It seems that there is a small typo in your answer for the location for the charge. In the video you say that the location is .37l, not .27l. (~4:10 in video)

Hi ceward, thanks for the comment. I'm pretty sure in this case the solution in the back of the book is not correct. I double checked the calculations in the video, and it all checks out. Let me know if you notice any error in the working.

Unless I am missing something, you have a silly math error here in comparing the value predicted from the gas law, 0.588, to the real value, 0.598. The difference is 0.01, not 0.02 as you wrote it.

Hi ctlawson, as part of our job in this question we need to break the electric fields into 'x' and 'y' components. It would be perfectly OK to use $60^\circ$. The difference is that you would have to think about which trig. function is the right one to use. For example, when I calculate $E_{Bx}$ using $\cos(30^\circ)$ you could instead use $\sin(60^\circ)$. It's just a matter of personal preference.

Hi jamesswaggernaut, darn, you're right. The process to solve it for part b) is exactly the same as what's shown for part a). Since the internal resistances are in series with the resistor in the same branch as the battery, the only difference compared with a) is to add one more ohm to $R_1$ and to $R_3$. I've made a note about this in the quick answer, and thanks for spotting it.

How do we know that E1 and E2 have the same magnitude? Is that always the case, even though one is closer and one has a stronger charge than the other?

Hi magiemay480, thanks for your question. E1 and E2 having equal magnitude is a special case for this question. The reason for it is that the electric field at point "P" is zero, and the only way for that to be possible is if E1 and E2 are of equal magnitude in opposite directions.

Hi mschick16, thanks for the question. I suppose it was unnecessary to take the absolute value of KE. You can safely ignore that part. It's just the way I was thinking about the problem in terms of magnitudes, rather than bothering with direction. Consider this different approach that I think you'll prefer in which I'm following the negative signs properly:

I think you'll agree that $\Delta KE = -\dfrac{1}{2}mv_i^2$, and the video shows that $\Delta KE = F_{net}d$, so this means $-\dfrac{1}{2}mv_i^2 = F_{net}d$. This means $F_{net}$ is negative, which we expect from the diagram since it's the friction force directed opposite to the direction of the initial velocity. $F_{net} = 0 - F_{fr}$ where I'm taking all the forces to the right (of which there are none, hence the zero), minus the forces to the left (just friction). Therefore $F_{net} = -F_{fr} = - \mu mg$, and then substituting above gives $-\dfrac{1}{2}mv_i^2 = -\mu mg d$, and this leads to the final expression in the video for initial velocity. That's a lot of extra trouble to follow the negative signs, which is why I skipped it, but following the negatives is more rigorous, so it's good stuff.

hello :).., i didnt understand why we take the final position as a reference instead of the initial point ..i thought that we always take initial point as a referance

I apologize but I made a mistake with my original question. I multiplied 1/2 K value (31.3 N m) by the amplitude squared (.65), not by the equilibrium value x=.36 m.
Sorry about that!

Hi Professor Ditchko,
For part c, (solving for the total energy), I set the force of gravity equal to the spring force (with x = 0.36 m) and solved for k (31.3 N m). I then used the total energy equation 1/2(K A^2) and came up with 6.6 J. Why is this wrong? Can I not use the given x=0.36 as the equilibrium position here?

Hi moonpen, thanks for the question. A couple of ways to think about this. First is that the total forces directed down need to equal the total forces up in order for these to be balanced. If there was a 'net' force either up or down then the pole would move up or down. This means the 'y-component' of the hinge upward needs to equal the total weight down of the pole and light.

The second way to think about this is that the resultant hinge force is at some angle to the vertical. The resultant isn't straight up nor straight sideways since it has both 'x' and 'y' components. I think what you're asking about is why not multiply the resultant hinge force by the angle between the pole and the vertical? It's important to notice that the angle of the resultant hinge force is *not along the pole*, so it has a different angle. We don't know what that angle is (although it could be calculated since we know the 'x' and 'y' components). Hope this helps.

Hi dsgarrido17, I can see what you mean since the video suddenly has a different sound and brightness at that point. It's the result of correcting a typo in the video where I stitched two different 'takes' together. It's still talking about part (a) at 3:40.

