### 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

Hi,
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?

Cheers,
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

### Giancoli 7th Edition, Chapter 13, Problem 14

By saie.joshi on Thu, 11/24/2016 - 08:42

Why do you subtract the values instead of adding them together? Don't both the water and the glass reduce volume?

By saie.joshi on Thu, 11/24/2016 - 08:43

never mind I got it!

### Giancoli 7th Edition, Chapter 8, Problem 22

By taylorreilley on Mon, 11/14/2016 - 08:32

Why did you plug the Wf value in for b when you wrote Wi

By Mr. Dychko on Wed, 11/23/2016 - 00:32

Hi taylerreilley,

Thanks for the question. "initial" and "final" is a bit confusing in this questions since there are more than one time periods to consider, each with their own "initial" and "final" moment. For part a), the "final" moment is when the car has reached $55 \textrm{ km/h}$, making $\omega_f = 38.194 \textrm{ rad/s}$. However, for part b), the initial moment is when the car is at $55 \textrm{ km/h}$, so for part b) $\omega_i = 38.194 \textrm{ rad/s}$. The initial moment of part b) is the final moment of part a).

I can see where you're coming from, and technically it would be more correct for me to put a subscript on the variables to assign them to parts of the question, such as for the angular speed for part a) written as $\omega_{iA}$ and $\omega_{fA}$, and then say $\omega_{fA} = \omega_{iB}$, and so on, but I think that also creates it's own source of confusion by presenting so many more subscripts.

All the best,
Mr. Dyckho

### Giancoli 7th Edition, Chapter 7, Problem 41

By sheumangutman on Fri, 11/11/2016 - 16:12

Why is this collision deemed "inelastic" in terms of the treatment of kinetic energy conservation. Seems based on the diagram, the objects are not sticking together, which would suggest its inelastic.

By Mr. Dychko on Sat, 11/12/2016 - 13:16

Hi sheumangutman,

Thanks for the question. The term "inelastic" means there is less total kinetic energy after a collision compared with before a collision. Since that's the case here, we can call it inelastic. It is not "completely inelastic", that's true. A completely inelastic collision involves the two particles sticking together, and this type of collision results in the maximum loss in kinetic energy that's possible while still conserving momentum. Seeing two particles bounce apart means either the collision is "inelastic" (but not "completely inelastic"), or maybe the collision is "perfectly elastic", in which case the total kinetic energy after collision is the same as it was before. In this case, the collision is a regular inelastic collision since some kinetic energy was lost, but not the maximum loss that would have otherwise been possible had they stuck together.

Best,
Mr. Dychko

### Giancoli 7th Edition, Chapter 6, Problem 9

By sheumangutman on Thu, 11/10/2016 - 08:28

Could you also approach the problem by first solving for final velocity by using V=a*t, and then plugging this velocity into v^2=vo^2+2AD to solve for displacement?

Thanks.

By Mr. Dychko on Sat, 11/12/2016 - 13:03

Hi sheumangutman,

Yes you can! Substituting $v=at$ info $v^2 = v_o^2 + 2aD$ gives $a^2t^2 = v_o^2 + 2aD$, which, with $v_o = 0$ rearranges to $D = \dfrac{1}{2}at^2$ after you divide both sides by $2a$. $D = \dfrac{1}{2}at^2$ is the formula I used in the video. Good work!

All the best,
Mr. Dychko

### Giancoli 7th Edition, Chapter 4, Problem 52

By schamorg on Wed, 11/09/2016 - 07:09

Thank you so much. This was very helpful actually.

By Mr. Dychko on Wed, 11/09/2016 - 07:11

Super! I'm glad, and thanks for the feedback. :)

### Giancoli 7th Edition, Chapter 3, Problem 37

By kmoons25 on Mon, 11/07/2016 - 20:00

why do you do these things?

