1) Place the answer in proper amount of significant digits and unit: K = [6.63x10^(-34) J.s] (7.09x10^14 s) - [2.17 x 10^(-19) J] Note: J.s = Joule x sceonds Can someone help me with this question, please? I don't know how to write the answer with the proper amount of significant digits and unit..... I got: K=4.70x10^(-19) J.s^2 - 2.17x10^(-19) J Note: J.s^2 = Joule x seconds^2 They have different units, how can I subtract this and what are the final units? What would be the correct number of significant digits in the final answer?
I think your 7.09 ought to have units of s^-1 rather than s. You can't add or subtract numbers with different units, so you can't subtract Joules from Joules.second^2.
I think so! I have another qeustion regarind sine law 2) "The sine law can yield an acute angle rather than the correct obtuse angle when solving for an angle grater than 90 degrees. This problem occurs because for an angle A between 0 degrees and 90 degrees, sin A = sin (A + 90 degrees). To avoid this problem, always check the validity of the angle opposite the largest side of a triangle." Why sin A = sin (A + 90 degrees) if A is between 0 and 90 degrees? I don't get it......
That's because you're quoted material is wrong. sin(90 degrees+A) = cos(A), not sin(A). That should have been sin(180 degrees-A) = sin(A).
Hi kingwinner, What do you know about the sine function so far? It's usually introduced as the ratio of opposite/hypotenuse for a right-angle triangle, but there's a lot more to it. Have you seen a graph of y = sin (x)? Have you seen how sin and cos relate to a graph of a unit circle? Try exploring here: analyzemath.com PS - D H is right.
It can be shown that the following formula is true: sin(A+B) = sin A cos B + cos A sin B........... (*) If we put B=90 degrees we get: sin(A + 90) = sin A cos 90 + cos A sin 90. Now, cos 90 = 0, and sin 90 = 1, so sin(A + 90) = cos A. If, instead, we put into the formula (*) that A=180 degrees and B=-X degrees, we get sin(180+(-X)) = sin 180 cos(-X) + cos 180 sin (-X). But sin 180 = 0 and cos 180 = -1, so sin(180 - X) = - sin(-X). We also know that sin(-X) = -sin(X), so sin(180 - X) = sin(X). It is this formula which causes the problem when you try to work backwards from sin X to the angle X, in a triangle. For example, suppose you know sin(X) = 1/2. That might mean that X=30 degrees. But, according to the formula above: sin(30) = sin(180 - 30) = sin (150) so it might be that X=150 degrees instead.
I have seen the cos and sin graphs, but I forgot most trig. stuff... Please Register or Log in to view the hidden image! "The sine law can yield an acute angle rather than the correct obtuse angle when solving for an angle grater than 90 degrees. This problem occurs because for an angle A between 0 degrees and 90 degrees, sin A = sin (A + 90 degrees). To avoid this problem, always check the validity of the angle opposite the largest side of a triangle." So is this quote wrong? It is right off my textbook...... ========================================== Besides, is the following also correct? cos(A) = sin(90-A) sin(A) = cos(90-A)
The statement that sin A = sin (A + 90 degrees) is wrong. See my post above. But it is true that cos A = sin(90 - A) and sin A = cos(90 - A). For example: sin(90-A) = sin(90+(-A)) = sin 90 cos(-A) + cos 90 sin(-A), and since cos 90 = 0 and sin 90 = 1, we have: sin(90-A) = cos(-A) = cos(A).
