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.

Hi mafiadx83, the rule for all force questions (and this rule goes by the name Newton's Second Law) is that net force equals mass * acceleration. In this particular question there is only one force, namely friction, and for this reason the net force is made up only of the friction force. Since the friction force is the net force in this particular question, that's why friction force equals mass * acceleration for this particular question.

why initial velocity equal zero ? is it suppose that final velocity equal zero because he is saying in the question what speed did the sprinter leave the starting block

Hi mafeadx83, yes, that's right. It's just implied that the sprinter is initially at rest on the starting block, and the question is asking for the speed with which the sprinter launches off the starting block after applying a force.

Hi merkinthedark, I ran the final numbers shown in the video through the calculator again and confirmed the answer, so please let me know if you find any specific error. It looks fine so far.

why does friction force opposite to the applied force?
In question 8 , applied force and friction force are in the same direction /
can you please elaborate this difference?

Hi mikepeter, thanks for the good question. A sliding friction force is always opposes motion. Sliding friction slows things down. This means the direction of a sliding friction force is in the opposite direction as the direction of motion. The direction of the applied force in problem 8, or here in problem 10, actually doesn't matter at all. It's only the direction of motion that matters in determining the direction of the sliding friction force.

Hi bmuniz8219, that's a good question. Each current "starts" at a junction. While the diagram draws $I_1$ to the left along the very top, it's understood (even though not explicitly drawn) that the current originates at the junction below the $12\Omega$ resistor where the $I_2$ and $I_3$ currents also meet. So $I_1$ initially goes up through the $12\Omega$ resistor, and THEN, left along the top.

Why can't we use the conservation of energy law here? Wext= Change in KE + Change in PE + Ethermal where we can take everything as the system so that no external work is done and Etherm is zero.

Hi hsumail, thanks for the question, and apologies for taking so long to get back. Conservation of energy is being used here, though it could be expressed as the work-energy theorem as well and get the same result. The idea is that the total energy before a process is the same as the total energy after the process, and the formulae surrounding this subject serve just to help account for all the different forms of energy before and after the processes. Conservation of energy problems are book-keeping exercises.

Now I'll get to your comment. You mentioned the work-energy theorem, so let's clarify that it's $W_{ext} = \Delta KE$, not the expression written in your comment. [edit: earlier comments about the Earth don't apply. I hadn't actually looked up the problem.] You could consider the car-spring as a single system, which is all well and good, but this isn't strategic for problem solving since your only conclusion would be that $W_{ext} = 0$. This problem is asking for the spring constant, and conservation of energy is the ticket to solving for it. Maybe you would prefer the following expression, rather than the one I wrote in the video:
$KE_1 + PE_1 + W_{NC} = KE_2 + PE_2$. $W_{NC}$ typically represents friction, which is zero in this case, so we plug different types of energy into this expression and get $\dfrac{1}{2}mv_i^2 + 0 + 0 = 0 + \dfrac{1}{2}kx^2$ where I've substituted specific expressions in the same order as the equation before that has more generic expressions. This will arrive at $\dfrac{1}{2}mv_i^2 = \dfrac{1}{2}kx^2$, which is the same as in the video.

roynunez273 is correct about B, however I took a different approach to the answer and ended up getting it right. Need to remove pi from your calculation:

So, F=PA= (7.3479x10^-7 N/m^2) (1.0x10^-4) should give you the correct answer.

Hi jaclynrgile, thanks for jumping in. I love the discussion this solution is making. In order to duplicate the answer in the back of the text, both you and roynunez273 are correct that you need to multiply the radiation pressure by a figure tip area approximated as $1.0 \textrm{ cm}^2$. Since part B is an estimate, and the figure tip area isn't specified, this creates some room for interpretation. While it's fine to estimate the finger tip area as $1.0 \textrm{ cm}^2$, as the text does, it's also OK to estimate a finger tip radius as $1.0 \textrm{ cm}^2$ and then calculate the area using $\pi r^2$, because who's finger tip is it? Each student, looking at their own finger tips, should get slightly different answers, and the only important point is that the estimate really is something close to the area of a finger tip.

Hi johncondon101, thanks for the question. The choice is kind of strategic. Either way will lead to the answer, but the algebra just looks a little tidier one way vs. the other, and the choice is more a matter of intuition. In problem #5 it made sense to have $f_2$ in the numerator since it was the unknown, and we ended up multiplying by the denominator ($f_1$ in that case) to isolate $f_2$ on one side of the equation. In this problem it's a bit less clear which choice to make, but since $m$ is the unknown and it will appear in both the numerator and denominator (eventually, after a couple of steps, which is why, again, setting things up in this case is more a matter of intuition), it's better to have it by itself in the denominator instead of adding to a term since we'll be multiplying by the denominator, and it's easier to multiply by a single factor.

