Recent content by stunner5000pt

This isn't a homework question per se but I wanted to understand how integration can connect things. If we integrated the area under the graph, would this give us the total charge to charge the capacitor? My logic here is purely based on units - if we integrate current on a current-time...

I see it now - testing the upper bound of the summand in the 'x' term gives pi/2 Thanks very much for your help!

Just looking at the summand, I can see that the function is ln(pi/4 + x^2) as the (i pi/2n) term is the 'x' term. How do I determine the limits of the integral, however? I was thinking about using the lower bound of the summation --> this given the (pi / 2n)^2 term, implying that nothing was...

Whoops! thanks for catching this! Something seemed off initially when i wrote it down too Just want to clarify.. did I not include the spring extending the mass when it first loaded?

Included a diagram as well... forgive me... I cannot seem to 'uninvert' the attached picture final velocity is zero if we set the lowest point that the mass reaches as zero, then the final height is zero let H be the 'extra' length that the spring reaches over and above the initial stretch...

Ok that does help - so we should be looking at bisulfate rather than sulfuric acid but wouldn't the concentration of bisulfate be the same i.e. the second part of the henderson hasselbach would still be for a. log (0.05/0.1) = log(0.5) and for b. log(0.1/0.5) = log(0.2) I feel like i'm...

THe answer is supposed to be (B) but I'm not understanding properly I initially chose A - to form an alkaline buffer, the # mol of acid should be half of this. But this doesn't seem to be the case I don't understand why the mol of base in this case should be 5x the mole of acid your help is...

Aha! missed that, thank you ok so then that means the distance travelled according to the traveller is: 6.5c \times 24 \times 365 \times 60^2 = 6.14 \times 10^{16} m when the person is in motion, how is the 6.5 ly being interpreted ? Is the noninertial observer seeing a shorter'length that...

If we took the perspective of the space traveller themselves, they are stationary and the whole universe goes past them at 0.7c. THen th elapsed time of 6.5 yr looking outside is \Delta t =6.5 \frac{1}{\sqrt{1-0.7^2}} = 9.11 yrs THen, when the traveller looks at the person travelling at 0.9c...

If we draw a punnet square, it would look like this Male on horizontal Female on vertical Cb Cb Cw CbCw CbCw Cw CbCw CbCw Does this mean that the offspring will have a blend of both black & white hair i.e. grey? Or is there something that is being missed here?

Sorry about this. Added that question as well Further to Mark44's comment, the answer I got still seems to be double counting the 7x7, 14x14, and so on Is it as simple as dividing the result by two ?

There are 14 integers between 1 to 100 that are divisible by 7 If we used indirect approach, then we would do the total number of possibilities C(100,2) = 4,950 and subtract all nubmers that are indivisible by 7 : C(100-14, 2) = C(86,2) = 3,655 the difference of these two numbers is 4,950 -...

I'm sort of stumped here , do i do this? (1+3x) \left( \frac{1+3x}{1+2x} \right)^2 = (1+3x) \left( \frac{3}{2} - \frac{1}{2(2x+1)} \right)^2 (1+3x) \left( \frac{3}{2} \right)^2 \left( 1 + \frac{-1}{3(2x+1)} \right)^2 and then apply the binomial theorem formula on the squared term above...

Thanks a lot - it certainly seems much easier Out of curiosity, is the 'direct' approach that I was taking above even solvable?

We don't need to worry about the n = -1 so we can assume that the function is continuous on any interval [a,b] where a, b are real numbers if I separate my interval into N partitions, then the right side values in my interval are a + \frac{b-a}{N}, a + 2 \frac{b-a}{N}, ... , a + k...