Hmm for example the given is:
Newtons per centimeter , what will be the consistent unit?
I think km/min is not the consistent one? I think it should be km/h? What do you think?
Yes,yes! Sorry I wasn't able to define.
This is what is EXACTLY written :)
Unit Consistency
whenever you make calculations involving measurements, always remember to check for unit consistency; that is, the units should belong in the same system. Generally, if you are using meter for...
Ok, this is not a homework. But we have an exam later 8 hours from now.
On our book there is an activity called the Unit Consistency.
The instruction says: Which of the units below do no exhibit unit consistency? Give the equivalent consistent units for those that are not consistent.
1)...
hi Cepheid, I was wrong and i was able to get your point. We were taught about that transposition thingy but not at that case. Sorry, my bad. But I was able too see my mistake and I got a perfect score in my homework.
d= (gt^2)/2
my solution is:
i multiplied both sides by two to eliminate...
LOL. 2d - g = gt^2 -g < you i agree with that. my bad. the g's on the right side should not be cancelled. but on our algebra that's the way we were taught. We have Adv. Algebra and Analytic Geometry and I really don't have a problem with that. But the one you're telling me made me confused coz...
so ya. i get it. it needs to be equal so:
2d - g = gt^2 -g
and on the right side g-g is cancelled
the eq will be
t^2=2d-g
to eliminate the exponent, ill extract a square root
so the final answer is:
t=sqrt 2d-g?
so is it correct? :D:D:D:D
oops.this equation "-t^2 = [g(-d)]/2" is quite confusing. i was confused with that too. hehe.sorry
so it has to be equal. hmm. so i won't be applying transposition? sorry but i don't get it.
right i need to get rid of the g so what i did is transpose it to the left which caused its sign to...
oops. I'm sorry. hmm. do i need to apply rationalization? i don't know, i was lost on that part. T__T coz what i did is.
d= (gt^2)/2
then i transposed t^2 it became
-t^2 = [g(-d)]/2
then i divided both sides by -1 to eliminate the negative sign on t
so it became
(-t^2)/-1 = [g(-d)]/2/-1
so i...
yea i think that's wrong too. I'm a little bit confused with that. i was comparing #4 from the 2nd problem. but i think i wasn't able to get it. :D what do you think?
yes!certainly. and sure. i already have answers from the problems tho I am not sure if those are correct.:
1) Vl =v2 - 2ad
Vl= sqrt(v2-2ad)
2) v2 = (2KE)/m
V= sqrt(2KE)/m
3) Vf = Vl + at
Vl = V - at
4) d= (gt^2)/2
t^2= 2(gd)
t=sqrt2(gd)
5) m=v^2 - acr
:) hope you could correct if...