Rick S.

asked • 12/23/12

Is it legal to add a unitless number to a number with units?

If not what is the procedure for solving for r in the equation:(35m/s)sqd/r = -0.0021-0.0026m/(s)sqd. The answer should be in m for meters. S=seconds. Thank you! Rick

Bill F.

I assume the (35m/s)sqd = (35m/s)2, and that is divided by r.  But I don't understand the right side of the equation:  Is the -0.0021 also m/s, and is the -0.0026m/(s)sqd actually (-0.0026m/s)??

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12/23/12

Mykola V.

Yes, this problem doesn't make sense unless 0.0021 has units and the last term is probably m/s2 Bill. 

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12/23/12

Rick S.

Thank you for your reply.

-0.0021 was arrived at by -0.83 m3/kg * 0.0025 kg/m3. The first term here is 1/? where ?= the density of air at the surface = 1.2 kg/m3. The second term is horizontal pressure gradient dP/dH or 30.6 kg/m2/ 12000m. So the units cancel or am I missing something?

Rick

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12/24/12

Rick S.

The ? was supposed to be rho. Somehow it became garbled.

Bill: Your first assumption is correct.

Rick

 

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12/24/12

Bill F.

Allow me to attempt to reconstruct the original equation; please correct any errors, or confirm if it's correct:

(35m/s)2/(1.2 kg/m3) = [(-0.83 m3/kg)(0.0025 kg/m3)] - [(30.6kg/m2)/12000m]

 

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12/24/12

Bill F.

The "trick" in working with units is to treat them just like the values (numbers) themselves: multiply and divide them in the same way - otherwise it's all too easy to "lose" one or drop an exponent, etc.   When you divide by units, invert them and multiply.

Example: 120 miles / 60 miles per hour =

(120 * miles/1) * (1/60 * hours/mile) = 2 hours, since the miles cancel out.

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12/24/12

Rick S.

Hi Bill:

Almost correct. The equation is:

[(35m/s)2 / (r)]  + [(7.29*10-5rad/s) * (35 m/s)] = {[(-1) / (1.2 kg/m3)] * [(30.6 kg/m2) / (12000m)]}

Solve for r. r =radius in meters
7.29*10-5 is the earth's angular velocity in radians. I understand the radian unit drops out so that's why I didn't mention it sooner. What is left is per sec.

Thank you for continuing to work on this.

Rick

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12/24/12

Bill F.

Merry Christmas Rick!

My only remaining question about the equation itself:  is the (35m/s)2 (at the beginning) actually 35m/s2 ?  It looks like it might be an acceleration component, measured in meters per second squared.

Here''s what I have so far (assumes the late equation you wrote is complete and accurate):

(35 m/s)2 = 1225m2/s2

[(-1 / 1.2 kg/m3) * (30.6 kg/m2)] / 12000 m = -25.5 m / 12000 m = 0.002125

Subtract [(7.29*10-5 rad/s) * (35 m/s)] from both sides:

(1225m2/s2) / r =  0.002125 - [(7.29*10-5 rad/s) * (35 m/s)]

= (1225 m2/s2) = r * {0.002125 - [(7.29*10-5 rad/s) * (35 m/s)]}

r = 1225 m2/s2 / {0.002125 - [(7.29*10-5 rad/s) * (35 m/s)]}

ok, so now what???

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12/24/12

Rick S.

I can't believe we are working on this on Christmas eve, but I'm liking it!

Confirmed. (35 m/s)2 = 1225m2/s2. Is actually velocity squared.

Proceeding to your last equation on the last line.......
r = 1255m2/s2 / {-0.002125 - [(7.29*10-5 rad/s) * (35 m/s)]}
we may neglect rad and multiply the last two terms to arrive at
-0.0026m/s2. Unless this assumption is wrong.
The first term in the demoninator is negative becasue -1/density
So to simplfy.
r = 1255m2/s2 / (-0.002125 - 0.0026m/s2).
So now my original question is how do you solve for r with a unitless number? Unless I'm making a mistake and the first term in the denominator actually has units.
Thank you so much! Rick

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12/24/12

Bill F.

I'm at my end of what I know...  I see your problem and question.  I know that the "unitless" number (0.002125) became that way merely because an earlier division canceled out the units, so what we are left with is a constant.  And as with quadratic equations that include both and unknown and a constant, there's no way to solve it without factoring or using some other equations in which to substitute (like the quadratric equation).  I don't see how you can take your final equation any further, at least not without substituting part of it with an equivalency or definition.  Good luck, Rick!

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12/25/12

2 Answers By Expert Tutors

By:

Daniel O. answered • 12/26/12

Math and Physics Tutor, with a math and physics degree

Sung taee L. answered • 12/26/12

Teach Concepts, Thinking methods, Step by Step (Physics & Math)

Rick S.

The answer is finally solved! To convert mb to MKS, one may write:

1 mb = 100 N/m2 ; noting that N = kg m/sec. The dimensions work out to meters when solving for r. Which is solved as 52,350 meters.

Thank you all for your help over the Christmas Holiday!

Have a safe 2013!  Rick

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12/27/12

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