Percent Uncertainty in Volume of Spherical Beach Ball

In summary, the conversation is about finding the percent uncertainty in the volume of a spherical beach ball with a given radius of 2.86m +/- 0.08m. The equation V = (4/3)pi r^3 is used to calculate the volume, and the resulting values are 97.9911 and 106.4463. The difference between these two values is 8.4552, which when divided by the original value of 97.9911 and multiplied by 100, gives a percent uncertainty of 8.6285%. However, due to the given significant figures of one, the final answer is rounded to 9%. The conversation also includes
  • #1
Struggling
52
0
hi didnt know wether to put this in phyics or maths but I am studying for physics so ill put it here.

Q: What is the percent uncertainty in the volume of a spherical beach ball whos radius is r = 2.86 +or-0.08m?

A: V= 4/3 pie r^3

V = 4/3 pie(2.94)^3 = 106.4463
V = 4/3 pie(2.86)^3 = 97.9911

106.4463-97.9911 = 8.4552

8.4552/97.991 x 100 = 8.6285%

shouldnt 8.6285% be the answer? the book has 9% which i understand could be rounded but considering I am doing the scientific notation chapter and it talks about how important it is to be clear with your answer.

i also tried the equation using only 3 significant numbers but i turned out further away from the answer getting something like 8.1%.

have i done something wrong in my calculations? this question really is simple but i can't think of any other way.

thanks
 
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  • #2
You might want to watch out for significative figures.What about the [itex]2.78 \mbox{cm} [/itex] value...?

And it's "pi" and it has no taste,it's a greek letter.

Daniel.
 
  • #3
yeh i tried it as well i get:

v=4/3 pi (2.78)^3 = 90.0
v=4/3 pi (2.94)^3 = 106
v=4/3 pi (2.86)^3 = 98.0

and i get lost here?

are my significant figures right?
 
  • #4
You need to use the extra digits to do the calculation, just as you have done, but then your final answer needs to match the least number of significant digits given in the problem; in this case, +or- 0.08m has only one significant digit. Therefore, the answer 8.6285 gets rounded to 9%.
 
  • #5
i need help on this problem also. Is there a method in solving this
 

1. What is percent uncertainty in volume?

Percent uncertainty in volume refers to the amount of uncertainty or potential error in the measurement of a spherical beach ball's volume. It is usually expressed as a percentage and can be calculated by taking the difference between the maximum and minimum possible values for the volume and dividing it by the average value.

2. Why is percent uncertainty in volume important?

Percent uncertainty in volume is important because it gives an indication of the accuracy of the measurement. The higher the percent uncertainty, the less precise the measurement is. This information can be useful in determining the reliability of data and making decisions based on the measurement.

3. How is percent uncertainty in volume calculated?

Percent uncertainty in volume is calculated by taking the difference between the maximum and minimum possible values for the volume and dividing it by the average value. This value is then multiplied by 100 to get the percentage. For example, if the maximum volume is 100 cm^3 and the minimum volume is 90 cm^3, and the average value is 95 cm^3, the percent uncertainty would be (100-90)/95 x 100 = 10.5%.

4. What factors can contribute to percent uncertainty in volume?

There are several factors that can contribute to percent uncertainty in volume, including human error in measurement, variations in the shape or size of the beach ball, and limitations of the measuring equipment. Environmental factors such as temperature and pressure can also affect the volume of the beach ball and contribute to uncertainty.

5. How can percent uncertainty in volume be reduced?

Percent uncertainty in volume can be reduced by using more precise measuring equipment, taking multiple measurements and averaging them, and minimizing external factors that could affect the volume. It is also important to follow proper measurement techniques and minimize any potential sources of human error.

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