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Sample solution

Giancoli 7th Edition, Chapter 5, Problem 12


Recent questions and answers

Giancoli 7th Edition, Chapter 2, Problem 42

By idan on Sun, 02/26/2017 - 14:42

By dividing by "t" instead of factoring aren't we losing a physically meaningful answer t=0 (the time the ball was hit)?

By Mr. Dychko on Mon, 02/27/2017 - 00:40

Hi idan, you are making a valid point. You have a sharp eye! Mathematically, it would be more correct to factor and then find the roots of the resulting equation, thereby discovering the answer $t=0$. However, since this question is asking for the "time in the air", $t=0$, as you know, is an extraneous solution, so the effort of finding the solution $t=0$ and then discarding it as extraneous isn't worth the effort. The technique in the video is perfectly fine for a physics class, since math is just a tool for finding solutions to the physical problem, which in this case is "how much time is the ball in the air", not "at what times is the ball at height zero". For the latter, one must follow the technique you suggest by factoring since $t=0$ would be a non-trivial solution, but as it is, $t=0$ is not a solution for determining the "time in the air" problem. In a math class there's no question that to answer "solve this equation", one must factor and include $t=0$ as a solution.

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 2, Problem 26

By idan on Sun, 02/26/2017 - 12:28

I noticed you called "d" distance, but aren't all the kinematic equations actually only referring to displacement?

By Mr. Dychko on Mon, 02/27/2017 - 00:30

Hi idan, you're quite right that the kinematics equations contain displacement, rather than distance. Often times, such as in this question, the two quantities have the same magnitude. The distance and the magnitude of the displacement are the same, in other words, but your comment and awareness that d is in fact always displacement is definitely important in questions for which that's not the case. Technically it's a mistake for me to refer to the factor d as a distance, but I fear that I probably do it quite often, making the distinction mostly just where it's necessary to do so. Hopefully this isn't too confusing!

All the best,
Mr. Dychko

Giancoli 6th Edition, Chapter 8, Problem 64

By thesouthportschool on Sat, 02/18/2017 - 15:37

Is the answer not 453.6926 J or 5*10^2 J, I copied the exact calculations from your final answer of part A (3:49) and did not get 5*10^16 J.

By thesouthportschool on Sat, 02/18/2017 - 15:43

And I also you changed the angular velocity from 3.333*10^-4 to 3.333*10^-11 when doing the calculations to work out the KE

By Mr. Dychko on Wed, 02/22/2017 - 00:37

Hello student from thesoutportschool, thank you for spotting the error. At 3:23 I mention correctly that the angular velocity is $3.333 \times 10^{-4}$, whereas at 3:46 I incorrectly wrote $3.333 \times 10^{-11}$. My calculations were nevertheless done using $3.333 \times 10^{-4}$, so the results are correct. I'll put a note above the video to use $3.333 \times 10^{-4}$ for the angular velocity.

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 12, Problem 15

By margolinw on Fri, 02/17/2017 - 08:04

What is your rationale for deciding to multiply by (r^2/10^beta/10) in part B?

By Mr. Dychko on Wed, 02/22/2017 - 00:14

Hi margolinw, I think you're referring to the algebra step at 3:36, right? Multiplying by $\dfrac{r^2}{10^{\beta/10}}$ is a way to isolate the unknown $r$ by itself on one side of the equation. Multiplying by $r^2$ cancels it on the left side, whereas dividing by $10^{\beta/10}$ cancels that term on the right side, and this results in $r^2$ by itself on the right side. Mission accomplished! Then I switched the sides around and took the square root of $r^2$.

Hope that helps,
Mr. Dychko

Giancoli 7th Edition, Chapter 16, Problem 5

By julia.wolfe on Sun, 02/12/2017 - 12:41

How are you supposed to know that the comb gained (rather than lost) electrons? I know that the chapter mentions that plastic items develop negative charges when rubbed, but I thought the problem would specify that the charge acquired was -3 microcoulombs if the comb had gained (vs. lost) electrons?

By Mr. Dychko on Wed, 02/22/2017 - 00:24

Hi julia.wolfe,

I think the question would be worded better if it said $-3.0 \textrm{ } \mu \textrm{C}$. I don't think you're meant to know that plastic gains electrons, although in this case, without the question specifying a negative charge, knowing the properties of plastic is the only way to know. I would presume that in any real test question that the negative sign would be given.

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 11, Problem 28

By donnarrnc on Fri, 02/10/2017 - 18:10

Is this calculation correct? It shows 4/pi^2 on the calculator display.

By Mr. Dychko on Fri, 02/10/2017 - 23:42

Hi donnarrnc,

Thanks for the question. Yes, things are all good here. The calculator display shows /4/pi^2, which is another way of saying what you might have been expecting: /(4pi^2).

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 18, Problem 36

By tuh20232 on Thu, 02/09/2017 - 15:50

I got confused with the P= 75 w and the P = the power consumed would you please explain the difference? Thank you

By Mr. Dychko on Fri, 02/10/2017 - 23:48

Hi tuh20232, thanks for the question. P = the power consumed, always. On the packaging for a bulb, the manufacturer will give the answer to the power calculation, assuming a particular voltage. For a bulb sold in the US, the manufacturer will assume 120 V in the calculation for the bulb power, where for a bulb sold in the EU, the manufacturer will assume 240 V.

P = the power consumed, whereas the P = ## rating for a bulb depends on where the bulb is sold.

Hope that helps,
Mr. Dychko

Giancoli 7th Edition, Chapter 19, Problem 18

By saie.joshi on Sun, 02/05/2017 - 11:21

do the nodes initially not mean anything then? I thought the nodes would make the resistors not in parallel or series.

By Mr. Dychko on Sun, 02/05/2017 - 12:24

Hi saie.joshi,

Thanks for the question. This diagram is a bit tricky since some resistors are in series, whereas others are in parallel. We break it down step by step. If current reaching a node has a choice between multiple paths, then the resistors along the two (or more) paths are in parallel. If current reaching a node has only one option, and that is to proceed through another resistor, then that next resistor is in series. Does that help?

All the best,
Mr. Dychko