Thanks for the good question. Choosing whether the upward or downward direction is positive is called "choosing a coordinate system". Choosing a coordinate system is something you do for every solution, and the choice is a matter of personal preference. The only rule is that you're consistent, which means if you choose the downward acceleration to be positive, then it means downward velocity also must be positive, and the same goes for downward displacement. I choose down to be the positive direction since it meant there were no negative signs in my work, which I think looks just a little cleaner. The question is asking only for distances anyhow, which are positive, so I might as well make my calculations result in the final answer, rather than finding a negative downward displacement (which would have happened if I made the more conventional choice of upward being positive for the coordinate system) and then taking it's "magnitude" to get the final positive answer for height of the cliff.

Well I double checked and I do indeed have a part c on #42 chapter 2. It is the 7th Edition as well. The question also matches up with the solution provided.

Thanks for the question. I think you mean to suggest $v_f = v_o + at$, correct? That approach would be equally fine, provided you establish that $v_f = -v_o$, which is true since it returns back to the original launch height. This means your suggested formula, with a substitution for $v_f$, would become $-v_o = v_o + at$ which rearranges to $t = \dfrac{-2v_o}{a}$ which is the same formula shown at 1:20 in the video. You have to be a little cautious with your suggested formula, however, since it works only when you can tell how the final velocity compares with the initial velocity. $v_f = -v_o$ only when something returns to it's original height. If something fell into a hole after being launched upwards, meaning it returns to a different height, you would be better off using the $d=v_ot + \dfrac{1}{2}at^2$ formula. For this problem, however, the formula you suggest would be just fine.

Hi zzdawit, choosing y as negative is a valid thing to do, but it should lead to the same answer. You have to take care that you distinguish between the acceleration of the two blocks. The accelerations are not the same anymore. They have opposite signs, and I would guess that you might consider carefully examining your work to see that you took their opposite signs into account. My comment above explains a bit more about this, but let me know if you have more questions.

Hi zzdawit, and thanks for the question. $9.8 \textrm{ N/kg}$ is an alternative way of writing what you might be used to seeing as $9.8 \textrm{ m/s}^2$. The two forms are equivalent, and the one you choose to use is just personal preference. The two forms have different names. $9.8 \textrm{ N/kg}$ is called the Gravitational field strength, whereas $9.8 \textrm{ m/s}^2$ is named the acceleration due to gravity. When solving problems that require the force due to gravity I prefer to write it as $9.8 \textrm{ N/kg}$ since it's easy for me to see the $\textrm{kg}$ cancel when multiplying it by $\textrm{kg}$, leaving $\textrm{N}$, whereas for problems involving the speed or acceleration of something falling I prefer to write $9.8 \textrm{ m/s}^2$.

To illustrate how the units are the same even though they look different, consider this (writing units here, not variables, for this "dimensional analysis"): $\textrm{N/kg} = \textrm{kg} \times \textrm{m/s}^2 / \textrm{kg} = \textrm{m/s}^2$ where the Newtons were expanded into kilograms times meters per second squared since $F = ma$.

The short answer: $9.8 \textrm{ N/kg} = 9.8 \textrm{ m/s}^2$. Use them interchangeably as you prefer.

Thank you for your question. I'm not really sure how to answer it though since I need some more details about what you're asking about. What are you proposing to multiply? The solution video appears to be correct, unless I'm missing something. We determine the net force on the ball as a result of it's change in kinetic energy, and that net force turns out to be the force on the ball due to the glove. The force on the glove due to the ball is what the problem asks us for, and that is the Newton's Third Law "reaction" counterpart to the force on the ball due to the glove. Taking the "negative" (multiplying by negative one, in other words) gives the answer.

Hi rohanblazers, thanks for reminding me about this error. You're correct that the part b) answer is $9.8 \times 10^4 \textrm{ Pa}$. I'm away from my recording equipment at the moment, but I'll make a note in the quick answer that the working in the video is correct, but I forgot to include the $\times 10^3$ after the $13.6 \textrm{ kg/m}^3$ when plugging into my calculator for the final answer.

Hi giancoli4747, that's no bother at all, but quite the opposite! Thank you for pointing out the typo in the quick answer. I've updated it to $Q = -5.11 \times 10^{-11} \textrm{ C}$, as it shows in the video.

