How do I convert 1.13 x 10^5 km/h^2 to m/s^2?

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AI Thread Summary
To convert 1.13 x 10^5 km/h² to m/s², it is essential to address the units correctly, particularly the h². The conversion involves recognizing that 1 hour equals 3600 seconds, which affects both the numerator and denominator. The correct conversion process requires multiplying by the appropriate factors to change hours to seconds, specifically using 1 hr² = (3600 sec)². The final conversion should yield the result in m/s². Understanding unit conversions is crucial for accurate calculations in physics.
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Homework Statement


1.13 * 10^5 km/h^2 to km/h * s
1.13 * 10^5 km/h^2 to m/s^2


Homework Equations


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The Attempt at a Solution


I'm not sure how to convert these because of the h^2 that's really screwing me up.
 
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Never mind I think I've figured it out anyway
 
Well in case you still didn't get it:

\frac{1}{hr^2} * \frac{1 hr}{3600 sec}
 
Feldoh said:
Well in case you still didn't get it:

\frac{1}{hr^2} * \frac{1 hr}{3600 sec}

If you post help, please get it right, should read:

\frac{1}{hr^2} * \frac{1 hr^2}{3600 sec^2}
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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