anonymous
  • anonymous
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.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
ΔV / V specified ..it also has this in the question I am not sure how to incorporate it either
zepdrix
  • zepdrix
Mmm lemme see if I've got the right idea here.
zepdrix
  • 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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zepdrix
  • zepdrix
\[\Large \Delta r=\pm 0.07\]
zepdrix
  • zepdrix
So plugging that value in, what do we get for DeltaV ?
anonymous
  • anonymous
is it 0.29 ?
anonymous
  • 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
  • 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
  • anonymous
so I get 83.36
zepdrix
  • zepdrix
`ΔV / V specified` What does that mean?
anonymous
  • anonymous
I really am not sure either it says ΔV / V specified = ____%
anonymous
  • anonymous
the teacher said that delta doesnt always mean change but I dont get what he was trying to say
zepdrix
  • zepdrix
Ok ok ok ok ok I think this is what we do. Lemme see if this makes sense.
zepdrix
  • zepdrix
Ahhh crap i gotta go :( SOrry sorry
zepdrix
  • zepdrix
@.Sam. @thomaster
anonymous
  • 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
  • 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
  • anonymous
so 8% is % uncertainty
anonymous
  • anonymous
thank you very much, thank you thank you!
anonymous
  • anonymous
If you use the second approach, it is 8%.
anonymous
  • anonymous
oh do you know what you got for Vo in the 1st approach ?
anonymous
  • anonymous
I do not understand your question
anonymous
  • anonymous
never mind but thank you!
anonymous
  • anonymous
ok
anonymous
  • anonymous
do you mind helping me on this other question I have
anonymous
  • anonymous
shoot
anonymous
  • 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
  • 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
  • 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
  • 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
  • anonymous
would it also be same?
anonymous
  • 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
  • anonymous
the other one is asking for distance though?
anonymous
  • 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
  • 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
  • anonymous
ok so the motorcycle has the greater distance than the bike but in acceleration there both the same?
anonymous
  • anonymous
Exactly!
anonymous
  • anonymous
It is just a problem, I guess it is not like this in real life (the acceleration thing, I mean)
anonymous
  • anonymous
can you also think of as the motorcycle is faster so it goes a longer way than the bike
anonymous
  • 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
  • anonymous
ohh okay thank you! are sticking around for a little bit I had a few more questions if you dont mind
anonymous
  • 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
  • anonymous
oki thank you very much
anonymous
  • anonymous
talk to you tomorrow!
anonymous
  • anonymous
thanks bye GN!
anonymous
  • anonymous
GN!

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