Respuesta :
Answer:
The mass of the liquid = 10538 kg
Explanation:
The pressure in a liquid is
P = ρgh ......................... Equation 1
ρ = P/gh ...................... Equation 2
Where P = pressure, ρ = density, g = acceleration due to gravity, h = height.
Given: p = 1.3 atm, h = 2.0 m, g = 9.81 m/s²
If, 1 atm = 1.013×10⁵ N/m²
Then, P = 1.3×1.013×10⁵ N/m² = 1.3169×10⁵ N/m²
Substituting in equation 2,
ρ = 1.3169×10⁵/(9.81×2)
ρ = 1.3169×10⁵/19.62
ρ = 6712.03 kg/m³.
But Density,
ρ = m/v
m = ρ × v........................ Equation 3
Where m = mass of the liquid, v = volume of the liquid in the vat
v = πd²h/4, where d = diameter = 1.0 m, h = 2.0 m.
v = 3.14(1)²×2/4
v = 1.57 m³ also, ρ = 6712.03 kg/m³.
Substituting into equation 3
m = 1.57×6712.03
m = 10537.887 kg
m ≈ 10538 kg.
Thus the mass of the liquid = 10538 kg
The mass of the liquid in the vat is 2435.54 kg
Suppose we consider [tex]\mathbf{\rho}[/tex] to be the density of the liquid;
Then the pressure situated at the bottom with a depth (h) of the vat can now be:
[tex]\mathbf{P = P_o + \rho gh }[/tex]
where;
- [tex]\mathbf{ P_o }[/tex] = atmospheric pressure.
[tex]\mathbf{P -P_o = \rho gh }[/tex]
[tex]\mathbf{ \rho = \dfrac{P -P_o }{gh }}[/tex]
We know that:
[tex]\mathbf{density (\rho) = \dfrac{mass}{volume}}[/tex]
The mass of the liquid in the vat can be determined by the formula:
[tex]\mathbf{m = \rho V}[/tex]
where;
[tex]\mathbf{ \rho = \dfrac{P -P_o }{gh }}[/tex]
Then;
[tex]\mathbf{m = \Big( \dfrac{P -P_o }{gh } \Big)\times Area (A) \times h}[/tex]
[tex]\mathbf{m = \Big( \dfrac{P -P_o }{g } \Big)\times Area (A) }[/tex]
[tex]\mathbf{m = \Big( \dfrac{P -P_o }{g } \Big)\times \pi \Big(\dfrac{d}{2}\Big )^2}[/tex]
replacing the values, we have:
[tex]\mathbf{m = \Big( \dfrac{1.3 \ atm - 1.0 \ atm }{9.8 \ m/s^2 }\Big)\times \pi \Big(\dfrac{1.0}{2}\Big )^2}[/tex]
[tex]\mathbf{m = \Big( \dfrac{1.3 \ atm - 1.0 \ atm }{9.8 \ m/s^2 } \Big) \times \Big ( \dfrac{1.013 \times 10^5 \ Pa}{1.0 \ atm } \Big) \times \pi \Big(\dfrac{1.0}{2}\Big )^2}[/tex]
[tex]\mathbf{m =(3101.020408 \times 0.7853981634) \ kg}[/tex]
[tex]\mathbf{m \simeq 2435.54 \ kg}[/tex]
Therefore, we can conclude that the mass of the liquid in the vat is 2435.54 kg
Learn more about the relation of density here:
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