Again, there's nothing much really here to worry about, it's just that one way might lead to neater work than the other. Either way gets to the answer after some algebra.

If in the previous problem we discovered that currents I1 and I2 go to the left then why is the junction rule applied so that I1= I2+I3? Is it not I3=I1+I2?

Hi merkinthedark, thanks for the question. Yes, we could have used knowledge from problem 31 to choose $I_2$ to the left, and this would be fine, but we don't have to do that. There's no particular reason why I didn't. It isn't important to be correct with the initial guess about the current direction. One just has to make a choice and then construct the equations accordingly. You'll notice that the final answer in this problem has $I_2$ going to the left after all, as we expect from our work in problem 31, so we arrive at the correct answer despite using an initial guess in the other direction.

Hi roynunez273, thanks for the comment. In part B we're calculating force by multiplying pressure by the area of the finger tip. The finger tip area is $\pi r^2$, and we're estimating the radius to be $1 \times 10^{-2} \textrm{ m}$, which is already in meters so that no conversion is necessary. When we multiply the pressure by that area we get the force shown in the solution, so please just let me know if I'm missing something, and please be specific.

Hi dennis.colon.jr, thanks for the comment. I can't see where the mistake is to arrive at the $1.7 \times 10^{-55}$ figure, but I've doubled checked that the answer I've provided is correct. If you have a more specific question about the solution, just let me know.

Hi moopen, thanks for the question. When calculating the net force on charge 2 we're interested only in forces exerted on charge 2. The charge 1 is exerting a force to the right on charge 2, so that force is taken as positive. It's true that charge 2 exerts a force to the left on charge 1 (this is the Newton's 3rd Law counterpart to the force exerted on charge 2 by charge 1), but this force isn't relevant since it isn't exerted on charge 2.

Hi elizabeth, since the question is asking for speed, the answer will always be positive since speed is the magnitude of the velocity. Put in other words, speed doesn't care about direction, it's always a positive number.

How can the final velocity be equal to both the positive and negative initial velocity? Shouldn't the final velocity be only equal to the negative initial velocity due to direction? Thank you.

Hi elizabeth, thanks for the question. The formula $v_f^2 = v_i^2 + 2ad$ is more generic than you're imagining. I think what you have in mind is that $v_f = -v_i$ at the time when the ball has returned to the thrower's hand, and that's true. But this formula works for all times when displacement is zero ($d=0$), and there's one other time when that's the case, namely, in the beginning! This "in the beginning" solution is kind of trivial since it's so obvious that the final velocity after basically no time has elapsed is the same as the initial velocity. This is what the solution $v_f = +v_i$ is telling us, that there is a moment when the final velocity is equal to the positive of the initial velocity, and it's our job to make the physical interpretation of that mathematical solution to understand that it's referring to the moment when the ball is launched. The formula gives all solutions when the displacement is zero, and there are two moments when that's the case: initial launch, and the much later return to the hand.

Thanks for the question. That's a pretty common one, but it would be very time consuming to also cover the MisConceptual Questions, so I'm sticking just to the regular Problems, which I hope meets most of the needs of most of the students.

Hi celloplayer091, thanks for the question. There is a factor of $3$, because of the 3 rear seat passengers, multiplying by their distance from the front of the car (3.90m), and then a factor $2$ multiplying by the distance to the two front seat occupants (2.80m). It sounds like you were expecting the factors of $3$ and $2$ to multiply by masses instead of distances, and that would work fine if you didn't "factor out" the $m_p$. For example, we could re-write the second line of algebra as $x_{cm} = \dfrac{m_cx_c + 2m_px_F + 3m_px_B}{m_c + 5m_p}$, and this would be the same as the version shown in the video with the $m_p$ factored out.

Hi coolkiddonald101, thanks for getting in touch. In order to find the moment of inertia of the helicopter rotor we need to use one of the formulas in Figure 8-20 on pg. 210 which gives a list of formulas for different shapes and positions of the axis of rotation. The question tells us to treat each blade as a rod with an axis of rotation at the end (the end in this case being the center of the rotor where it connects to the axel that goes to the motor). The moment of inertia formula for a rod with an axis of rotation at the end is $\dfrac{1}{3}ML^2$, so that's where the $\dfrac{1}{3}$ comes from: it's part of the appropriate formula. If you're asking why that formula has the $\dfrac{1}{3}$, then don't worry about it since it's a non-obvious result of using calculus, which is beyond the scope of a course in algebraic physics.

