What, roughly, is the percent uncertainty in the volume of a spherical beach ball whose radius is r=2.64±0.07m?
Express your answer using one significant figure.

- anonymous

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- anonymous

ΔV / V specified ..it also has this in the question I am not sure how to incorporate it either

- zepdrix

Mmm lemme see if I've got the right idea here.

- zepdrix

\[\Large V=\frac{4}{3}\pi r^3 \qquad\to\qquad \Delta V=\frac{4}{3}\pi (\Delta r)^3\]

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## More answers

- zepdrix

\[\Large \Delta r=\pm 0.07\]

- zepdrix

So plugging that value in, what do we get for DeltaV ?

- anonymous

is it 0.29 ?

- anonymous

this is what I did before
v= 4/3 (pi) (2.64 + 0.07) = 11.35162
&
v= 4/3 (pi) ( 2.64 - 0.07) = 10.76519
now do I subtract these two answers from each other
like this
11.352-10.765 = 0.587
then do this
0.0587 / 2.64 x 100 = 22%... something seems wrong

- zepdrix

Mmmm maybe that's the right approach.
But shouldn't there be a 3 exponent when you do that first step?\[\Huge V= 4/3 (\pi) (2.64 + 0.07)^{\color{red}{3}}\]

- anonymous

so I get 83.36

- zepdrix

`ΔV / V specified`
What does that mean?

- anonymous

I really am not sure either it says
ΔV / V specified = ____%

- anonymous

the teacher said that delta doesnt always mean change but I dont get what he was trying to say

- zepdrix

Ok ok ok ok ok I think this is what we do.
Lemme see if this makes sense.

- zepdrix

Ahhh crap i gotta go :( SOrry sorry

- zepdrix

@.Sam. @thomaster

- anonymous

You need to calculate three volumes:
when R=2.64-0.7=2.57----->V1=71.10--->(V1-Vo)/Vo=-7.7%
when R=2.64------>Vo=77.07
when R=2.64+0.07=2.71-------->V2=83.37--->(V2-Vo)/Vo=+8.2% Then
V=77.07 (+8.2%/-7.7%)

- anonymous

Another way to calculate it\[V=\frac{ 4 }{ 3 }\pi r^3\rightarrow \Delta V=\frac{ \delta V }{ \delta r }\Delta r=4 \pi r^2 \Delta r \rightarrow \frac{ \Delta V }{ V } =3\frac{ \Delta r }{ r }\]for r=2.64 and ∆=0.07 you get:\[\frac{ \Delta V }{ V }=\frac{ 3·0.07 }{ 2.64 }=0.080\] and that correspoonds to an 8.0%(+/-)

- anonymous

so 8% is % uncertainty

- anonymous

thank you very much, thank you thank you!

- anonymous

If you use the second approach, it is 8%.

- anonymous

oh do you know what you got for Vo in the 1st approach ?

- anonymous

I do not understand your question

- anonymous

never mind but thank you!

- anonymous

ok

- anonymous

do you mind helping me on this other question I have

- anonymous

shoot

- anonymous

Compare the acceleration of a motorcycle that accelerates from 80 km/h to 90 km/h with the acceleration of a bicycle that accelerates from rest to 10 km/h in the same time.
Part A
Compare the distance traveled between the motorcycle and the bicycle
-- The motorcycle and bicycle travels the same distance.
-- The motorcycle travels a greater distance than the bicycle.
-- The motorcycle travels a smaller distance than the bicycle.

- anonymous

For the motorcycle:\[ V_m=v_{0m}+a_m·t \rightarrow t=\frac{ V_m-v_{0m} }{ a_m }=\frac{ 90-80 }{ a_m }\] For the bike\[V_b=v_{0b}+a_b·t \rightarrow t=\frac{ V_b-v_{0b} }{ a_b }=\frac{ 10-0 }{ a_b }=\frac{ 10 }{ a_b }\]

- anonymous

As all that happens in the same time:\[\frac{ 10 }{ a_m}=\frac{ 10 }{ a_b }\]and that means accelerations are the same

- anonymous

this is Part B
Compare the acceleration between the motorcycle and the bicycle.
Compare the acceleration between the motorcycle and the bicycle.
The motorcycle has a greater acceleration than the bicycle.
The motorcycle has a smaller acceleration than the bicycle.
The motorcycle and bicycle have the same acceleration.

- anonymous

would it also be same?

- anonymous

yes, both increase the velocity by the same amount (10 Km/h) in the same period of time, then accelerations are the same

- anonymous

the other one is asking for distance though?

- anonymous

That is easy, distance is given by the following formula:\[D=v_0t+\frac{ 1 }{ 2 }a t^2\]And accelerations are the same and time is the same, then the only thing that will make both distances (the motorcycle and the bike) is their initial velocity. Since the only one that has initial velocity is the motorcycle, then its distance is longer than the bike's
Does it make sense to you?

- anonymous

Let me rephrase it:
And accelerations are the same and time is the same, then the only thing that will make both distances (the motorcycle and the bike) DIFFERENT is their initial velocity. Since the only one that has initial velocity is the motorcycle, then its distance is longer than the bike's Does it make sense to you

- anonymous

ok so the motorcycle has the greater distance than the bike but in acceleration there both the same?

- anonymous

Exactly!

- anonymous

It is just a problem, I guess it is not like this in real life (the acceleration thing, I mean)

- anonymous

can you also think of as the motorcycle is faster so it goes a longer way than the bike

- anonymous

Is faster because it had an initial velocity. If they had both started from rest they would have run the same distance because they had the same acceleration. Then what makes the difference is the initial velocity of the motorbike =80 Km/h

- anonymous

ohh okay thank you! are sticking around for a little bit I had a few more questions if you dont mind

- anonymous

Well, can we leave them for tomorrow? I am writing from Spain and it is 05:00. I would love to hit the hay!

- anonymous

oki thank you very much

- anonymous

talk to you tomorrow!

- anonymous

thanks bye GN!

- anonymous

GN!

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