Answer:
ρ = 1.13 10⁴ km/m³
Explanation:
For this exercise we use Newton's equilibrium equation
B –W + W_scale = 0
Where B is the thrust and W_scale is the balance reading
The push is given by Archimedes' law
B = ρ_water g V
B = W- W_scale
B = m g - m_scale g
Let's calculate
B = 14.7 9.8 - 13.4 9.8
B = 12.74 N
ρ_water g V = 12.74
V = 12.74 / ρ_water g
V = 12.74 / 1000 9.8
V = 0.0013 m³
Let's use density
ρ = m / V
We replace
ρ = 14.7 / 0.0013
ρ = 1.13 10⁴ km/m³
Mass= density x volume
1.3 kg/m^3 x ( 2.5x4x10) m^3
= 130 kg
<span>it either changes the direction of the object or it stops the object in motion.</span>
Complete question:
It is measured that 3/4 of a body's volume is submerged in oil of density 800kg/m³. What is the specific gravity of oil?
Answer:
The specific gravity of the oil is 0.8.
Explanation:
Given;
density of the oil,
= 800 kg/m³
density of water,
= 1000 kg/m³
The specific gravity of any substance is the ratio of the substance density to the density of water.
Specific gravity of the oil = density of the oil / density of water
Specific gravity of the oil = 800/1000
Specific gravity of the oil = 0.8
Therefore, the specific gravity of the oil is 0.8.
Not sure if this it completely right but here’s what I got: 77.16050617284 J