## 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

## Giancoli 7th Edition, Chapter 4, Problem 40

By mafiadx83 on Sat, 04/30/2016 - 06:49why friction force equal mass * acceleration ?

Hi mafiadx83, the rule for all force questions (and this rule goes by the name Newton's Second Law) is that

net forceequals mass * acceleration. In this particular question there is only one force, namely friction, and for this reason the net force is made up only of the friction force. Since the friction force is the net force in this particular question, that's why friction force equals mass * acceleration for this particular question.All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 4, Problem 29

By mafiadx83 on Fri, 04/29/2016 - 10:45why initial velocity equal zero ? is it suppose that final velocity equal zero because he is saying in the question what speed did the sprinter leave the starting block

Hi mafeadx83, yes, that's right. It's just implied that the sprinter is initially at rest on the starting block, and the question is asking for the speed with which the sprinter launches off the starting block after applying a force.

Best wishes,

Mr. Dychko

## Giancoli 7th Edition, Chapter 20, Problem 31

By merkinthedark on Mon, 04/25/2016 - 13:39can you check your calculations because I get my answer in the order of 5.0437x10^-7

Hi merkinthedark, I ran the final numbers shown in the video through the calculator again and confirmed the answer, so please let me know if you find any specific error. It looks fine so far.

Best,

Mr. Dychko

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

By mikepeter on Sun, 04/24/2016 - 15:23why does friction force opposite to the applied force?

In question 8 , applied force and friction force are in the same direction /

can you please elaborate this difference?

Hi mikepeter, thanks for the good question. A sliding friction force is always opposes motion. Sliding friction slows things down. This means the direction of a sliding friction force is in the opposite direction as the direction of motion. The direction of the applied force in problem 8, or here in problem 10, actually doesn't matter at all. It's only the direction of motion that matters in determining the direction of the sliding friction force.

All the best with your studies,

Mr. Dychko

## Giancoli 7th Edition, Chapter 19, Problem 34

By bmuniz8219 on Fri, 04/15/2016 - 14:16Hi i have a quick question. If I1 points towards the left how does it in include 12OHMs

Hi bmuniz8219, that's a good question. Each current "starts" at a junction. While the diagram draws $I_1$ to the left along the very top, it's understood (even though not explicitly drawn) that the current originates at the junction below the $12\Omega$ resistor where the $I_2$ and $I_3$ currents also meet. So $I_1$ initially goes up through the $12\Omega$ resistor, and THEN, left along the top.

All the best with your studies,

Mr. Dychko

## Giancoli 6th Edition, Chapter 6, Problem 29

By hsumal on Wed, 04/13/2016 - 22:37*Work-Energy Theorem

## Giancoli 6th Edition, Chapter 6, Problem 29

By hsumal on Wed, 04/13/2016 - 22:33Why can't we use the conservation of energy law here? Wext= Change in KE + Change in PE + Ethermal where we can take everything as the system so that no external work is done and Etherm is zero.

Hi hsumail, thanks for the question, and apologies for taking so long to get back. Conservation of energy is being used here, though it could be expressed as the work-energy theorem as well and get the same result. The idea is that the total energy before a process is the same as the total energy after the process, and the formulae surrounding this subject serve just to help account for all the different forms of energy before and after the processes. Conservation of energy problems are book-keeping exercises.

Now I'll get to your comment. You mentioned the work-energy theorem, so let's clarify that it's $W_{ext} = \Delta KE$, not the expression written in your comment. [edit: earlier comments about the Earth don't apply. I hadn't actually looked up the problem.] You could consider the car-spring as a single system, which is all well and good, but this isn't strategic for problem solving since your only conclusion would be that $W_{ext} = 0$. This problem is asking for the spring constant, and conservation of energy is the ticket to solving for it. Maybe you would prefer the following expression, rather than the one I wrote in the video:

$KE_1 + PE_1 + W_{NC} = KE_2 + PE_2$. $W_{NC}$ typically represents friction, which is zero in this case, so we plug different types of energy into this expression and get $\dfrac{1}{2}mv_i^2 + 0 + 0 = 0 + \dfrac{1}{2}kx^2$ where I've substituted specific expressions in the same order as the equation before that has more generic expressions. This will arrive at $\dfrac{1}{2}mv_i^2 = \dfrac{1}{2}kx^2$, which is the same as in the video.

Hope this helps,

Mr. Dychko

## Giancoli 6th Edition, Chapter 11, Problem 8

By johncondon101 on Sun, 04/10/2016 - 15:49Thanks a lot

My pleasure. You're welcome.

