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December 19, 2014

December 19, 2014

Posted by **galileo** on Monday, January 15, 2007 at 8:48pm.

My answer: delta x = ct^3

m = c s^3 (s: seconds)

m/s^3= c

The SI units for c: m/s^3 correct?

2) Determine if dimensions are correct:

K = 1/2mv2

kgm^2/s^2 = kgm^2/s^2 * 1/2

kgm^2/s^2 = kgm^2/2s^2

The above equation is not dimensionally correct. However, I am somewhat confused as this is suppose to be the kinetic energy formula, and I just found that it is not dimensionally correct. Did I go wrong somewhere?

3) The mass of the Earth is 5.98E24 kg and its average density is 5.520kg/m^3. Assuming the Earth is spherical, find (a)its volume in m^3, (b) its radius in m, and (c) its radius in mi (miles).

My half solution:

Although I know the volume of a sphere is 4/3 pi r^3, here it throws me off.

I can find the volume simply by:

5,520kg = m^3/5.98E24kg; cross multiply and kg cancel(?) leaving me with V(earth)= 2.15E6 m^3(?)

Here, I'm lost about finding the radius in m and radius in miles. Should I use formula 4/3 pi r^3?

Thank you so much.

<<I'm lost about finding the radius in m and radius in miles. Should I use formula 4/3 pi r^3?

>>

Since you know V, rearrange that equation to get

r = (cube root of)[3 V/(4 pi)]

To convert meters to miles, divide by 1609.3 meters/mile

<<I'm lost about finding the radius in m and radius in miles. Should I use formula 4/3 pi r^3?

>>

Since you know V, rearrange that equation to get

r = (cube root of)[3 V/(4 pi)]

To convert meters to miles, divide by 1609.3 meters/mile

(1)Your answer is correct

(2) 1 kg m^2/s^2 is the same as 1 Joule, or 1 Newton-meter. The KE equation IS dimensionally correct

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