Is there anyway the solutions can be laid out onto one sheet so there isn't much scrolling? The calculations are nice and detailed but I lose my train of thought every time the scrolling happens. Makes it hard to follow the solutions -Alice Zhang

Hi tongmeng.zhang, yes I can see what you mean, especially on a 10min video. The video is a screen capture of my computer screen. I wonder if in the future I should change the orientation of the screen to be in "portrait" mode rather than "landscape"? It would sacrifice some width, but would keep more of the work on the screen at once. Unfortunately for the videos that are already made it would be difficult to change things. I could "zoom out", but then it would be too small to read. Thanks for the feedback though, since I'll try using "portrait" mode on future videos.

In the beginning when we solved for the Vx Initial and got 2.3077 why did we also use that in the magnitude calculation in the end? We solved for a final velocity of y, but we reused the velocity of x, could you please explain. Thanks.

Hi Icbishop, thanks for the question. Just to be clear, we're talking about part a), right? In part a) we solved for the resultant velocity, which is to say we solved for the vector sum of the x and y components. The answer for part a) is not the final velocity of y, but rather the final resultant velocity, which is normally just referred to as the velocity, but I'm including the word resultant to answer your question by explaining that the velocity is arrived at by using the pythagorean theorem to get it's magnitude by taking the square root of the sum of the squares of the x and y components of the velocity, and using trigonometry to get the direction.

Hi idan, I think it's a matter of personal opinion. Yes, using a graphing calculator is fine, but an "analytical solution" (ie. with algebra) can be useful too. The advantage of an algebra solution is that you can learn from an algebraic solution, and more generally characterize the behaviour of a system. Sorry to be so abstract, but we're talking about solutions in general, so it's difficult to be concrete. But consider the solution to problem 32 in chapter 4: $a = \dfrac{gm_B}{m_A + m_b}\textrm{ , } F_T = \dfrac{gm_Am_B}{m_A + m_B}$. That's an analytical solution, and it wouldn't be possible to use a graphing calculator since there are no numbers given in the problem. By looking at that analytical solution you can say things like "the acceleration decreases as $m_A$ increases", or "the tension force increases in proportion to both $m_A$ or $m_B$". Being able to create an analytical solution is important, so I guess my answer to your question is: yes, it's OK to use a graphing calculator, but make sure you could create an analytical solution if you had to, since that's your only option for problems with no numbers. It's like driving a race car: it's OK to drive it fast if you're a skilled driver, but not OK to drive fast if you're doing so just because you're unable to drive slow.

Thanks for the good question.
All the best,
Mr. Dychko

Hi Icbishop, thanks for the question. It's fine to choose the coordinate system you prefer (such as up as positive, which is conventional), and this will result in a correct calculation regardless of which choice you made, but you need to be aware of what the question is asking for. In this case the question is asking for the height of the cliff. Heights are always positive. They're a magnitude. This means that regardless of whether your calculation results in a positive or negative, you would give a positive as your final answer. The calculation gives you the displacement of the stone, but "displacement of the stone" is not what the question is asking for. The cliff height is the magnitude of the stone's displacement, which is to say the "absolute value", which are two fancy ways of saying "make it positive".

Thanks sanghoonwilliamk, you're totally right. $0.63 \textrm{ }\mu\textrm{F}$ is correct. While the working is correct in the video, I made a careless error turning the scientific notation into a decimal, and I put a note about this in the quick answer.

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.

## Giancoli 7th Edition, Chapter 4, Problem 66

By tongmeng.zhang on Sat, 10/15/2016 - 11:43Is there anyway the solutions can be laid out onto one sheet so there isn't much scrolling? The calculations are nice and detailed but I lose my train of thought every time the scrolling happens. Makes it hard to follow the solutions -Alice Zhang

Hi tongmeng.zhang, yes I can see what you mean, especially on a 10min video. The video is a screen capture of my computer screen. I wonder if in the future I should change the orientation of the screen to be in "portrait" mode rather than "landscape"? It would sacrifice some width, but would keep more of the work on the screen at once. Unfortunately for the videos that are already made it would be difficult to change things. I could "zoom out", but then it would be too small to read. Thanks for the feedback though, since I'll try using "portrait" mode on future videos.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 3, Problem 33

By lcbishop on Thu, 10/13/2016 - 21:31In the beginning when we solved for the Vx Initial and got 2.3077 why did we also use that in the magnitude calculation in the end? We solved for a final velocity of y, but we reused the velocity of x, could you please explain. Thanks.

