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Astronauts visiting a new planet find a lake filled with an unknown liquid. They have with them a plastic cube, 8.0 cm on each side, with a density of 840 kg/m3. First they weigh the cube with a spring scale, measuring a weight of 25 N. Then they float the cube in the lake and find that two-thirds of the cube is submerged.At what depth in the lake will the pressure be twice the atmospheric pressure of 85 kPa?

Respuesta :

Answer:

depth is 1.16 m

Explanation:

given data

length = 8 cm

density = 840 kg/m³

weight = 25 N

atmospheric pressure = 85 kPa

to find out

what depth in the lake will the pressure be twice the atmospheric pressure

solution

we know here that two thirds of the cube is submerged

we get first mass of cube that is

mass = 0.08³ × 0.08 = 0.43008 kg

and g = weight ÷ mass

g = [tex]\frac{25}{0.43008}[/tex]

g = 58.12 m/s²

and we get now ρ for two thirds of the cube is submerged that is

weight = 0.08³ × 58.12 × ρ × [tex]\frac{2}{3}[/tex]

25 = 0.08³ × 58.12 × ρ × [tex]\frac{2}{3}[/tex]

ρ = 1260.18 kg/m³

pressure = ρgd   ..................1

pressure = 1260.18 × 58.12 × d

85 × 10³ = 1260.18 × 58.12 × d

d = 1.16 m

snehashish65

The pressure will be twice the atmospheric pressure at the depth of 2.31  m below the lake.

Given data:

Length of side of plastic cube is, [tex]d=8.0 \;\rm cm =0.08\;\rm m[/tex].

The density of cube is, [tex]\rho =840 \;\rm kg/m^{3}[/tex].

The weight of cube is, [tex]W=25\;\rm N[/tex].

Atmospheric pressure is, [tex]P_{a}=85 \;\rm kPa =85 \times 10^{3} \;\rm Pa[/tex].

For twice the atmospheric pressure, the expression is given as,

[tex]P'= \rho \times g\times h \\2P_{a}= \rho \times g\times h .....................................(1)[/tex]

here, g is the gravitational acceleration and h is the depth of lake.

From the weight of cube,

[tex]W=m \times g[/tex]

m is the mass of cube. And for two-third submerged volume,

[tex]\rho =\dfrac{m}{V'} =\dfrac{m}{(\dfrac{2}{3}V)}[/tex]

Here, V is the volume of cube. Solving as,

[tex]\rho \times \dfrac{2}{3} \times d^{3}=m}\\m=840 \times \dfrac{2}{3} \times 0.08^{3}\\m=0.286 \;\rm kg[/tex]

Then gravitational acceleration is,

[tex]25=0.286 \times g\\g=87.41 \;\rm m/s^{2}[/tex]

Substituting values in equation (1) as,

[tex]2 \times 85 \times 10^{3}= 840 \times 87.41\times h\\h =\dfrac{2 \times 85 \times 10^{3}}{840 \times 87.41} \\h =2.31 \;\rm m[/tex]

Thus, we can conclude that at a depth of 2.31 m, the pressure will be twice the atmospheric pressure.

Learn more about atmospheric pressure here:

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