By Mr. Dychko on Tue, 11/08/2016 - 00:54

Hi kmoons25, Well, I know physics can be confusing. When you have a more specific question, just let me know.

All the best,
Mr. Dychko

### Giancoli 7th Edition, Chapter 6, Problem 53

By taylorreilley on Sun, 11/06/2016 - 15:46

Why is there still potential energy at the end?

By Mr. Dychko on Tue, 11/08/2016 - 00:53

Hi taylorreilley,

Thanks for the question. One thing to understand about potential energy is that you can never know how much there is. :) There's no such thing as an "absolute" potential energy. It's possible only to tell what is the change in potential energy. Sometimes we speak as though there is an absolute potential energy, such as "the book is 3m above the ground, so it has $mg(3 \textrm{ m})$ of potential energy, but don't be fooled by that as it's just a concise (maybe lazy?) way of speaking, when in fact it means the difference in potential energy of the book with respect to the ground is $mg(3 \textrm{ m})$. With that in mind, the skier has a change in potential energy resulting from their drop in altitude of $230 \textrm{ m}$, and this change in potential energy is turned into other forms of energy, namely kinetic energy and heat. Subtracting away the kinetic energy gained from the loss of potential energy, we're left with how much of that potential energy loss was turned into heat.

To put it in one sentence, the "potential energy" term you see in the final calculation is not the "potential energy at the end", but rather, it is the change in potential energy when the skier descends.

All the best,
Mr. Dychko

### Giancoli 7th Edition, Chapter 9, Problem 28

By leopinzon18 on Sat, 11/05/2016 - 10:02

what formulas are being used

By Mr. Dychko on Tue, 11/08/2016 - 00:42

Hi leopinzon18, for formulas being used are $\tau = Fl_{\bot}$ which says torque equals force times the perpendicular component of the lever arm, and the other formula is that $\Sigma \tau = 0$ for static equilibrium, which is also to say $\Sigma \tau_{ccw} = \Sigma \tau_{cw}$ meaning the counter-clockwise torques equal the clockwise torques. If you have a more specific question, just let me know and I'll see if I can answer it.

Best,
Mr. Dychko

### Giancoli 7th Edition, Chapter 4, Problem 52

By schamorg on Thu, 11/03/2016 - 19:43

Hello!
I am still confused as to why the diagrams for a) and b) are the same. I don't understand how in a) the box has static friction, but b) has kinetic friction and their diagrams are the same. Shouldn't b) have a velocity vector pointing downward?
Thank you so much for your time!

By Mr. Dychko on Tue, 11/08/2016 - 00:37

Hi schamorg, thanks for asking this question. Understanding free body diagrams is really important, so it's good your digging into the topic. A free body diagram does not include a velocity vector. I sketched one in for part "c", but that was just to emphasize which direction the box is moving in order to explain why the force arrows are pointing the way they are. Notice that the velocity vector does not begin on the center of the box, since I didn't want to make is seem like another force vector. Force vectors are always written with their beginning a the center of mass of the object (usually the geometric center... just ignore this point if it's confusing). Technically, velocity vectors do not belong on free body diagrams. OK, so to answer your question, for parts "a" and "b" there is intentionally no velocity vector at all since free body diagrams only show force vectors. In both parts "a" and "b", the friction force is directed up the ramp, and so the arrow representing these frictions forces point in the same direction. I suppose there is a case for making separate diagrams, but the distinction is only this: the kinetic force vector of part "b" should be shorter than the static friction force vector of part "a" since kinetic friction is always weaker than static friction.

Hope this helps,
Mr. Dychko

### Giancoli 7th Edition, Chapter 7, Problem 10

By lcbishop on Wed, 11/02/2016 - 11:51

Why is the negative sign from the v2 prime early not considered in the final calculation. Thanks.

