Hey thanks!
Problem #2, I believe, is assuming an object is on perhaps an air track...(frictionless)...and moving horizontally...
So then for #2, it is answer c...??
And for #6...then then mass does not affect the frequency of the period T?? so then the correct answer for #6 would be...
Going through these last few multiple choice HW problems tonight (about 50 of them total to finish tonight and I am down to these alst few)...and wondering if these are correct:
1. Newton's First law states...
a) An object can only move in a circle if external forces act on it.
b) An...
Prove this identity:
sin (2 X)
----------------- = 2 sin (X) - 2 sin^2 (X)
sec (X) + tan (X)
How would you do it? It was suggested in another thread that this can't be proved or that there is an error in the problem...but I think it is a hard problem to figure out. Anyeays, the...
Thanks! Appreciate your help!
I know that last one is a real tough one for sure...I wonder if I could us the quadratic formula to solve it?? Thanks! :approve:
Thanks! I tried working both sides and got this:
2 - 2sin^2 X = 2 cot X sin X cos X
Left side of 2 - 2sin^2 X
= 2-1 (1-cos 2 X)
=2-1 + cos 2 X
=1 + cos 2 X
For the right side, I worked it like this:
2 cot X sin X cos X
= 2 (cos X /sin X) (sin X) (cos X)
= 2 (cos X) (cos...
ok, I think I may have one of the last ones...
1. 2 - 2sin^2 X = 2 cot X sin X cos X
Working on the Right side,
I replaced cot X with (cos X/sin X)
Which makes...
2 (cos X/ sin X) sin X cos X
the two cos X's cancel, leaving
2 (cos X) (cos X)
=2 (cos^2 X)
and since...
Awesome! Thanks so much for your help! Really appreciate it.
I have been working on the rest of my homework problems and have only three left to go, but two of them are totally different than the other problems he assigned, but I'd still like to finish them tonight, eventhough he said we'd...
Ok, now for trig identity #3.
cot^2 X = (cot^2 X+1)(cos^2 X)
Here's what I did:
Working on the right side,
replace (cot^2 X + 1) with csc^2 X
which gives me...
csc^2 X cos^2 X
then I rewrite it as ...
1/sin^2 X (cos^2 X)
which is...
cos^2 X
--------
sin^2 X...
Thanks for the tips! I'll be back in a bit and post what I have done :smile:
OK, for #1.
cos^3 X-cos X = - cos X sin X
---------------
sin X
working on left side, I get:
cos X (cos^2 X - 1)
--------------------
sin X
and that gives me...
cos X (-sin^2 X)
----------------...
Trig Identities Help...Arrrggghhh! :)
I typed all of this out but the system then said I was not logged in and I lost it all...don't have time to retype everything.
note:
the ^ symbol is like "squared"
for example, if I have 4^2 , that would be "four squared" or 4 times 4 = 16...
Here is a homework problem I am working on:
1. Suppose 50 cm^3 of water at 85 degrees C is addedto 100 cm^3 of water at 10 degrees C. What is the equilibrium temperature of the mixture, assuming there is no heat lost to the environment?
* = multiplied by
T = Temperature of water
m = mass...
ok , gotcha...
I reworked it as follows:
nRT=45.6 atm L
n=3.00 moles :smile:
R=.0821
and solve for T
(3.00 moles)(.0821)(T)=45.6
DIVIDE both sides by (3.00)(.0821) OR (.2463)
and solve for T
T=45.6/.2463
T=185.140073082 Kelvin or approximately 185 Kelvin
So, as...