## Giancoli 7th Edition, Chapter 22, Problem 31

By jaclynrgile on Sun, 04/10/2016 - 06:13roynunez273 is correct about B, however I took a different approach to the answer and ended up getting it right. Need to remove pi from your calculation:

So, F=PA= (7.3479x10^-7 N/m^2) (1.0x10^-4) should give you the correct answer.

Hi jaclynrgile, thanks for jumping in. I love the discussion this solution is making. In order to duplicate the answer in the back of the text, both you and roynunez273 are correct that you need to multiply the radiation pressure by a figure tip area approximated as $1.0 \textrm{ cm}^2$. Since part B is an estimate, and the figure tip area isn't specified, this creates some room for interpretation. While it's fine to estimate the finger tip area as $1.0 \textrm{ cm}^2$, as the text does, it's also OK to estimate a finger tip radius as $1.0 \textrm{ cm}^2$ and then calculate the area using $\pi r^2$, because who's finger tip is it? Each student, looking at their own finger tips, should get slightly different answers, and the only important point is that the estimate really is something close to the area of a finger tip.

Best wishes,

Mr. Dychko

## Giancoli 6th Edition, Chapter 11, Problem 8

By johncondon101 on Fri, 04/08/2016 - 17:05in number 5 it was f2/f1, why is this problem f1/f2? How do you tell which way to do it, or is it arbitrary?

Hi johncondon101, thanks for the question. The choice is kind of strategic. Either way will lead to the answer, but the algebra just looks a little tidier one way vs. the other, and the choice is more a matter of intuition. In problem #5 it made sense to have $f_2$ in the numerator since it was the unknown, and we ended up multiplying by the denominator ($f_1$ in that case) to isolate $f_2$ on one side of the equation. In this problem it's a bit less clear which choice to make, but since $m$ is the unknown and it will appear in both the numerator and denominator (eventually, after a couple of steps, which is why, again, setting things up in this case is more a matter of intuition), it's better to have it by itself in the denominator instead of adding to a term since we'll be multiplying by the denominator, and it's easier to multiply by a single factor.

Again, there's nothing much really here to worry about, it's just that one way might lead to neater work than the other. Either way gets to the answer after some algebra.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 19, Problem 32

By merkinthedark on Wed, 04/06/2016 - 13:02If in the previous problem we discovered that currents I1 and I2 go to the left then why is the junction rule applied so that I1= I2+I3? Is it not I3=I1+I2?

Hi merkinthedark, thanks for the question. Yes, we could have used knowledge from problem 31 to choose $I_2$ to the left, and this would be fine, but we don't have to do that. There's no particular reason why I didn't. It isn't important to be correct with the initial guess about the current direction. One just has to make a choice and then construct the equations accordingly. You'll notice that the final answer in this problem has $I_2$ going to the left after all, as we expect from our work in problem 31, so we arrive at the correct answer despite using an initial guess in the other direction.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 22, Problem 31

By roynunez273 on Sun, 04/03/2016 - 20:00I think part B is wrong. We are given the area of the finger. All we have to do is just convert the area from cm^2 to m^2

Hi roynunez273, thanks for the comment. In part B we're calculating force by multiplying pressure by the area of the finger tip. The finger tip area is $\pi r^2$, and we're estimating the radius to be $1 \times 10^{-2} \textrm{ m}$, which is already in meters so that no conversion is necessary. When we multiply the pressure by that area we get the force shown in the solution, so please just let me know if I'm missing something, and please be specific.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 16, Problem 5

By dennis.colon.jr on Thu, 03/31/2016 - 07:53I got 1.708x10^-55

Hi dennis.colon.jr, thanks for the comment. I can't see where the mistake is to arrive at the $1.7 \times 10^{-55}$ figure, but I've doubled checked that the answer I've provided is correct. If you have a more specific question about the solution, just let me know.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 16, Problem 11

By moonpen on Wed, 03/23/2016 - 11:47when calculating the Force of 2, why isn't the force from 2 to 1 negative?

Hi moopen, thanks for the question. When calculating the net force on charge 2 we're interested only in forces exerted

oncharge 2. The charge 1 is exerting a force to the right on charge 2, so that force is taken as positive. It's true that charge 2 exerts a force to the left on charge 1 (this is the Newton's 3rd Law counterpart to the force exerted on charge 2 by charge 1), but this force isn't relevant since it isn't exertedoncharge 2.Hope that helps,

Mr. Dychko

## Giancoli 7th Edition, Chapter 6, Problem 29

By daniel.weiss1 on Mon, 03/21/2016 - 18:43Did a decimal place get lost in the answer for problem a? I believe that the answer is 9.06 x 10^5 for both a and b

Hi daniel, yes indeed, thank you for spotting that missing decimal. I have updated the quick answer.

