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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The plastic tube in Figure has a cross-sectional area of5.00cm2. The tube is filled with water until the short arm (of lengthd=0.800m) is full. Then the short arm is sealed and more water is gradually poured into the long arm. If the seal will pop off when the force on it exceeds9.80N, what total height of water in the long arm will put the seal on the verge of popping?

In Figure, water stands at depth D=35.0mbehind the vertical upstream face of a dam of width W=314m .(a)Find the net horizontal force on the dam from the gauge pressure of the water.(b)Find the net torque due to that force about a horizontal line through parallel to the (long) width of the dam. This torque tends to rotate the dam around that line, which would cause the dam to fail. (c)Find the moment arm of the torque.

Fresh water flows horizontally from pipe section 1 of cross-sectional area A1into pipe section 2 of cross-sectional area A2. Figure gives a plot of the pressure difference p2-p1versus the inverse area squared A,-2that would be expected for a volume flow rate of a certain value if the water flow were laminar under all circumstances. The scale on the vertical axis is set by ps=300kN/m2. For the conditions of the figure,

(a) what is the value of A2and

(b) what is the value of the volume flow rate?

Water is moving with a speed of 5.0m/sthrough a pipe with a cross-sectional area of 4.0cm2. The water gradually descends 10mas the pipe cross-sectional area increases to 8.0\mathrm{~cm}^{2}.

(a) What is the speed at the lower level?

(b) If the pressure at the upper level is1.5\times10^{5}\mathrm{~Pa},What is the pressure at the lower level?

A 8.60kgsphere of radius 6.22cmis at a depth of 2.22kmin seawater that has an average density of 1025kg/m3.What are the (a) gauge pressure, (b) total pressure, and (c) corresponding total force compressing the sphere’s surface? What are (d) the magnitude of the buoyant force on the sphere and (e) the magnitude of the sphere’s acceleration if it is free to move? Take atmospheric pressure to be 1.01×105Pa.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free