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Chapter 14: Q. 18 (page 384)

Astronauts visiting a new planet find a lake filled with an unknown liquid. They have with them a plastic cube, 6.0cmon each side, with a density of 840Kg/m3. First they weigh the cube with a spring scale, measuring a weight of 21N. 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 of85kPa ?

Short Answer

Expert verified

The depth value for the lake will the pressure be twice is 0.5827 m

Step by step solution

01

Given information

Side of the cube is6cm.

Density value is 840Kg/m3.

Weight is 21N.

Atmospheric pressure is 85kPa.6cm

02

Calculation for depth

Mass of the cube is,

=(0.06×(840)

=0.18144Kg

It weighs 21 N so

we can find the value of g

mg = 21 N

=> 0.18144 kg x g = 21 N

=> g = 115.740 m/s2

The cube is immersed two-thirds of the way.

So,

(0.06m)323×(115.740m/s2)ρ=21N

ρ=1260.18kg/m3

Depth is,

p=ρgd

localid="1651160111576" ⇒1260.18×(115.740)×d=85×103

From solving that,

localid="1651160159914" d=0.5827m

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Most popular questions from this chapter

In addition to the buoyant force, an object moving in a liquid experiences a linear drag force F→drag=(bv, direction opposite the motion), where bis a constant. For a sphere of radius R, the drag constant can be shown to be b=6πηR, where ηis the viscosity of the liquid. Consider a sphere of radiusRand density ρthat is released from rest at the surface of a liquid with density ρf.

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