Calculating Volume Change of a Partially Inflated Yoga Ball

  • Thread starter Thread starter lalahelp
  • Start date Start date
  • Tags Tags
    Volume
AI Thread Summary
To calculate the volume change of a partially inflated yoga ball, the surface area of the ball is initially measured and then found to be 1.5 times larger when fully inflated. The relevant formulas for surface area (S = 4*π*r^2) and volume (V = [4*π*r^3]/3) are established. By setting up the equation S2 = 1.5*S1, the relationship between the radii of the partially inflated and fully inflated balls can be derived. This relationship allows for the calculation of the volume change factor (F = V2/V1). The discussion highlights the need for clarification on the steps to derive the volume change from the surface area increase.
lalahelp
Messages
74
Reaction score
0

Homework Statement



Initially, A yoga ball is partially blown up and its surface area measured. Afterwards, it is filled completely and its surface area is found to be 1.50 times as big as it was recorded earlier. By what factor has its volume changed?



Im not sure how to solve for this? can someone explain please


The Attempt at a Solution

 
Physics news on Phys.org
First write down any formulas that you think might help,

Surface area of sphere S = 4*∏*r^2

Volume of a sphere V = [4*∏*r^3]/3

Write down what you know, S2 = 1.5*S1 or

4*∏*r2^2 = 1.5*4*∏*r1^2 simplify and use this to find,

what you want to know, F*V1 = V2 or F = V2/V1 = ? Use the above expression for volume to write out and simplify V2/V1.

Does this get you started?
 
No, I don't understand your steps of what to do.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top