Why don't you subtract the potential due to the inductance of the coil? I thought the minus sign in the E = -L (dI/dt) formula was meant to indicate that the voltage would be in the opposite direction following Lenz's law (current increasing > flux increasing > induced current needs to decrease flux to compensate > current flows in other direction > voltage drop in other direction). What am I missing?

Hi julia.wolfe, thanks for the question. What's important to notice is that the sign of the potential difference will be the same across the resistor as it is across the inductor. In both cases, the circuit element causes a potential drop. (I say "circuit element" since it's OK to model this circuit as containing a single resistor and a single inductor connected by zero-resistance wires). Recall that when using Kirchoff's Laws to analyze a circuit, you make the potential across a resistor negative (it's a potential drop, in other words) when traversing the resistor in the same direction as the current. Because the inductor opposes the change in current, as you correctly mentioned, this inductor also causes a potential drop to oppose the change (note that, contrary to what you mentioned, the current does not end up flowing in the other direction. Rather, it's increase is opposed, and the inductor would disappear once the current stops changing). It would be fine to call both potential drops negative if you wish, but the question isn't specific about direction, so calling both positive is fine when answering a question asking for a "potential difference".

Hi elkinsk, thank you for noticing that! You're quite correct that the temperature should be $22^\circ\textrm{C}$ instead of $20^\circ\textrm{C}$. It turns out that the final answer is the same in either case since the 22 or 20 get added to 273 in order to convert to Kelvin, and the percent difference between 295 and 293 is small. I'll put a note for other students about this, and thanks again.

Bonjour girondebordeaux92, thanks for the question. The 4.186 figure you're quoting is, with units, $4.186 \dfrac{\textrm{kJ}}{\textrm{kg} \cdot \textrm{C}^\circ}$. Understanding units answers your question, since if you consult table 14-1 you'll notice that an alternative way to mention the specific heat of water is to write $1.00 \dfrac{\textrm{kcal}}{\textrm{kg} \cdot \textrm{C}^\circ}$. It turns out that the "kcal" way of writing specific heat, as shown in the video, if just convenient in this case since the energy of the candy bar is giving in kcal.

Hi cj_vana, thanks for pointing this out. It turn out that I did, in fact, square the $120 \textrm{ V}$ when doing the calculation, even though I forgot to write that the number is squared in the working. I'll make a note about this for other students.

Hi brian7989, thanks for the question. It's important to combine series resistances before parallel resistances. From step 2 to step 3, Req2 needs to be combined wi the R on the right since it's in series with it. It isn't possible to combine Req2 with the resistor below since it's in parallel with the R below, and it's necessary to know the total resistance of each branch when combining them in parallel. We don't yet know the total resistance of the branch containing Req2 until we add it to the R on the right.

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.

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?

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.

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 $50^\circ\textrm{ C}$.

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.

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?

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.

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 https://status.aws.amazon.com/, 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.

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.

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!

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.

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 $r^2$.

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?
Thanks,
Julia

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.

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).

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.

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?

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?

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.

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.

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?

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.

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?

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.

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.

## Giancoli 7th Edition, Chapter 21, Problem 44

By julia.wolfe on Sun, 04/16/2017 - 09:00Why don't you subtract the potential due to the inductance of the coil? I thought the minus sign in the E = -L (dI/dt) formula was meant to indicate that the voltage would be in the opposite direction following Lenz's law (current increasing > flux increasing > induced current needs to decrease flux to compensate > current flows in other direction > voltage drop in other direction). What am I missing?

Hi julia.wolfe, thanks for the question. What's important to notice is that the sign of the potential difference will be the same across the resistor as it is across the inductor. In both cases, the circuit element causes a potential

drop. (I say "circuit element" since it's OK to model this circuit as containing a single resistor and a single inductor connected by zero-resistance wires). Recall that when using Kirchoff's Laws to analyze a circuit, you make the potential across a resistor negative (it's a potential drop, in other words) when traversing the resistor in the same direction as the current. Because the inductor opposes the change in current, as you correctly mentioned, this inductor also causes a potential drop to oppose the change (note that, contrary to what you mentioned, the current does not end up flowing in the other direction. Rather, it's increase is opposed, and the inductor would disappear once the current stops changing). It would be fine to call both potential drops negative if you wish, but the question isn't specific about direction, so calling both positive is fine when answering a question asking for a "potential difference".Hope that helps,

Mr. Dychko

## Giancoli 7th Edition, Chapter 13, Problem 47

By elkinsk on Sat, 04/15/2017 - 18:32i think the temperature is 22° not 20°?