Hi Icbishop, thanks for the question. Just to be clear, we're talking about part a), right? In part a) we solved for the

resultant velocity, which is to say we solved for the vector sum of thexandycomponents. The answer for part a) is not the final velocity of y, but rather the finalresultant velocity, which is normally just referred to as thevelocity, but I'm including the wordresultantto answer your question by explaining that the velocity is arrived at by using the pythagorean theorem to get it's magnitude by taking the square root of the sum of the squares of the x and y components of the velocity, and using trigonometry to get the direction.Hope that helps,

Mr. Dychko

## Giancoli 7th Edition, Chapter 2, Problem 49

By idan on Thu, 10/13/2016 - 17:13Do you think it's acceptable to use a graphing calculator to find the zeros of these functions? We're studying physics, not math.

Hi idan, I think it's a matter of personal opinion. Yes, using a graphing calculator is fine, but an "analytical solution" (ie. with algebra) can be useful too. The advantage of an algebra solution is that you can learn from an algebraic solution, and more generally characterize the behaviour of a system. Sorry to be so abstract, but we're talking about solutions in general, so it's difficult to be concrete. But consider the solution to problem 32 in chapter 4: $a = \dfrac{gm_B}{m_A + m_b}\textrm{ , } F_T = \dfrac{gm_Am_B}{m_A + m_B}$. That's an analytical solution, and it wouldn't be possible to use a graphing calculator since there are no numbers given in the problem. By looking at that analytical solution you can say things like "the acceleration decreases as $m_A$ increases", or "the tension force increases in proportion to both $m_A$ or $m_B$". Being able to create an analytical solution is important, so I guess my answer to your question is: yes, it's OK to use a graphing calculator, but make sure you

couldcreate an analytical solution if you had to, since that's your only option for problems with no numbers. It's like driving a race car: it's OK to drive it fast if you're a skilled driver, but not OK to drive fast if you're doing so just because you're unable to drive slow.Thanks for the good question.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 2, Problem 39

By lcbishop on Sat, 10/08/2016 - 17:17If for instance we considered down not to be positive and ended up with a negative. Does this essentially make the answer wrong?

Hi Icbishop, thanks for the question. It's fine to choose the coordinate system you prefer (such as up as positive, which is conventional), and this will result in a correct calculation regardless of which choice you made, but you need to be aware of what the question is asking for. In this case the question is asking for the

heightof the cliff. Heights are always positive. They're amagnitude. This means that regardless of whether your calculation results in a positive or negative, you would give a positive as your final answer. The calculation gives you the displacement of the stone, but "displacement of the stone" is not what the question is asking for. The cliff height is themagnitudeof the stone's displacement, which is to say the "absolute value", which are two fancy ways of saying "make it positive".Hope that helps,

Mr. Dychko

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

By daniel.weiss1 on Fri, 09/30/2016 - 11:02Great video, small note: I think you mistakenly have coulombs in your answer above the video instead of amps.

Thanks daniel.weiss1! I just fixed it.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 2, Problem 41

By sheumangutman on Tue, 09/27/2016 - 06:57Why cant the formula v1=vo+At be used for this problem assuming v1=0?

## Giancoli 7th Edition, Chapter 3, Problem 42

By julia.wolfe on Mon, 09/26/2016 - 12:54At 1:45 why is the x component 4.24 + 1.7 shown as equaling 2.124? Shouldn't it be 5.94 since the stair climb and boat direction are the same?

Oops nevermind! I had copied some numbers wrong!

## Giancoli 7th Edition, Chapter 17, Problem 39

By sanghoonwilliamk on Sat, 09/24/2016 - 21:45i think this should be .63 microfarads

Thanks sanghoonwilliamk, you're totally right. $0.63 \textrm{ }\mu\textrm{F}$ is correct. While the working is correct in the video, I made a careless error turning the scientific notation into a decimal, and I put a note about this in the quick answer.

Thanks for the sharp eye!

Mr. Dychko

Okay gotcha~ thanks for the clarification.

## 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.