I am a little confused, in part A are we not solving for V^2? in which case we should solve for by square root of 90^2+(-65^2)? this comes out with the same answer posted but the math works out correctly.

Hi tthorne15, thanks for the question. If I don't quite understand your question, please just leave another comment. I think what you're asking about is the minus sign under the square root? Are you proposing $\sqrt{90^2 + 65^2}$ rather than the $\sqrt{90^2 - 65^2}$ shown in the video? The plus is used in the Pythagoras theorem when solving for the longest side of a right triangle (that side being known as the hypotenuse). In this case we're solving for one of the shorter sides (known as the "legs") instead, and after an algebraic manipulation (shown in the video) we end up with a minus.

Hi Angie, thanks for the request. Unfortunately I'm away from my recording equipment at the moment. Vector addition is definitely an important technique. I use it in many other solution videos, so hopefully you can see the technique there?

Right at the end of the video you said 5.2m instead of 2.5m (which you wrote down). All good no biggie just a simple error and just pointing it out in case you didn't realise!

Hi Physics 120, thanks for the question. You're asking how we know the electric field is directed South, I assume. A bit of background first:

Electric field points in the direction in which it would push a positive charge.

Note the emphasis on positive. Since this question concerns an electron, the charge is negative. That doesn't change the convention about electric field direction: it still points in the direction of force that would be applied on a positive charge. You could also say the electric field always points in the opposite direction to the force it applies on a negative charge. In this question we have a negative charge experiencing a force North, and so the electric field is pointing in the opposite direction: South.

Hope that helps, and good luck with "Physics 120"!
Mr. Dychko

Can you explain how M(0.10g) is equal to Force up - Force mg = ma?
I get the equation: F=ma and sum of F =Fup-Fmg=ma. I don't get how M(0.10g) is equal to all that.

Hi trandzuyhieu, thanks for the question. $M(0.10g)$ replaces the $ma$ in $F_{up} - mg = ma$. Data from the question say the helicopter mass is $M$, and it's acceleration is $(0.10g)$, so we substitute for each of these in $ma$: $M$ replaces $m$, and $(0.10g)$ replaces $a$.

Two quick questions:
1) Why is the answer for part b different from what you would find by doing Req calculations based on the rules for parallel and in-series resistors? (I got 5R/8 that way).
2) By considering all three resistors that are arranged in series as having the same current by definition, you could solve for I with 3 equations and 3 unknowns. Would this be a correct approach?

1) The series/parallel formulas don't apply here. Looking at the circuit diagram, the junction with $I_2$ going in and $I_3$ and $I_4$ coming out makes all the difference. None of the resistors in this picture are in series. Surprising, I know! When resistors are in series it means that all the current going through one resistor must also go through the next, but with these junctions interfering, some of the current through one is being diverted among multiple resistors downstream, and for this reason they're not in series. None of the resistors are in parallel either because of these pesky junctions, which makes it a clever circuit diagram.
2) See (1) :)

Hi kniffin.1, thanks for spotting that I missed that part. I'll try to fill the gap with a typed answer. Since $R = \rho \dfrac{l}{A}$ and we're told that $R = R_1 + R_2$ and $R = R_1 + 4R_1$. We can re-write that in terms of length by substituting for each resistance: $\rho \dfrac{L}{A} = \rho \dfrac{L_1}{A} + 4\rho \dfrac{L_1}{A}$. All the common factors cancel leaving $L = L_1 + 4L_1 = 5L_1$. So the total length of the wire is five times that of the first segment, or written another way after dividing both sides by 5: $L_1 = \dfrac{L}{5}$ or $L_1 = 20\%L$. The short wire is 20% of the total length of the wire.

Hi williams.dpw, thanks for the question. Yes, I suppose that's an interesting point to consider, although things are not quite as simple as they might seem. Taking the two points in isolation (considering a case that is simple first), it's true that $E = \dfrac{kQ}{r^2}$ which means the field strength is directly proportional to the charge. This means that a point with three times the charge of another point will have an electric field strength 3 times as strong, and yes, in that case the number of field lines should be three times as well. When the two charges are near each other, as in this question, and their fields are interacting, then this interaction changes the number of field lines you would expect compared to the case when they're in isolation. Nevertheless I would agree that the picture could use more field lines going into the $-3Q$ charge. When considering a position very close to a point charge, the effect of the other point charge becomes negligible since their field strengths are inversely proportional to distance from the point charge. What this means is that, close to one point charge, the other point charge doesn't matter, and if you zoom in on any point charge it will be possible to find a "zoom level" at which the field lines look the same as a point charge in isolation.