Hi Professor Dychko,
You said that theta and X are similar and therefore interchangeable which I understand; but why can we sub out amplitude and replace it with theta max? How are the two related?

Hi aheumangutman, thanks for the question. Seeing as $x$ and $\theta$ are interchangable, let's rewrite the position formula for a simple harmonic oscillator in a way that might answer your question. Normally this position is written as $x = A \cos (2 \pi ft)$, which means the position at a particular time is some fraction (since cos is always less than 1) of the maximum position. So let's rewrite that as $x = x_{max} \cos (2 \pi ft)$. If $x=\theta$ then it stands to reason that $x_{max} = \theta_{max}$ because we're considering $\theta$ to represent a 'position' now, and if the formula is asking for the maximum position (normally called the amplitude), then that's where we write $\theta_{max}$.

Hello Mr. Giancoli . Could you help me please i have final tomorrow and i cant solve one problem ., i have no one to ask

A 30-g mass moving in positive direction collides head-on with a 10-g mass moving at 60 cm/s in the negative direction. The two masses stick together after the collision, and their velocity is 0.10 m/s. Find the velocity of 30-g mass before collision.

I converted : 30 g = 0.03 kg ; 10 g= 0.01kg ; V2 = - 0.6 m/s ; V prime = 0.1 m/s

I found P = 0.004, i have problem to find V1 of mass 0.03 kg

I thought I would let you know, for some reason you wrote the answer for b) is 1.95 rad, however in the textbook and in your video it is stated the answer is 1.05 rad.

Hi moopen, thanks for letting me know about this typo. The '9' is right beside the '0' so that must be why I accidentally entered 1.95 instead of 1.05.

Hi merkinthedark, thank you for spotting this error. To get the final answer I should have solved for $\dfrac{v_{rms_1}}{v_{rms_2}}$ which would give $\sqrt{\dfrac{m_2}{m_1}}$, and then plugged the numbers as shown in the video. The final answer would be correct for $\dfrac{v_{rms_1}}{v_{rms_2}}$. If we pause to apply the "reality check" to the answer (by asking yourself "self, does this number make sense?") we would expect the isotope with less mass to be faster than the heavier isotope since the speed depends only on temperature (which is the same for each in this question), and inversely proportional to mass (meaning smaller mass gives higher speed) according to $v_{rms}=\sqrt{\dfrac{3kT}{m}}$. This means, with the numbers plugged in the way they are in the video, this should be solving for $\dfrac{v_{rms_1}}{v_{rms_2}}$, not $\dfrac{v_{rms_2}}{v_{rms_1}}$. I've flagged this video for a retake one day and made a note in the quick answer section.

Giancoli,
The correct answer on your drawing shows to the 4th power but the correct answer is to the third power. It shows this on your calculator too.

Hi andtorres, thanks for spotting that. Yes indeed, the final answer should be times ten to the third power. I've updated the quick answer and made a note for students.

in the blue equation it shows the square of m1/m2 but when you solve it you put the square of m2/m1...can you explain why you did not flip Vrms2/Vrms1 but did so for the left side of the equation?

Hello Mr. Dychko,
I just would like to point out that while you entered the correct formula into the calculator, in the written equation you forgot to square 3.0m in the last part of part "B".

Hi markinthedark, yes, nice catch. I recalculated the length with helium rather than recalculating the fundamental frequency. To find the fundamental frequency of an open ended tube, with the length calculated in part (a) as $0.58537 \textrm{ m}$, and the speed of sound in helium as $1005 \textrm{ m/s}$ (this is a number I just had to look up), the fundamental frequency is $f_1 = \dfrac{v}{2l} $$= \dfrac{1005 \textrm{ m/s}}{2 * 0.58537 \textrm{ m}}$$=858.43 \textrm{ Hz} = 858 \textrm{ Hz}$. I've flagged the video to update it one day, and I'll post a note in the quick answer section. Let me know if you still have any questions about this problem.

