Giancoli 6th Edition, Chapter 21, Problem 6

By thesouthportschool on Wed, 03/15/2017 - 18:34

I believe you may have gotten the answer wrong and it is 0.036V and not 0.048V.

By Mr. Dychko on Fri, 03/17/2017 - 08:04

Hi thesouthportschool, things look fine after double checking, so please let me know if you're still noticing a discrepancy.

Best wishes with your studies,
Mr. Dychko

Giancoli 7th Edition, Chapter 4, Problem 60

By suriyak786 on Sun, 03/12/2017 - 15:45

If the box is moving up and my positive x component is to the right. Then the Fgx would be going up?

By Mr. Dychko on Fri, 03/17/2017 - 07:58

Hi suriyak786, thanks for this question. The direction of gravity is unaffected by the direction of motion of the box, and it's also unaffected by the coordinate system (whether right is positive or negative). The component of gravity along the ramp will always point down the ramp. Whether that component gets a negative or positive sign is determined by your personal choice of coordinate system (whether "right" is positive or negative, in other words), but the arrow will always point down the ramp.

Hope that helps,
Mr. Dychko

Giancoli 7th Edition, Chapter 3, Problem 37

By suriyak786 on Sat, 03/11/2017 - 20:04

For part b why didn't we also find out Vx. Why did we find Vy

By Mr. Dychko on Sun, 03/12/2017 - 13:17

Hi suriyak786, thank you for your question. At 3:20 we made use of $v_x$ to create an expression for $t$ in terms of things that we know, such as $x$ and $\theta$, and the one thing we don't know, $v$. That was then substituted into the vertical displacement formula, which only has $v_y$ in it since only the vertical component of velocity affects the vertical displacement of the car. Does that help?

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 13, Problem 10

By rdattafl on Sat, 03/11/2017 - 18:00

Mr. Dychko,

The range of temperatures stated in the problem are from -30 degrees C to 50 degrees C. Thus, shouldn't the change in temperature, or delta(T), be 80 degrees C?
With this value of delta(T), the width of the expansion cracks become 12 mm.

By Mr. Dychko on Sun, 03/12/2017 - 13:49

Hi rdattafi, thank you for your question. It turns out that this question calls for a very careful reading since the way it's worded is a bit sneaky. It mentions wanting to know the expansion gap needed at $15^\circ\textrm{ C}$. This expansion gap is needed to deal with the concrete slabs becoming bigger as they get hotter on hot days. The temperatures cooler than $15^\circ\textrm{ C}$ don't concern us since the slabs will only get smaller as they get cooler. $-30^\circ\textrm{ C}$ is a red herring and we can ignore it. The question is not asking "by how much will the concrete expand when changing temperature from $-30^\circ\textrm{ C}$ to $50^\circ\textrm{ C}$". Rather, it's asking for how much of a gap should be left between slabs that are $15^\circ\textrm{ C}$ so that they don't touch and then buckle when they reach $50^\circ\textrm{ C}$.

Best wishes with your studies,
Mr. Dychko

Giancoli 7th Edition, Chapter 2, Problem 36

By suriyak786 on Fri, 03/10/2017 - 21:59

why is there Vmax? Can we just write normal V

By Mr. Dychko on Sun, 03/12/2017 - 13:39

Hi suriyak786, yes, you could write $v$ instead of $v_{max}$, just so long as there's an understanding of what $v$ (or $v_{max})$ is: it's the speed the car reaches after the period of acceleration.

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 3, Problem 37

By idan on Fri, 03/03/2017 - 13:55

I originally tried using the Range formula which gave me a slightly wrong answer. I'm assuming it's not applicable in this situation because of the 1.5 m drop?

By Mr. Dychko on Sun, 03/12/2017 - 12:53

Hi idan, yes you're exactly right. The range formula was derived using the assumption that the final and initial heights are the same. Since that's not the case here, as you say, the range formula doesn't apply.

