How do you know the angle theta used to resolve the components is the same as the angle the incline makes with the horizontal? Please refresh my geometry!

Hi Professor Dychko,
I'm confused on why you would not solve for the x and y components of both electric fields and then take the magnitude of the net electric field.
Thanks!

Hi Professor Dychko,
I'm confused on why you would not solve for the x and y components of both electric fields and then take the magnitude of the net electric field.
Thanks!

Hi kbick, this is a good question since the small angle approximation is often a very useful trick to make the algebra much easier. Yes, I know it seems unbelievable that you can just replace $sin(x)$ with simply $x$, so what you should do to convince yourself that it's acceptable when $x$ is small (I'm saying $x$, but $\theta$, or whatever variable represents the angle in your equation) is this: plot the graph of $y=x$ and $y=sin(x)$ on the same graph, but there's a catch. Zoom in on the graph so that you can only see $x$ between, say, -1 and 1. You'll notice the graphs look the same! This is the reason the approximation works, is that within this restricted domain of $-1 \leq x \leq 1$, both functions $y=x$ and $y=sin(x)$ give the same result. When you zoom out and look at larger values of $x$, then of course the graphs do not look the same, and you can no longer claim that $x \approx sin(x)$, like you can for small $x$.

Your video says released and the answer says required. Just wanted to point it out was confused at first but then after watching the video I understood what happened.

Hi flsktkd, thanks for the comment. Putting angular position in radians is often the better choice, and it's the official best unit for expressing angular position, that's true. However, revolutions are still used frequently since it's a unit that people can better understand. When looking at the tachometer in a car, it shows the engine speed in revolutions per minute, for example. Since this question asks for the final angular speed in rpm, it makes sense to use revolutions throughout, although it doesn't specify units for part (a) so I suppose it would be fine to use radians per second squared there.

Hi suriyak786, thanks for the question. Yes, since we haven't specified the coordinate system, it's true that up is the positive direction, and down is negative. The elevator would have a negative velocity while it's falling. However, it's initial displacement is positive since we've done one unconventional thing here: the origin (in other words the location where the vertical position is zero) is not where the elevator starts. Instead we've chosen the origin, also called the reference level when talking about vertical position, to be at the height of the fully compressed spring, which is below where the elevator starts. This choice of reference level simplified our algebra a bit. $h$ is the displacement of the elevator compared to the reference level, and it is positive since it's above the reference level, and we have the conventional coordinate system with up being positive.

My answer comes out to 316N/m can you show how to input into the calculator? I like when you show the calculator because this assists me in using the calculator and making sure I have the equation right.

Hi dshadowalker, thanks for the question. I get 316 N/m too! :) The difference is that I've rounded it to two significant figures, which explains why the answer is $320 \textrm{ N/m}$. Yes, showing calcs. with the on-screen calculator is nice, but it was getting very time consuming doing that so I included it only for a few chapters.

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).

## Giancoli 7th Edition, Chapter 23, Problem 51

By arielle.lab on Wed, 07/05/2017 - 06:01Hi! How do you know that do is positive?

## Giancoli 7th Edition, Chapter 4, Problem 57

By idan on Thu, 06/22/2017 - 14:40I found a good explanation online so I'm all set!

## Giancoli 7th Edition, Chapter 4, Problem 57

By idan on Thu, 06/22/2017 - 14:16How do you know the angle theta used to resolve the components is the same as the angle the incline makes with the horizontal? Please refresh my geometry!

## Giancoli 7th Edition, Chapter 16, Problem 30

By aheumangutman on Mon, 06/05/2017 - 19:27Hi Professor Dychko,

I'm confused on why you would not solve for the x and y components of both electric fields and then take the magnitude of the net electric field.

Thanks!

## Giancoli 7th Edition, Chapter 16, Problem 30

By aheumangutman on Mon, 06/05/2017 - 19:27Hi Professor Dychko,

I'm confused on why you would not solve for the x and y components of both electric fields and then take the magnitude of the net electric field.

Thanks!

## Giancoli 7th Edition, Chapter 28, Problem 2

By kbick on Sun, 05/07/2017 - 11:33Hi! Just wondering when we can use the small angle approximation - you say here that it is when theta is less than zero, but how can that be?

Hi kbick, this is a good question since the small angle approximation is often a very useful trick to make the algebra much easier. Yes, I know it seems unbelievable that you can just replace $sin(x)$ with simply $x$, so what you should do to convince yourself that it's acceptable when $x$ is small (I'm saying $x$, but $\theta$, or whatever variable represents the angle in your equation) is this: plot the graph of $y=x$ and $y=sin(x)$ on the same graph, but there's a catch. Zoom in on the graph so that you can only see $x$ between, say, -1 and 1. You'll notice the graphs look the same! This is the reason the approximation works, is that within this restricted domain of $-1 \leq x \leq 1$, both functions $y=x$ and $y=sin(x)$ give the same result. When you zoom out and look at larger values of $x$, then of course the graphs do not look the same, and you can no longer claim that $x \approx sin(x)$, like you can for small $x$.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 31, Problem 14

By fridley.26 on Wed, 04/26/2017 - 11:46Your video says released and the answer says required. Just wanted to point it out was confused at first but then after watching the video I understood what happened.

Ah, thanks fridley.26! Sorry for the confusion. I've updated the quick answer to say

released.Cheers,

Mr. Dychko

## Giancoli 7th "Global" Edition, Chapter 8, Problem 19

By lina09037788 on Tue, 04/25/2017 - 06:04angular position should be in radians, so it's 23*2*3.14 not 23 revolutions.

Hi flsktkd, thanks for the comment. Putting angular position in radians is often the better choice, and it's the

officialbest unit for expressing angular position, that's true. However, revolutions are still used frequently since it's a unit that people can better understand. When looking at the tachometer in a car, it shows the engine speed in revolutions per minute, for example. Since this question asks for the final angular speed inrpm, it makes sense to use revolutions throughout, although it doesn't specify units for part (a) so I suppose it would be fine to use radians per second squared there.Best wishes,

Mr. Dychko

## Giancoli 7th Edition, Chapter 6, Problem 43

By suriyak786 on Sat, 04/22/2017 - 21:22if the elevator is falling, wouldn't H be negative?

Hi suriyak786, thanks for the question. Yes, since we haven't specified the coordinate system, it's true that up is the positive direction, and down is negative. The elevator would have a negative velocity while it's falling. However, it's initial displacement is positive since we've done one unconventional thing here: the origin (in other words the location where the vertical position is zero) is not where the elevator starts. Instead we've chosen the origin, also called the

reference levelwhen talking about vertical position, to be at the height of the fully compressed spring, which is below where the elevator starts. This choice of reference level simplified our algebra a bit. $h$ is the displacement of the elevator compared to the reference level, and it is positive since it's above the reference level, and we have the conventional coordinate system with up being positive.All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 11, Problem 4

By dshadowalker on Thu, 04/20/2017 - 10:14My answer comes out to 316N/m can you show how to input into the calculator? I like when you show the calculator because this assists me in using the calculator and making sure I have the equation right.

Hi dshadowalker, thanks for the question. I get 316 N/m too! :) The difference is that I've rounded it to two significant figures, which explains why the answer is $320 \textrm{ N/m}$. Yes, showing calcs. with the on-screen calculator is nice, but it was getting very time consuming doing that so I included it only for a few chapters.

Best wishes with your studies,

Mr. Dychko

## 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