I suppose I'm arguing both for and against your excellent point. The only way to get a quantitatively correct picture is to use a computer to create the vector field by calculating the resultant electric field (which is not a trivial calculation since these fields add as vectors) at every point. The hand drawn picture is meant just to show an approximation. I think an important feature of the drawing is to show the field emanating left of the left point charge bending around to the right hand charge, which would happen only if the right hand charge has a greater magnitude of charge. Also, yes, I would agree for the most part that the density of field lines around the right hand charge should be nearly 3 times that of the left hand charge, but not exactly so since the left hand charge reduces the field strength around the right hand charge.

For part b why can't you use the initial kinetic energy of the ball using the velocity you found from part to find the potential energy (mgh) and then solve for h? Why do you need to use the potential energy of the spring?

Hi poojamukund10, thanks for the question. The tricky part is in the meaning of "h". You could do what you're suggesting, and calculate "h" to be the height that gives an amount of potential energy that is equivalent to the kinetic energy that the ball has when it is at the tip of the extended spring, but in this approach "h" will be the height above the extended spring, not the height above the initial starting point. As an interesting exercise (I haven't tried this, but it should work), why not add the answer for "h" in your approach to the amount by which the spring was compressed ($0.160 \textrm{ m}$) and see if you get the same answer? Let me know how it goes.

Sharp eye thesoutportschool. Thanks for spotting that. When that is corrected to use 0.04m instead of 4cm it turns out the answer becomes $0.1369 \textrm{ s}$ which rounds to $0.14 \textrm{ s}$, rather than the $0.13 \textrm{ s}$ given. I've made a note about the error in the Quick Answer.

Hi kingrhino, thanks for the question. Writing $2.6 \times 10^2 \textrm{ m/s}^2$ is the safer way to write the answer if you were, for example, concerned that someone marking a test paper was looking closely at significant figures. Strictly speaking $260 \textrm{ m/s}^2$ also has only two significant figures, and the trailing zero does not count as a significant figure in this case. The trailing zero serves as a "place holder" in order to position the "2" in the hundreds place, and the "6" in the tens place. I've seen $260.$ with a barely visible "dot" following the zero to indicate three significant figures, but that's not a good idea, so don't do that since it's too difficult to see. When in doubt, use scientific notation as you suggested, but know that you can write answers with "place holder" zeros as well.

At the beginning of Part B (0:50 mark) you mentioned that positive y is in the down direction. When setting up the solution, how did you know/ why did you decide to make the positive y going downwards?
Thank you!

Sorry for taking so long to get back. You've asked a good question. Deciding on which direction is positive is a matter of convenience or personal preference. In this case I chose down to be positive because that would make the acceleration of the falling block $m_B$ have the same sign as the acceleration of the sliding block $m_A$. Since we know the blocks will have the same magnitude of acceleration since they're connected by the rope, the fact that they also have the same sign of their acceleration meant that I could simplify the algebra a little bit by denoting the acceleration of each block with the same variable $a$. If the accelerations had different signs (by choosing the conventional "up" as positive) I would have had to denote the acceleration of block A as $a_A$ and that of block B as $a_B$, and then establish that $a_A = -a_B$, and then substitute that into the equations... and this is not less correct, but just a little bit of messiness I could avoid by choose down as positive in this case.

Hi mafiadx83, thanks for the question. The scenario in this problem is that a pitcher is throwing a baseball. In that scenario, the ball begins at rest in the pitcher's hand, and then they throw their arm forward to throw the ball. The problem says to assume that their throwing action takes place over a distance of $3.5 \textrm{ m}$. The final velocity of the ball is whatever velocity with which it leaves their hand at the end of the pitching action, and the problem tells us this is $43 \textrm{ m/s}$. That's how we make our assumptions, and the videos explains how to get the acceleration given the information provided, but please let me know if it isn't clear.