Best wishes with your studies,

Mr. Dychko

## Giancoli 6th Edition, Chapter 2, Problem 47

By elizabeth on Fri, 03/18/2016 - 13:22For b), shouldn't the answer be negative as the direction of the velocity is downwards? Thank you.

Hi elizabeth, since the question is asking for

speed, the answer will always be positive since speed is themagnitudeof the velocity. Put in other words, speed doesn't care about direction, it's always a positive number.Best wishes,

Mr. Dychko

## Giancoli 6th Edition, Chapter 2, Problem 41

By elizabeth on Thu, 03/17/2016 - 22:02How can the final velocity be equal to both the positive and negative initial velocity? Shouldn't the final velocity be only equal to the negative initial velocity due to direction? Thank you.

Hi elizabeth, thanks for the question. The formula $v_f^2 = v_i^2 + 2ad$ is more generic than you're imagining. I think what you have in mind is that $v_f = -v_i$ at the time when the ball has returned to the thrower's hand, and that's true. But this formula works for all times when displacement is zero ($d=0$), and there's one other time when that's the case, namely, in the beginning! This "in the beginning" solution is kind of trivial since it's so obvious that the final velocity after basically no time has elapsed is the same as the initial velocity. This is what the solution $v_f = +v_i$ is telling us, that there is a moment when the final velocity is equal to the positive of the initial velocity, and it's our job to make the physical interpretation of that mathematical solution to understand that it's referring to the moment when the ball is launched. The formula gives all solutions when the displacement is zero, and there are two moments when that's the case: initial launch, and the much later return to the hand.

Hope that helps,

Mr. Dychko

## 7th Edition is complete

By hoga0114 on Sun, 03/13/2016 - 17:29*Dychko

## 7th Edition is complete

By hoga0114 on Sun, 03/13/2016 - 17:28Hi Mr. Dychoko,

Do you plan to add the MisConceptual questions? I think it would be really useful, and I can't find those answers anywhere. Thanks!

Anna

Hi Anna,

Thanks for the question. That's a pretty common one, but it would be very time consuming to also cover the MisConceptual Questions, so I'm sticking just to the regular Problems, which I hope meets most of the needs of most of the students.

Best wishes with your studies,

Mr. Dychko

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

By celloplayer091 on Thu, 03/10/2016 - 18:49Why did you not multiply three times 65 for the three passengers?

Hi celloplayer091, thanks for the question. There is a factor of $3$, because of the 3 rear seat passengers, multiplying by their distance from the front of the car (3.90m), and then a factor $2$ multiplying by the distance to the two front seat occupants (2.80m). It sounds like you were expecting the factors of $3$ and $2$ to multiply by masses instead of distances, and that would work fine if you didn't "factor out" the $m_p$. For example, we could re-write the second line of algebra as $x_{cm} = \dfrac{m_cx_c + 2m_px_F + 3m_px_B}{m_c + 5m_p}$, and this would be the same as the version shown in the video with the $m_p$ factored out.

Hopefully that helps,

Mr. Dychko

## Giancoli 7th Edition, Chapter 19, Problem 29

By chuy4espindola4 on Wed, 03/09/2016 - 08:51How does the -2.5263 I3 added to the -1.4118 I3 equal 4.981 I3? Should it not be 3.981 I3?

Hi chuy4espindola4, that's a good question. Another student also asked about that step, and here was my reply: https://www.giancolianswers.com/giancoli-physics-7th-edition-solutions/c.... Please just ask further if something there isn't clear.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 8, Problem 43

By coolkiddonald101 on Tue, 03/08/2016 - 12:56please explain why you have (1/3)

Hi coolkiddonald101, thanks for getting in touch. In order to find the moment of inertia of the helicopter rotor we need to use one of the formulas in Figure 8-20 on pg. 210 which gives a list of formulas for different shapes and positions of the axis of rotation. The question tells us to treat each blade as a rod with an axis of rotation at the end (the end in this case being the center of the rotor where it connects to the axel that goes to the motor). The moment of inertia formula for a rod with an axis of rotation at the end is $\dfrac{1}{3}ML^2$, so that's where the $\dfrac{1}{3}$ comes from: it's part of the appropriate formula. If you're asking why that formula has the $\dfrac{1}{3}$, then don't worry about it since it's a non-obvious result of using calculus, which is beyond the scope of a course in algebraic physics.

Best wishes with your studies,

Mr. Dychko