Hi elkinsk, thank you for noticing that! You're quite correct that the temperature should be $22^\circ\textrm{C}$ instead of $20^\circ\textrm{C}$. It turns out that the final answer is the same in either case since the 22 or 20 get added to 273 in order to convert to Kelvin, and the percent difference between 295 and 293 is small. I'll put a note for other students about this, and thanks again.

Best wishes with your studies,

Mr. Dychko

## Giancoli 7th Edition, Chapter 14, Problem 3

By girondebordeaux92 on Sun, 04/09/2017 - 03:19I think specific heat capacity of 4.186 is missing in the calculation? Can you please clarify

Bonjour girondebordeaux92, thanks for the question. The 4.186 figure you're quoting is, with units, $4.186 \dfrac{\textrm{kJ}}{\textrm{kg} \cdot \textrm{C}^\circ}$. Understanding units answers your question, since if you consult table 14-1 you'll notice that an alternative way to mention the specific heat of water is to write $1.00 \dfrac{\textrm{kcal}}{\textrm{kg} \cdot \textrm{C}^\circ}$. It turns out that the "kcal" way of writing specific heat, as shown in the video, if just convenient in this case since the energy of the candy bar is giving in kcal.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 18, Problem 35

By cj_vana on Mon, 04/03/2017 - 01:41For part b, should the 120V be squared?

Hi cj_vana, thanks for pointing this out. It turn out that I did, in fact, square the $120 \textrm{ V}$ when doing the calculation, even though I forgot to write that the number is squared in the working. I'll make a note about this for other students.

All the best,

Mr. Dychko

## Giancoli 6th Edition, Chapter 5, Problem 21

By thesouthportschool on Wed, 03/29/2017 - 22:17FN = ... +mgcosO

Hi thesouthportschool, I'm not really sure what the comment is here. I could maybe help if the comment was more specific.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 19, Problem 19

By brian7989 on Wed, 03/29/2017 - 17:06From step 2 to step 3, how do you know whether you should combine Req2 with the R on the right of Req2 or R that's below Req 2?

Hi brian7989, thanks for the question. It's important to combine series resistances before parallel resistances. From step 2 to step 3, Req2 needs to be combined wi the R on the right since it's in series with it. It isn't possible to combine Req2 with the resistor below since it's in parallel with the R below, and it's necessary to know the total resistance of each branch when combining them in parallel. We don't yet know the total resistance of the branch containing Req2 until we add it to the R on the right.

Best wishes,

Mr. Dychko

## Giancoli 6th Edition, Chapter 21, Problem 6

By thesouthportschool on Wed, 03/15/2017 - 18:34I believe you may have gotten the answer wrong and it is 0.036V and not 0.048V.

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 - 15:45If the box is moving up and my positive x component is to the right. Then the Fgx would be going up?

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 - 20:04For part b why didn't we also find out Vx. Why did we find Vy

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 - 18:00Mr. 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.

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 - 21:59why is there Vmax? Can we just write normal V

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 - 13:55I 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?

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.

Cheers,

Mr. Dychko

## Giancoli 7th Edition, Chapter 3, Problem 27

By shichunye on Tue, 02/28/2017 - 13:371. 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 - 10:24During 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.

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 https://status.aws.amazon.com/, 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 - 14:42By dividing by "t" instead of factoring aren't we losing a physically meaningful answer t=0 (the time the ball was hit)?

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 - 12:28I noticed you called "d" distance, but aren't all the kinematic equations actually only referring to displacement?

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 thatdis 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 factordas 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 - 15:37Is 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.

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

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 - 08:04What is your rationale for deciding to multiply by (r^2/10^beta/10) in part B?

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 - 12:41How 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?

Thanks,

Julia

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 - 18:10Is this calculation correct? It shows 4/pi^2 on the calculator display.

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 - 15:50I got confused with the P= 75 w and the P = the power consumed would you please explain the difference? Thank you

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 - 11:21do the nodes initially not mean anything then? I thought the nodes would make the resistors not in parallel or series.

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 - 11:00In 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?

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 - 19:20Hi,

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

Hi m_iqbal, thanks for the question. The negative sign for velocity indicates

directiononly. 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 - 15:56Why are you multiplying by 1/100? Shouldn't you be multiplying by 100 to get the percentage?

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 - 23:29Hi! 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?

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 - 17:16In 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.

Hi sophiaswimgirl,

In question 13 the vector

Bis 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 vectorBis 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

Bis $360 - 56 = 304^\circ$, 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 - 00:23I 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.

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 - 12:21hi .. 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 - 06:51in the answer box, don't you mean 2.72 X 10^-17 J?