Hi kingrhino, yes, you're quite right that it does ask for meters. I've updated the quick answer and made a note about the mistake in the video. Thanks for the sharp eye.

Hi carolsilber, thanks for asking. $30,000$ is written as $3.0 \times 10^4$, so it doesn't look like there's any problem here. Let me know if you suspect any other errors though.

Hi daniel.weiss1, thanks for the question. Since the grinding wheel is rotating with a constant angular velocity (2200 rpm) this means there is no tangential acceleration of any point on the edge of the wheel. Points on the edge (or anywhere since there's nothing special about the edge) are all going at a constant speed, in other words. The direction of their velocity is changing, despite the constant speed, and it's the radial acceleration which causes this change in direction.

## Giancoli 7th Edition, Chapter 3, Problem 18

By sheumangutman on Sat, 09/24/2016 - 19:00Can you please explain why the acceleration in this question is assumed to be positive while it was negative in question #17? Thanks.

Hi sheumangutman,

Thanks for the good question. Choosing whether the upward or downward direction is positive is called "choosing a coordinate system". Choosing a coordinate system is something you do for every solution, and the choice is a matter of personal preference. The only rule is that you're consistent, which means if you choose the downward acceleration to be positive, then it means downward velocity also must be positive, and the same goes for downward displacement. I choose down to be the positive direction since it meant there were no negative signs in my work, which I think looks just a little cleaner. The question is asking only for distances anyhow, which are positive, so I might as well make my calculations result in the final answer, rather than finding a negative downward displacement (which would have happened if I made the more conventional choice of upward being positive for the coordinate system) and then taking it's "magnitude" to get the final positive answer for height of the cliff.

Hope that helps,

Mr. Dychko

## Giancoli 7th Edition, Chapter 2, Problem 42

By lcbishop on Sat, 09/24/2016 - 12:13For part c) What specific factors make this an estimate?

Hi Icbishop, did you post this question on the wrong video? I don't see a part c) for this problem.

Cheers,

Mr. Dychko

Well I double checked and I do indeed have a part c on #42 chapter 2. It is the 7th Edition as well. The question also matches up with the solution provided.

## Giancoli 7th Edition, Chapter 2, Problem 42

By sheumangutman on Sat, 09/24/2016 - 10:44For the 2nd part of the question, why can't you use the formula Vf=Vo+2at?

Thanks.

Hi sheumangutman,

Thanks for the question. I think you mean to suggest $v_f = v_o + at$, correct? That approach would be equally fine, provided you establish that $v_f = -v_o$, which is true since it returns back to the original launch height. This means your suggested formula, with a substitution for $v_f$, would become $-v_o = v_o + at$ which rearranges to $t = \dfrac{-2v_o}{a}$ which is the same formula shown at 1:20 in the video. You have to be a little cautious with your suggested formula, however, since it works only when you can tell how the final velocity compares with the initial velocity. $v_f = -v_o$ only when something returns to it's original height. If something fell into a hole after being launched upwards, meaning it returns to a different height, you would be better off using the $d=v_ot + \dfrac{1}{2}at^2$ formula. For this problem, however, the formula you suggest would be just fine.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 4, Problem 32

By zzdawit@yahoo.com on Thu, 09/22/2016 - 22:04but i got a different value when I chose to use y as negative . what can I do about it ? are both yours and mine correct ?

Hi zzdawit, choosing y as negative is a valid thing to do, but it should lead to the same answer. You have to take care that you distinguish between the acceleration of the two blocks. The accelerations are not the same anymore. They have opposite signs, and I would guess that you might consider carefully examining your work to see that you took their opposite signs into account. My comment above explains a bit more about this, but let me know if you have more questions.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 4, Problem 13

By zzdawit@yahoo.com on Wed, 09/21/2016 - 07:55what is 9.8N/kg

Hi zzdawit, and thanks for the question. $9.8 \textrm{ N/kg}$ is an alternative way of writing what you might be used to seeing as $9.8 \textrm{ m/s}^2$. The two forms are equivalent, and the one you choose to use is just personal preference. The two forms have different names. $9.8 \textrm{ N/kg}$ is called the