Mr. Dychko

Giancoli 7th Edition, Chapter 3, Problem 27

By shichunye on Tue, 02/28/2017 - 13:37

1. why is it 36.6m/s(sin42)-(9.80)(1.50), doesnt the equation say V0sin+ayt?

2. how did you obtain 27.113m/s^2 for Vx?

Giancoli 7th Edition, Chapter 19, Problem 10

By elisabeth.burnor on Tue, 02/28/2017 - 10:24

During my free trial, I had no trouble viewing these videos, but now I cannot get any of the explanation videos to load on my PC Chrome Browser.

By Mr. Dychko on Tue, 02/28/2017 - 11:31

Hi elisabeth.burnor, thank you very much for reporting this. There is a temporary issue currently with the system, called Amazon S3, that hosts the videos and thumbnail images. I'm keeping an eye on, and I would imagine it won't take them too long to sort thing out. When they do, Giancoli Answers will be back to normal.

Best wishes,
Mr. Dychko

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

Giancoli 7th Edition, Chapter 10, Problem 22

By rdattafl on Thu, 02/02/2017 - 11:00

In the video, you wrote 1.29 * 10^5 Pa, but in the main answer section, you wrote 1.20 * 10^5 Pa. Merely for clarity, can you correct the second answer?

By Mr. Dychko on Sun, 02/05/2017 - 12:25

Thank you very much for spotting that rdattafl. I've updated the quick answer.

Giancoli 7th Edition, Chapter 2, Problem 30

By m_iqbal on Tue, 01/31/2017 - 19:20

I have a question that if the object has constant negative velocity, will the object speed up or slow down?

By Mr. Dychko on Sun, 02/05/2017 - 12:35

Hi m_iqbal, thanks for the question. The negative sign for velocity indicates direction only. It says nothing about whether the object is speeding up or slowing down. A negative velocity means only that the object is moving in the negative direction, which is typically to the left, although a person solving a problem is free to redefine the coordinate system by saying "I feel like making the negative direction to the right". It's a bit unusual to say "right is negative", but you do need to be prepared for this idea since occasionally I do say "let down be the positive direction", rather than the traditional "up is positive", when doing so simplifies the mathematics by avoiding so many negative signs in a formula.

A change in speed is given by acceleration, so your next question might be: does a negative acceleration mean the object is speeding up or slowing down. The answer here depends on whether the object currently has a negative velocity, or not. If the acceleration has the same sign as the velocity, it's speeding up. An acceleration with an opposite sign to the velocity means the object is slowing down. So, running through the list of possibilities: +a & +v => speeding up, -a & +v => slowing down, +a & -v => slowing down, -a & -v => speeding up. The last scenario is probably the one you'll want to take note of, where a negative acceleration means the object is speeding up when it already has negative velocity.

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 9, Problem 48

By jsh2672 on Tue, 01/31/2017 - 15:56

Why are you multiplying by 1/100? Shouldn't you be multiplying by 100 to get the percentage?

By Mr. Dychko on Sun, 02/05/2017 - 12:43

Thank you very much jsh2672 for noticing that! I've corrected the quick answer to show the result after multiplying by 100, and put a note about the error in the video.

Best wishes,
Mr. Dychko

Giancoli 7th Edition, Chapter 3, Problem 25

By sophiaswimgirl on Tue, 01/24/2017 - 23:29

Hi! Would you be able to explain something to me? I don't understand why solving for that x component gives the average horizontal speed. I thought that x component is only applicable to the instantaneous initial velocity?

By Mr. Dychko on Sun, 02/05/2017 - 12:50

Hi sophiaswimgirl,

Thanks a lot for the question, and I'm sorry for taking so long to get back. I hope you're still working on this unit....

Sure, it's true that the x-component of the velocity is that of the instantaneous initial velocity, as you say. However, the x-component of the velocity never changes. There is no horizontal acceleration. Since the x-component of the velocity never changes, this means that whatever value it has initially will also be the value it has at any other time, and so in this circumstance it's initial value is also it's average. The average of a value that's constant is whatever that value is, at any time.

