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Physics 21 Online
OpenStudy (iwanttogotostanford):

A bike has an inertia of 10.04 × 1237 mg•m3 and a mass of 7.2 × 1094 kg. Based on this, what is the bike rolling down's radius?

OpenStudy (iwanttogotostanford):

@Nnesha @pooja195

OpenStudy (iwanttogotostanford):

@Mashy

OpenStudy (anonymous):

Do you know what is the expression for Moment of inertia of a solid sphere?

OpenStudy (iwanttogotostanford):

I=2/5 MR^2 correct?

OpenStudy (anonymous):

that is correct. Mass is given to you and I is given to you so can't you just substitute and solve for R ?

OpenStudy (iwanttogotostanford):

i got R=93442.8, but i don't think thats right

OpenStudy (anonymous):

No, that isn't.. Just crunch in the numbers carefully

OpenStudy (iwanttogotostanford):

what?

OpenStudy (iwanttogotostanford):

i got 9.74x1037=2/5(r)^2

OpenStudy (iwanttogotostanford):

and mass in there

OpenStudy (anonymous):

it is 10 power 37, hope you plugged that in correctly and then took the square root!

OpenStudy (iwanttogotostanford):

ok ill try one more time and see what i get hold on please

OpenStudy (iwanttogotostanford):

do i have to use the m^2 in the moment of inertia??!

OpenStudy (iwanttogotostanford):

ok, this time i got R= 9.28, am i getting there? @Mashy

OpenStudy (anonymous):

Nope

OpenStudy (iwanttogotostanford):

ok well I obviously need help, could i get some please???

OpenStudy (anonymous):

Do step by step.. \[R^2 = \frac{5I}{2M} = \frac{5 \times 9.74 \times 10^{37}}{5.98 \times 10^{24}}\]

OpenStudy (iwanttogotostanford):

where'd you get the r^2

OpenStudy (anonymous):

Crunch in the numbers and tell me what number pops out for R square?

OpenStudy (anonymous):

In the moment of inertial equation buddy!

OpenStudy (iwanttogotostanford):

i got a really crazy number for my answer... for the second part of the equation only though i got: 8.14x10^64 :-(

OpenStudy (anonymous):

I forgto to put a 2 in the denominator But I guess your calculator is broken, or you don't know how to use it :P

OpenStudy (iwanttogotostanford):

you are sort of rude... you are here to help not insult me on how your physics are superior to mine, you are a teacher not showing off what you can do. i asked for help, obviously i don't know how to do it or i wouldn't be asking for your help!

OpenStudy (anonymous):

Oh, I am not trying to be rude, I am just not sure, how to show you to put numbers in the calculator

OpenStudy (iwanttogotostanford):

ok well I asked for help so obviously I am going to not know exactly how to do it, otherwise I wouldn't be asking. Sorry if me not knowing is annoying, but thats why Ia m here

OpenStudy (anonymous):

Tell me what answer you get for R^2???

OpenStudy (iwanttogotostanford):

I am not sure because thats what I am trying to figure out. first i did 5×9.74×10375.98×1024 part of the equation and got an absurd answer of : 8.14x10^61 and i don't know what to do next

OpenStudy (anonymous):

But that is wrong.. I forgot to put a 2 in the denominator and secondly the powers must be SUBTRACTED

OpenStudy (iwanttogotostanford):

oh ok, sorry!

OpenStudy (anonymous):

\[R^2 = \frac{5 \times 9.74 \times 10^{37}}{2 \times 5.98 \times 10^{24}}\]

OpenStudy (iwanttogotostanford):

now i got 1.32 x10 ^54...

OpenStudy (anonymous):

No it is \[4.07 \times 10^{13}\] :-/

OpenStudy (iwanttogotostanford):

oh, the two 10's simplify to one ok

OpenStudy (iwanttogotostanford):

so now what?

OpenStudy (anonymous):

Now take the root

OpenStudy (iwanttogotostanford):

how do i do that again? sorry....I am in middle school but have to take this

OpenStudy (anonymous):

Are you using a calculator?

OpenStudy (iwanttogotostanford):

yes

OpenStudy (anonymous):

Is there a button which says \[\sqrt .\]

OpenStudy (iwanttogotostanford):

yes

OpenStudy (anonymous):

That is the square root, but you got to make the ten power even. Can you try that?

OpenStudy (iwanttogotostanford):

do i hit square root button and then enter 4.07 x 10^13?

OpenStudy (anonymous):

That depends on the type of calculator And before you do, please convert that number into an even power of 10, PLEASE, PLEASE DO THAT!!!

OpenStudy (iwanttogotostanford):

how? sorry

OpenStudy (anonymous):

I ll just give you the answer. This planet is our home planet! So google the radius of earth, ,problem solved TADA!!!

OpenStudy (iwanttogotostanford):

please, i really do need help but this problem is really hard for me in only 7th grade but advanced

OpenStudy (anonymous):

You need to watch some Khan Academy videos on math. You have math problem, not physics. You need to work on that.

OpenStudy (iwanttogotostanford):

please i need help just really with this problem :-( I'm jut going to keep asking on here for help so its not helping by leaving me

OpenStudy (anonymous):

Parth, this is all yours!

OpenStudy (iwanttogotostanford):

:-(

OpenStudy (anonymous):

Don't worry, Parth is awesome! :P

OpenStudy (iwanttogotostanford):

he left@

Parth (parthkohli):

did you learn solving equations in math?

OpenStudy (anonymous):

okay \[300 = 3 \times 10 ^2 = 30 \times 10^1 = 300 \times 10^0 = 0.3 \times 10^4 = 0.03 \times 10^5\] does that make sense?

OpenStudy (anonymous):

oh pellet lol wait.. typo!!

OpenStudy (iwanttogotostanford):

yes

Parth (parthkohli):

\[I = \frac{2}5 MR^2\]\[9.74\times 10^{37} = \frac{2}5 \times 5.98 \times 10^{24} \times R^2\]Can you solve this equation?

OpenStudy (anonymous):

\[300 = 3 \times 10^2 = 0.3 \times 10^3\]

OpenStudy (iwanttogotostanford):

I subtract the exponents right? @Mashy @ParthKohli

OpenStudy (anonymous):

yes@!!!

OpenStudy (iwanttogotostanford):

ok so far I have- 9.74x10^37=23.92xr^2

OpenStudy (iwanttogotostanford):

@Mashy

OpenStudy (anonymous):

k

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