceleration
r is the distance from the centre of the merry go round
Since the girl is near the outer edge and the boy is closer to the centre, the value of r for the girl is larger than for the boy, so the girl has greater tangential acceleration.