Hope this helps,
Mr. Dychko

Giancoli 7th Edition, Chapter 3, Problem 13

By sophiaswimgirl on Fri, 01/20/2017 - 17:16

In part a) I'm confused as to why By is negative.
sin 56 = By/(-26.5) will result in a negative - I get that - but in question number 9, which uses the same vectors, the solution to By was positive, so why aren't they the same?

Is it because in the equation we're asked to solve for, we have to subtract vector B? Then why isn't Bx negative?

Thank you.

By Mr. Dychko on Sun, 02/05/2017 - 13:19

Hi sophiaswimgirl,

In question 13 the vector B is the negative of what it was in problem #9, which makes it point in the opposite direction. When I see a vector being subtracted, as vector B is in this case, I like to think of it as 'adding it's negative', which mean apply the usual rules for adding vectors, but flip around the one that's being subtracted. Instead of up and to the left in #9, it is now down and to the right here in #13. Since it's pointing downish, that makes the y-component, $B_y$, negative.

In both cases, the calculation for $B_y$ was $26.5 \sin(56^\circ)$ but here in #13 the result gets a negative in front if it despite the calculator saying 'positive'. This has to do with our understanding of the physical situation, in that the vector is pointing down, and so with that understanding we just put the negative there, regardless of what the calculator says. This might be going on a tangent, but just for arguments sake, if you wanted to always have the calculator say the correct sign, you would have to enter angles in standard position which is measured counter clockwise starting from the positive x-axis. In problem 13 the standard position angle of vector B is $360 - 56 = 304^\circ$, in which case $26.5 \times \sin(304^\circ) = -21.97...$ with the negative sign, rather than $26.5 \times \sin(124^\circ) = +21.97...$ (where $124^\circ$ is the standard position angle in #9). I don't recommend changing your angles to standard position, but I just mention it so that you're more comfortable placing negative signs where they belong 'by decrie', secure in knowing that there is a way to make the mathematics consistent with the physical reality if you felt like it. You're not doing anything wrong, in other words, by overriding the sign of what the calculator says and choosing the sign that fits the physical situation.

Maybe that was a bit long, but I hope it helps.

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 4, Problem 5

By sophiaswimgirl on Thu, 01/19/2017 - 00:23

I guess I missed this in chapter 2, but why do we square the magnitude of the velocity when using the kinematic equation (when trying to solve for the acceleration, blue writing, 2 minutes 3 seconds in)? It's a mistake I keep making but I can't figure it out why.

Thanks so much for your help.

By sophiaswimgirl on Thu, 01/19/2017 - 00:24

Wait, I figured it out. It was staring me right in the face - I had the wrong formula. Sorry, thanks!

Giancoli 6th Edition, Chapter 14, Problem 25

By sulaiman_gos on Tue, 01/10/2017 - 12:21

hi .. I have a question here .. why you keep solving with ΔT as it's ( Ti - Tf ) but it is actually ΔT = ( Tf - Ti ) !? please I need explanation!!

Giancoli 7th Edition, Chapter 17, Problem 2

By a_murayyan on Tue, 01/10/2017 - 06:51

in the answer box, don't you mean 2.72 X 10^-17 J?

Giancoli 6th Edition, Chapter 7, Problem 35

By upasna.us1 on Sun, 01/08/2017 - 20:28

I still don't understand why the initial velocity of the moving car is equal to the final velocity of both vehicles together? Could you explain this please?

Giancoli 6th Edition, Chapter 2, Problem 44

By dmorochnick on Sat, 12/31/2016 - 07:40

At about -2:05, shouldn't acceleration be negative: -9.8 yielding the solution d=4.3m instead of 2.1m?