By Mr. Dychko on Tue, 11/08/2016 - 00:29

Hi Ichbishop, thanks for the question. I suppose there's a bit of an oversight here, although it isn't a critical one since in the end we're looking for a ratio of masses, which we know must be positive. In the substitution shown in green, it would be slightly better to write $\dfrac{2V_2^{'2}}{v_1'} = \dfrac{-v_2'}{v_1'}$ in which case $\dfrac{v_2'}{v_1'} = \dfrac{-1}{2}$. Then the substitution into $\dfrac{m_1}{m_2} = \dfrac{-v_2'}{v_1'}$ would lead to the negatives cancelling, resulting in the expected positive final answer. Thanks for noticing that detail.

All the best,
Mr. Dychko

### Giancoli 7th Edition, Chapter 6, Problem 5

By lyonsca1 on Fri, 10/28/2016 - 08:23

Why don't you use cosine theta in the work formula at the very end of the problem if W=Fdcos(theta)? instead of just d (just displacement) in incline problems like this?

By Mr. Dychko on Sat, 10/29/2016 - 13:01

Hi lyonsca1,

Thanks for the question. I don't mean at all to contradict $W = Fd \cos(\theta)$ (which can also be written $W = F \cos(\theta) (d)$). That formula means that you multiply the force by the displacement, then multiply by cosine of the angle between them. In this particular solution, the force and displacement are actually parallel. The force I'm referring to is the force applied by some person or machine to make the car move up the ramp at constant speed. This applied force is directed along the ramp, so it's parallel to the length of the ramp. This means $W = F d \cos(0)$. Since $\cos(0) = 1$, it's common to not bother writing it at all, and to just write $W = F d$ for the special case where $\theta = 0$, which is the case where the force and displacement are parallel, as they are in this problem.

The applied force, according to the diagram, is $F_a = mg \sin(\theta)$, where theta here is referring to the angle of the incline. That's confusing, since the "theta" in the formula you wrote is referring to something different: namely the angle between the applied force and the displacement. If I were to write things out completely, using numbers in place of the angles, it would be $W = mg \sin(9.0^\circ) d \cos(0^\circ)$, where I added the $\cos(0^\circ)$ to try and be clear that that angle is the one in the $W = Fd\cos(\theta)$ formula, not the $9.0^\circ$ used to calculate the component of gravity along the incline.

Hope that helps,
Mr. Dychko

### Giancoli 7th Edition, Chapter 6, Problem 52

By lcbishop on Mon, 10/24/2016 - 11:35

Can you give a brief explanation of what you mean by "energy goes into heat"? Is it transformed into heat? Thanks.

By Mr. Dychko on Sat, 10/29/2016 - 11:49

Hi Icbishop, yes exactly. By "energy goes into heat" I mean to say "energy is transformed into heat energy". Rather than being gravitational potential energy, or kinetic energy, some of the energy is transformed into heat. This is the result of friction between particles in the ball when it gets squished upon colliding with the ground. The same could be said for particles in the ground internally "rubbing" and creating heat as the result of friction. Plus particles of the ground rub against particles of the ball. What I didn't mention is that some energy turns into sound energy, but this is typically very small compared to the amount turned into heat.

Hope that helps,
Mr. Dychko

### Giancoli 7th Edition, Chapter 6, Problem 31

By jessicasingh001 on Sun, 10/23/2016 - 18:46

Does anyone know why the 8 degree incline is ignored?

By Mr. Dychko on Sat, 10/29/2016 - 11:44

Hi jessicasingh001,

Thanks for the question. Only the change in height is important for questions involving gravitational potential energy, so the 8 degree incline is an extraneous detail. It's the vertical change in height that's important.

Hope that helps,
Mr. Dychko

### Giancoli 7th Edition, Chapter 7, Problem 28

By gillander97 on Sun, 10/23/2016 - 15:26

no sound ;-(

By Mr. Dychko on Sat, 10/29/2016 - 11:40

Hi gellander97, thanks for the comment. The sound plays fine on this end, so just let me know if that happens again.

All the best,
Mr. Dychko