A metal coin has a volume of 835 mm3 and a mass of 5.67 g. What is the density of the coin?
2 answers:
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Answer:
= 5/9
Explanation:
This is an exercise that we can solve using Archimedes' principle which states that the thrust is equal to the weight of the desalted liquid.
B = ρ_liquid g V_liquid
let's write the translational equilibrium condition
B - W = 0
let's use the definition of density
ρ_body = m / V_body
m = ρ_body V_body
W = ρ_body V_body g
we substitute
ρ_liquid g V_liquid = ρ_body g V_body
In the problem they indicate that the ratio of densities is 5/9, we write the volume of the bar
V = A h_bogy
Thus
we substitute
5/9 =
Answer:
The positive displacement from the midpoint of its motion at the speed equal one half of its maximum speed is 3.56 cm.
Explanation:
Maximum speed is :
v (max) = Aω
Speed v at any displacement y is given by
=
(
-
) ........................................................ i
And,
v = v (max)
or, 2 × v = Aω .................................................... ii
Eliminating ω from equations i and ii,
=
(
-
)
or, = (
)
=(
)
or, y = 3.56 cm.