If a bubble in sparkling water accelerates upward at the rate of0.225m/s2and has a radius of 0.500mm, what is its mass? Assume that the drag force on the bubble is negligible.

Short Answer

Expert verified

The mass of the bubble in sparkling water is5.11x10-7kg.

Step by step solution

01

Listing the given quantities.

  • The radius of the bubble

(r)=0.500mm

=0.500x10-3m

  • The acceleration is, a=0.225m/s2.
  • The density of water is ρw=1000kg/m3
02

Understanding the concept of buoyant force.

There are two forces acting on the bubble – buoyant force (Fb)and weight.As the bubble is accelerating, using Newton’s second law, we can find the mass of the bubble(Mb).

Formula:

Fnet=Ma

Fb=Mfg

W=Mbg

V=43πr3

03

Calculation of mass of the bubble.

The volume of the water displaced by the bubble (Vw)= Volume of the bubble (Vb)

Vw=Vb

=43πr3

Substitute the value in the above equation.

=43x3.14x(500x10-3m)3

role="math" localid="1657556016855" =1.57x10-9m3(1)

Using Newton’s second law,

Fb-W=Mba

ρwVbg-ρbgVb=ρbVba

Since the volume is the same, it would be canceled.

Rearranging the equation,

ρb=ρwgg+a

Substitute the values in the above equation.

=998m3x9.8m/s29.8m/s2+0.255m/s2

role="math" localid="1657556387852" =975.6kg/m3(2)

Mb=ρbxVb

Using the values from (1) and (2), we have

Mb=1.57x10-9kg/m3x975.6m3

=5.11x10-7kg

Mass of the bubble in sparkling water is5.11x10-7kg.

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