A 4.5 kg box slides down a 4.3-m-high frictionless hill, starting from rest, across a 2.0-m-wide horizontal surface, then hits a
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Answer:
h = 9.83 cm
Explanation:
Let's analyze this interesting exercise a bit, let's start by comparing the density of the ball with that of water
let's reduce the magnitudes to the SI system
r = 10 cm = 0.10 m
m = 10 g = 0.010 kg
A = 100 cm² = 0.01 m²
the definition of density is
ρ = m / V
the volume of a sphere
V =
V = π 0.1³
V = 4.189 10⁻³ m³
let's calculate the density of the ball
ρ =
ρ = 2.387 kg / m³
the tabulated density of water is
ρ_water = 997 kg / m³
we can see that the density of the body is less than the density of water. Consequently the body floats in the water, therefore the water level that rises corresponds to the submerged part of the body. Let's write the equilibrium equation
B - W = 0
B = W
where B is the thrust that is given by Archimedes' principle
ρ_liquid g V_submerged = m g
V_submerged = m / ρ_liquid
we calculate
V _submerged = 0.10 9.8 / 997
V_submerged = 9.83 10⁻⁴ m³
The volume increassed of the water container
V = A h
h = V / A
let's calculate
h = 9.83 10⁻⁴ / 0.01
h = 0.0983 m
this is equal to h = 9.83 cm