What will be the dimensions of temperature in a particular second, say the 10th?
Sorry, I wanted to say temperature change in the 10th second?
I have no idea what your question is! Is this part of a homework problem? If so, perhaps you could provide us with the entire problem. If not, do you have some context you can give us, that may better explain your question?
No, it is not part of any home work. I was reading units and dimensions chapter and found some thing interesting.
You know, what we mean dimensional formula of a physical quantity.
Dimensional formula of force for example, is M L T^(-2).
Dimensional formula of distance is L.
Now what is dimensional formula of the distance travelled by an object in the nth second? That is my question. Will it be L or some thing else?
A change in x will have the same dimension as x.
A change in x per unit y will have the same dimension as x/y.
This is what was a surprise to me.
I knew change in x will have the same dimension as x.
A chane in x per unit y will have the dimension of x/y.
But I expected that distance travelled in nth second will have the same dimension as distance (I considered it change in distance because distance in nth second is nothing difference between the distances travelled in nth second and (n-1)th second.)
But when I actually derived the equation for the distance travelled in the nth second, it turned out to be actually having the dimensions of LT^-1, the dimension of velocity.
Without looking exactly at the whole problem, or knowing what you did, there's no way for me to comment on that. As with our requirements in the HW/Coursework forum, to be able to clearly address an issue, the whole problem must be stated clearly. We do not like to be given stuff in bits and pieces, because it simply prolongs the agony (not to mention, people like me have very little patience for it).
You might also want to consider that a "velocity" is identical to "distance traveled in a unit time", which is x/y.
It does have the dimensions of L. The distance traveled during any time interval whatsoever has the dimensions of L.
The formula for what you are describing is:
| (Displacement at the END of the nth second)
- (Displacement at the BEGINNING of the nth second) |
This has dimensions of L.
For example, suppose an object moves from x = 3 cm to x = 5 cm, in some coordinate system, during the nth second. The distance traveled is then
|5 cm - 3 cm|
= |5-3| cm
= |2| cm
= 2 cm
Dimension = cm, a unit of length (L)
p.s. Might the derivation you refer to have involved dividing by a time quantity somewhere? In which case it is not a "distance traveled", but is a velocity instead.
What I am referring to is the equation I independently derived just to check the logic but that is available everywhere in kinematics literature.
Consider a body in linear motion with constant acceleration.
You know s = ut + 1/2 at^2.
And also is available the equation for the distance travelled in the nth second = distance travelled in nth second - distance travelled in (n-1)th second
s(n) = u + a/2 (2n-1)
On the left hand side we have distnace travelled in the nth second, and right hand side we have velocity.
This is the background, please give your comments.
That equation for s(n) is incorrect. While it gives the correct numbers when time is measured in seconds, the units are wrong.
You say the right hand side has units of velocity, but that is not true for the term
which has units of acceleration.
The equation should read:
s(n sec) - s((n-1) sec)
= u*(1 sec) + (a/2)*[2n*(1 sec) - (1 sec)^2]
since we are actually using t = n seconds, and not t = n.
If you use t = n, you are implying that time has no dimensions.
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