Gravitational field strength, whereas $9.8 \textrm{ m/s}^2$ is namedthe acceleration due to gravity. When solving problems that require the force due to gravity I prefer to write it as $9.8 \textrm{ N/kg}$ since it's easy for me to see the $\textrm{kg}$ cancel when multiplying it by $\textrm{kg}$, leaving $\textrm{N}$, whereas for problems involving the speed or acceleration of something falling I prefer to write $9.8 \textrm{ m/s}^2$.To illustrate how the units are the same even though they look different, consider this (writing units here, not variables, for this "dimensional analysis"): $\textrm{N/kg} = \textrm{kg} \times \textrm{m/s}^2 / \textrm{kg} = \textrm{m/s}^2$ where the Newtons were expanded into kilograms times meters per second squared since $F = ma$.

The short answer: $9.8 \textrm{ N/kg} = 9.8 \textrm{ m/s}^2$. Use them interchangeably as you prefer.

## Giancoli 7th Edition, Chapter 3, Problem 44

By clarkmg on Sat, 09/17/2016 - 21:09Your answer for question a is listed above in units of meters/second, but the work shows that the units are actually kilometers/hour.

Hi clarkmg,

Thank you very much for noticing that. I've fixed the Quick Answer to reflect the video.

Best wishes with your studies,

Mr. Dychko

## Giancoli 7th Edition, Chapter 6, Problem 20

By coralis.hernandez1 on Thu, 09/15/2016 - 09:39This answer is wrong? Its supposed to be a multiplication not a sum, right?

Hi coralis.hernandez1,

Thank you for your question. I'm not really sure how to answer it though since I need some more details about what you're asking about. What are you proposing to multiply? The solution video appears to be correct, unless I'm missing something. We determine the net force on the ball as a result of it's change in kinetic energy, and that net force turns out to be the force on the ball due to the glove. The force on the glove due to the ball is what the problem asks us for, and that is the Newton's Third Law "reaction" counterpart to the force on the ball due to the glove. Taking the "negative" (multiplying by negative one, in other words) gives the answer.

All the best,

Mr. Dychko

## Giancoli 6th Edition, Chapter 10, Problem 19

By rohanblazers on Mon, 09/12/2016 - 20:18The textbook denotes the answer as 9.8 x 10^4 Pa

Hi rohanblazers, thanks for reminding me about this error. You're correct that the part b) answer is $9.8 \times 10^4 \textrm{ Pa}$. I'm away from my recording equipment at the moment, but I'll make a note in the quick answer that the working in the video is correct, but I forgot to include the $\times 10^3$ after the $13.6 \textrm{ kg/m}^3$ when plugging into my calculator for the final answer.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 16, Problem 42

By giancoli4747 on Thu, 09/08/2016 - 17:06Hello!

Sorry to bother you, but shouldn't the answer to the problem be Q = -5.11 x 10^-11 C, instead of to the 4th power? Just thought I would help you out!

Hi giancoli4747, that's no bother at all, but quite the opposite! Thank you for pointing out the typo in the quick answer. I've updated it to $Q = -5.11 \times 10^{-11} \textrm{ C}$, as it shows in the video.

Best wishes with your studies,

Mr. Dychko

## Giancoli 7th Edition, Chapter 3, Problem 16

By tthorne15 on Tue, 09/06/2016 - 22:41I am a little confused, in part A are we not solving for V^2? in which case we should solve for by square root of 90^2+(-65^2)? this comes out with the same answer posted but the math works out correctly.

Hi tthorne15, thanks for the question. If I don't quite understand your question, please just leave another comment. I think what you're asking about is the minus sign under the square root? Are you proposing $\sqrt{90^2 + 65^2}$ rather than the $\sqrt{90^2 - 65^2}$ shown in the video? The plus is used in the Pythagoras theorem when solving for the longest side of a right triangle (that side being known as the hypotenuse). In this case we're solving for one of the shorter sides (known as the "legs") instead, and after an algebraic manipulation (shown in the video) we end up with a minus.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 16, Problem 12

By angie.carballo01 on Tue, 09/06/2016 - 21:37Hi Mr. Dychko,

Could you please also show the solution using vector addition?

Thank you so much!