By Mr. Dychko on Sun, 01/01/2017 - 01:15

Hi dmorochnick, thanks for the question. Earlier in the video I mentioned that the coordinate system was chosen so that down is positive. While the convention is that if one says nothing about the coordinate system, the assumption is that up is positive, is OK to explicitly change that. The choice is just personal preference. Where it's convenient, I sometimes prefer to make down positive to avoid substituting many negative values into the formula since I find so many negatives, and subtracting negatives, to be confusing. If you wish to have up as positive, it's totally OK to do so. You would arrive at the same answer (not $4.3 \textrm{ m}$), since keep in mind that the displacement past the window would also have to be negative since it's downward (which in your coordinate system is the negative direction if up is positive).

All the best,
Mr. Dychko

Giancoli 6th Edition, Chapter 6, Problem 3

By nalakija on Thu, 12/08/2016 - 15:52

Why are all the editions labeled as one??? Messing me up..

By Mr. Dychko on Thu, 12/08/2016 - 23:53

Hi nalakija, thanks for the comment. Could you please be more specific? I'm seeing editions labeled as "7th Edition", "6th Edition" and "Global Edition" in the menu bar. Where are you seeing them labelled as one?

Mr. Dychko

Giancoli 7th Edition, Chapter 6, Problem 37

By jmarra_villanova on Wed, 12/07/2016 - 20:31

Hi, how do you distinguish whether to use KEf + PEf = KEi + PEi on a spring question like this, rather than using Wext = DeltaPE in a question like #35?
Thank you

By Mr. Dychko on Thu, 12/08/2016 - 23:51

Hi jmarra_villanova,

Thanks for the question. The formula $KE_f + PE_f = KE_i + PE_i$ is a law, which means it's always valid. This is an algebraic way of saying energy is conserved after a movement of some kind. As a law, it could also be used in #35 instead of $W_{ext} = \Delta PE$. Doing so would have looked like this: $0 + mgy_2 = 0 + mgy_1 + \dfrac{1}{2}k(y_2 - y_1)^2$, where the zero's are substituting for kinetic energy, of which there's none since in #35 the block isn't moving (or the movement is so slow as to be negligible), and $(y_2 - y_1)$ is the amount the spring was stretched. Using $W_{ext} = \Delta PE$ is just a convenience since it reduces a couple of algebra steps. When in doubt, use conservation of energy.

Hope that helps,
Mr. Dychko

Giancoli 7th "Global" Edition, Chapter 12, Problem 35

By maia.fehr.oth23 on Thu, 12/01/2016 - 21:44

Why 331 and not 343?

Giancoli 7th Edition, Chapter 10, Problem 40

By odelay.chewy on Fri, 11/25/2016 - 00:47

I am struggling understanding the funky algebra used in this step. You kind of gloss over it. Could you show it step for step?

By Mr. Dychko on Sat, 11/26/2016 - 11:39

Hi odelay.chewy,

Thanks for the question. I'm away from my recording equipment at the moment, but I'll try to give some highlights here. We start with saying the total forces directed up equal the total forces directed down, then I make a substitution for each up force with the archimedes principle that the buoyant force is the weight of fluid displaced, which is $F = \rho V g$, and the down forces are the usual weight formula $F = mg$. Then, in green, the volume of the wood and lead is expressed in terms of their density and mass. This is in turn substituted (using red) for each volume factor in the up force terms. The specific gravity of wood replaces the ratio of the density of water to wood. Then comes a confusing line where the mass of the wood is shown moving to the left, but actually ends up on the right. The thing to notice with that line is that we're collecting like terms. We're placing the wood terms together on one side, and the lead terms on the other side. The mass of lead and mass of wood are factored out on their respective sides, leaving us with some bracket multiplied by a mass on each side. We're interested in knowing the mass of lead, so we divide both sides by the bracket by which the lead mass is multiplied. Then we plug and chug. Hopefully this video commentary is a bit helpful...

All the best,
Mr. Dychko