Answer:
0.10013 atm
Explanation:
Applying Boyle's Law,
P'V' = PV................... Equation 1
Where P' = Initial pressure of air, V' = Initial volume of air, P = Final pressure of air, V = Final volume of air.
make P the subject of the equation
P = P'V'/V..................... Equation 2
Given: P' = 0.355 atm, V' 0.110 m³, V = 0.390 m³
Substitute into equation 2
P = 0.355(0.11)/0.39
P = 0.10013 atm.
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
You have learned your lesson well, Suhay. Your statement is correct.
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