Hi Angie, thanks for the request. Unfortunately I'm away from my recording equipment at the moment. Vector addition is definitely an important technique. I use it in many other solution videos, so hopefully you can see the technique there?

Best wishes with your studies,

Mr. Dychko

## Giancoli 6th Edition, Chapter 20, Problem 63

By thesouthportschool on Fri, 09/02/2016 - 22:53Right at the end of the video you said 5.2m instead of 2.5m (which you wrote down). All good no biggie just a simple error and just pointing it out in case you didn't realise!

Hi thesouthportschool, thanks noticing that. I'll put a note above the video.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 16, Problem 27

By Physics 120 on Tue, 08/30/2016 - 14:24How did you determine that the direction is south?

Hi Physics 120, thanks for the question. You're asking how we know the electric field is directed South, I assume. A bit of background first:

Note the emphasis on

positive. Since this question concerns an electron, the charge isnegative. That doesn't change the convention about electric field direction: it still points in the direction of force that would be applied on apositivecharge. You could also say the electric field always points in the opposite direction to the force it applies on anegativecharge. In this question we have a negative charge experiencing a force North, and so the electric field is pointing in the opposite direction:South.Hope that helps, and good luck with "Physics 120"!

Mr. Dychko

## Giancoli 6th Edition, Chapter 20, Problem 1

By thesouthportschool on Mon, 08/22/2016 - 21:526.43 for part b

no its not my bad

No worries at all. Thanks for digging into the problem, and good job finding the solution.

## Giancoli 6th Edition, Chapter 17, Problem 8

By thesouthportschool on Tue, 08/02/2016 - 05:52For the answer for this question it is shown as -3.25x10^4 instead of 10^3!

Ah, thanks for spotting that Southport School! It has been fixed.

Best wishes,

Mr. Dychko

## Giancoli 7th Edition, Chapter 20, Problem 10

By livanessa98 on Tue, 07/12/2016 - 13:26Why would the magnetic field come to an before the electron reaches the bottom of its circular path?

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

By trandzuyhieu on Fri, 07/08/2016 - 21:01Can you explain how M(0.10g) is equal to Force up - Force mg = ma?

I get the equation: F=ma and sum of F =Fup-Fmg=ma. I don't get how M(0.10g) is equal to all that.

Thanks

Hi trandzuyhieu, thanks for the question. $M(0.10g)$ replaces the $ma$ in $F_{up} - mg = ma$. Data from the question say the helicopter mass is $M$, and it's acceleration is $(0.10g)$, so we substitute for each of these in $ma$: $M$ replaces $m$, and $(0.10g)$ replaces $a$.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 19, Problem 33

By livanessa98 on Tue, 07/05/2016 - 09:52Two quick questions:

1) Why is the answer for part b different from what you would find by doing Req calculations based on the rules for parallel and in-series resistors? (I got 5R/8 that way).

2) By considering all three resistors that are arranged in series as having the same current by definition, you could solve for I with 3 equations and 3 unknowns. Would this be a correct approach?

Hi livanessa98, good questions.

1) The series/parallel formulas don't apply here. Looking at the circuit diagram, the junction with $I_2$ going in and $I_3$ and $I_4$ coming out makes all the difference. None of the resistors in this picture are in series. Surprising, I know! When resistors are in series it means that

allthe current going through one resistormustalso go through the next, but with these junctions interfering, some of the current through one is being diverted among multiple resistors downstream, and for this reason they're not in series. None of the resistors are in parallel either because of these pesky junctions, which makes it a clever circuit diagram.2) See (1) :)

Best wishes,

Mr. Dychko

## Giancoli 6th Edition, Chapter 12, Problem 28

By kcecchini on Mon, 07/04/2016 - 16:23how do I do better and pass my last exam and not have to repeat physics?

thank you for your solutions, they are terrific.

Yay! I'm glad you don't have to repeat your course! :) I'm glad I could help, and thank you very much for the feedback.

Best wishes with your future studies,

Mr. Dychko

## Giancoli 7th Edition, Chapter 18, Problem 16

By kniffin.1 on Wed, 06/15/2016 - 07:06how would you find the length of the short wire?

