Question 958122
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How many revolution per second will an automobile wheel make at the rate of 63.4km per hour if the diameter is 5.4dm??p
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Circumference (curved-distance around the wheel) = {{{pi*Diameter = 5.4pi}}}
1 dm = .0001 km
1 hr = 3,600 secs

With R being the number of revolutions, we then get the following DIMENSION
ANALYSIS setup: {{{(matrix(1,2, 63.4, km)/matrix(1,2, 1, hr(s))) * (matrix(1,2, 1, hr(s))/matrix(1,2, "3,600", secs)) * (matrix(1,2, 1, dm)/matrix(1,2, .0001, km)) * (R/matrix(1,2, 5.4pi, dm))}}}.
                              {{{(matrix(1,2, 63.4, cross(km))/matrix(1,2, 1, cross(hr(s)))) * (matrix(1,2, 1, cross(hr(s)))/matrix(1,2, "3,600", secs)) * (matrix(1,2, 1, cross(dm))/matrix(1,2, .0001, cross(km))) * (R/matrix(1,2, 5.4pi, cross(dm)))}}} 
                              {{{(63.4/1) * (1/matrix(1,2, "3,600", secs)) * (1/.0001) * (R/(5.4pi))}}}
                              {{{("634,000"/1) * (1/matrix(1,2, "3,600", secs)) * (1/1) * (R/(5.4pi))}}} = {{{("1,585"/1) * (1/matrix(1,2, 9, secs)) * (R/(5.4pi))}}} = {{{matrix(1,2, "1,585"/9(5.4pi), (R/sec(s)))}}} = {{{highlight(matrix(1,3, cross(10.831), 10.381, (REVs/sec(s))))}}}
                               OR
                              {{{("634,000"/1) * (1/matrix(1,2, "3,600", secs)) * (1/1) * (R/(5.4pi))}}} = {{{matrix(1,2, "634,000"/"3.600"(5.4pi), (R/sec(s)))}}} = {{{highlight(matrix(1,3, cross(10.831), 10.381, (REVs/sec(s))))}}}

Thank you. @Ikleyn. The calculated amount was indeed 10.381, NOT 10.831. The error was corrected. Thanks again!</font></font></font></b></pre>