Hi kniffin.1, thanks for spotting that I missed that part. I'll try to fill the gap with a typed answer. Since $R = \rho \dfrac{l}{A}$ and we're told that $R = R_1 + R_2$ and $R = R_1 + 4R_1$. We can re-write that in terms of length by substituting for each resistance: $\rho \dfrac{L}{A} = \rho \dfrac{L_1}{A} + 4\rho \dfrac{L_1}{A}$. All the common factors cancel leaving $L = L_1 + 4L_1 = 5L_1$. So the total length of the wire is five times that of the first segment, or written another way after dividing both sides by 5: $L_1 = \dfrac{L}{5}$ or $L_1 = 20\%L$. The short wire is 20% of the total length of the wire.

Hope that helps,

Mr. Dychko

## Giancoli 7th Edition, Chapter 16, Problem 25

By williams.dpw on Mon, 06/13/2016 - 09:54Should there be 3x's as many field lines going into the -3Q charge compared to the +1Q charge?

Hi williams.dpw, thanks for the question. Yes, I suppose that's an interesting point to consider, although things are not quite as simple as they might seem. Taking the two points in isolation (considering a case that is simple first), it's true that $E = \dfrac{kQ}{r^2}$ which means the field strength is directly proportional to the charge. This means that a point with three times the charge of another point will have an electric field strength 3 times as strong, and yes, in that case the number of field lines should be three times as well. When the two charges are near each other, as in this question, and their fields are interacting, then this interaction changes the number of field lines you would expect compared to the case when they're in isolation. Nevertheless I would agree that the picture could use more field lines going into the $-3Q$ charge. When considering a position very close to a point charge, the effect of the other point charge becomes negligible since their field strengths are inversely proportional to distance from the point charge. What this means is that, close to one point charge, the other point charge doesn't matter, and if you zoom in on any point charge it will be possible to find a "zoom level" at which the field lines look the same as a point charge in isolation.

I suppose I'm arguing both for and against your excellent point. The only way to get a quantitatively correct picture is to use a computer to create the vector field by calculating the resultant electric field (which is not a trivial calculation since these fields add as vectors) at every point. The hand drawn picture is meant just to show an approximation. I think an important feature of the drawing is to show the field emanating left of the left point charge bending around to the right hand charge, which would happen only if the right hand charge has a greater magnitude of charge. Also, yes, I would agree for the most part that the density of field lines around the right hand charge should be nearly 3 times that of the left hand charge, but not exactly so since the left hand charge reduces the field strength around the right hand charge.

Hope that helps!

Mr. Dychko

## Giancoli 7th Edition, Chapter 6, Problem 39

By poojamukund10 on Sat, 06/11/2016 - 12:02For part b why can't you use the initial kinetic energy of the ball using the velocity you found from part to find the potential energy (mgh) and then solve for h? Why do you need to use the potential energy of the spring?

Hi poojamukund10, thanks for the question. The tricky part is in the meaning of "h". You could do what you're suggesting, and calculate "h" to be the height that gives an amount of potential energy that is equivalent to the kinetic energy that the ball has when it is at the tip of the extended spring, but in this approach "h" will be the height

above the extended spring, not the height above the initial starting point. As an interesting exercise (I haven't tried this, but it should work), why not add the answer for "h" in your approach to the amount by which the spring was compressed ($0.160 \textrm{ m}$) and see if you get the same answer? Let me know how it goes.All the best,

Mr. Dychko

Just for clarification the way I thought it would be was KEi = PEf and both energies were referring to the energy of the ball.

## Giancoli 7th Edition, Chapter 17, Problem 54

By kniffin.1 on Wed, 06/08/2016 - 15:36this is not the answer my homework is giving me...the answer they want is x*10^-6 J

Hi kniffin.1, thanks for the comment. If you have a specific question, please let me know. I haven't noticed any error in the solution here.

All the best,

Mr. Dychko

## Giancoli 6th Edition, Chapter 22, Problem 16

By thesouthportschool on Thu, 06/02/2016 - 18:27Shouldn't it be 0.04m instead of 4cm because you are working in meters?

Sharp eye thesoutportschool. Thanks for spotting that. When that is corrected to use 0.04m instead of 4cm it turns out the answer becomes $0.1369 \textrm{ s}$ which rounds to $0.14 \textrm{ s}$, rather than the $0.13 \textrm{ s}$ given. I've made a note about the error in the Quick Answer.

Best wishes with your studies,

Mr. Dychko

## Giancoli 7th Edition, Chapter 2, Problem 25

By kingrhino on Tue, 05/31/2016 - 02:44Should the answer be 2.6*10^2 m/s^2 rather than 260? Is putting the zero ok or is it assuming too much accuracy? Thanks

Hi kingrhino, thanks for the question. Writing $2.6 \times 10^2 \textrm{ m/s}^2$ is the safer way to write the answer if you were, for example, concerned that someone marking a test paper was looking closely at significant figures. Strictly speaking $260 \textrm{ m/s}^2$ also has only

twosignificant figures, and the trailing zero does not count as a significant figure in this case. The trailing zero serves as a "place holder" in order to position the "2" in the hundreds place, and the "6" in the tens place. I've seen $260.$ with a barely visible "dot" following the zero to indicate three significant figures, but that's not a good idea, so don't do that since it's too difficult to see. When in doubt, use scientific notation as you suggested, but know that you can write answers with "place holder" zeros as well.Best wishes,

Mr. Dychko

## Giancoli 7th Edition, Chapter 4, Problem 32

By sarahailu610 on Mon, 05/30/2016 - 07:54At the beginning of Part B (0:50 mark) you mentioned that positive y is in the down direction. When setting up the solution, how did you know/ why did you decide to make the positive y going downwards?

Thank you!

Hi sarahailu610,

Sorry for taking so long to get back. You've asked a good question. Deciding on which direction is positive is a matter of convenience or personal preference. In this case I chose down to be positive because that would make the acceleration of the falling block $m_B$ have the same sign as the acceleration of the sliding block $m_A$. Since we know the blocks will have the same

magnitudeof acceleration since they're connected by the rope, the fact that they also have the samesignof their acceleration meant that I could simplify the algebra a little bit by denoting the acceleration of each block with the same variable $a$. If the accelerations had different signs (by choosing the conventional "up" as positive) I would have had to denote the acceleration of block A as $a_A$ and that of block B as $a_B$, and then establish that $a_A = -a_B$, and then substitute that into the equations... and this is not less correct, but just a little bit of messiness I could avoid by choose down as positive in this case.All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 2, Problem 25

By mafiadx83 on Sat, 05/28/2016 - 08:31how did you know that inital velocity is zero why not the final is zero ?

Hi mafiadx83, thanks for the question. The scenario in this problem is that a pitcher is throwing a baseball. In that scenario, the ball begins at rest in the pitcher's hand, and then they throw their arm forward to throw the ball. The problem says to assume that their throwing action takes place over a distance of $3.5 \textrm{ m}$. The final velocity of the ball is whatever velocity with which it leaves their hand at the end of the pitching action, and the problem tells us this is $43 \textrm{ m/s}$. That's how we make our assumptions, and the videos explains how to get the acceleration given the information provided, but please let me know if it isn't clear.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 1, Problem 15

By kingrhino on Mon, 05/16/2016 - 03:48Is part A a mistake? It asked for 93 million miles in meters. I would think that it would be rounded to 1.5 x10^11 meters

Hi kingrhino, yes, you're quite right that it does ask for meters. I've updated the quick answer and made a note about the mistake in the video. Thanks for the sharp eye.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 8, Problem 18

By carolsilber on Wed, 05/04/2016 - 17:55wouldnt it be 3.0x10^5 because the answer is 30,000

Hi carolsilber, thanks for asking. $30,000$ is written as $3.0 \times 10^4$, so it doesn't look like there's any problem here. Let me know if you suspect any other errors though.

All the best,

Mr. Dychko

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

By daniel.weiss1 on Wed, 05/04/2016 - 07:22For part b, how do we know were are looking for the radial acceleration and not the tangential acceleration?

Thanks

Hi daniel.weiss1, thanks for the question. Since the grinding wheel is rotating with a constant angular velocity (2200 rpm) this means there is no tangential acceleration of any point on the edge of the wheel. Points on the edge (or anywhere since there's nothing special about the edge) are all going at a constant speed, in other words. The direction of their velocity is changing, despite the constant speed, and it's the radial acceleration which causes this change in direction.

Hope that helps,

Mr. Dychko