There isn't really an explicit equation mentioned in the textbook for this. Notice that the equation resembles an equation for a straight line $y = mx + b$. This means if you know the "y" intercept (zero in this case), then you can add that to the quantity that varies (the length of the alcohol column in this case) times the amount by which the temperature changes per change in length. That's how I think of it to myself, but that might not make a lot of sense as an explanation... Here's another take:
Consider the first line where we're calculating $\dfrac{\Delta T}{\Delta l}$. I wrote this with numbers, but it could have been written with variables. It would look like this: $\dfrac{\Delta T}{\Delta l} = \dfrac{T - T_o}{l - l_o}$. Multiply both sides by $l-l_o$ gives $\dfrac{\Delta T}{\Delta l}(l-l_o)$. Then add $T_o$ to both sides, and there you have it: the equation of interest. Hopefully one of those explanations helps.

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

By williams.dpw on Tue, 01/26/2016 - 22:55It seems that there is a small typo in your answer for the location for the charge. In the video you say that the location is .37l, not .27l. (~4:10 in video)

Hi williams.dpw, thank you very much for spotting that. I have updated the quick answer.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 12, Problem 33

By ceward on Tue, 01/26/2016 - 18:22The answer in the back of the book is 0.22m for part A

Hi ceward, thanks for the comment. I'm pretty sure in this case the solution in the back of the book is not correct. I double checked the calculations in the video, and it all checks out. Let me know if you notice any error in the working.

Best wishes,

Mr. Dychko

## Giancoli 7th Edition, Chapter 13, Problem 35

By jimh on Mon, 01/25/2016 - 14:37Unless I am missing something, you have a silly math error here in comparing the value predicted from the gas law, 0.588, to the real value, 0.598. The difference is 0.01, not 0.02 as you wrote it.

Thanks a lot jimh, you're quite right that there is a silly math error. The difference is $0.01 \textrm{ kg/m}^3$ as you said.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 16, Problem 33

By ctlawson on Wed, 01/20/2016 - 18:15Hi,

Could you explain why you used 30 degree angles instead of 60?

Thank you.

Hi ctlawson, as part of our job in this question we need to break the electric fields into 'x' and 'y' components. It would be perfectly OK to use $60^\circ$. The difference is that you would have to think about which trig. function is the right one to use. For example, when I calculate $E_{Bx}$ using $\cos(30^\circ)$ you could instead use $\sin(60^\circ)$. It's just a matter of personal preference.

Cheers,

Mr. Dychko

Thank you, Mr. Dychko.

## Giancoli 7th Edition, Chapter 19, Problem 28

By brandonsugarman on Mon, 01/18/2016 - 20:35theres a part b i think you neglected to answer ;)

Hi jamesswaggernaut, darn, you're right. The process to solve it for part b) is exactly the same as what's shown for part a). Since the internal resistances are in series with the resistor in the same branch as the battery, the only difference compared with a) is to add one more ohm to $R_1$ and to $R_3$. I've made a note about this in the quick answer, and thanks for spotting it.

Best wishes,

Mr. Dychko

## Giancoli 7th Edition, Chapter 16, Problem 32

By maggiemay480 on Mon, 01/18/2016 - 14:28How do we know that E1 and E2 have the same magnitude? Is that always the case, even though one is closer and one has a stronger charge than the other?

Hi magiemay480, thanks for your question. E1 and E2 having equal magnitude is a special case for this question. The reason for it is that the electric field at point "P" is zero, and the only way for that to be possible is if E1 and E2 are of equal magnitude in opposite directions.

Hope that helps,

Mr. Dychko

Thanks so much, Mr. Dychko! You were a lifesaver in my last course and are continuing to be invaluable in E&M as well! :)

## Giancoli 7th Edition, Chapter 6, Problem 27

By sueqrahn on Sat, 01/09/2016 - 20:15This answer should only have the units of meters, not meters per second

Hi sueqrahn, I've corrected the typo. Thanks for spotting that.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 6, Problem 23

By mschick16 on Fri, 01/08/2016 - 15:17Why are you allowed to take the absolute value of KE? Shouldn't it be -1/2mvi^2 instead of positive?

Hi mschick16, thanks for the question. I suppose it was unnecessary to take the absolute value of KE. You can safely ignore that part. It's just the way I was thinking about the problem in terms of magnitudes, rather than bothering with direction. Consider this different approach that I think you'll prefer in which I'm following the negative signs properly:

I think you'll agree that $\Delta KE = -\dfrac{1}{2}mv_i^2$, and the video shows that $\Delta KE = F_{net}d$, so this means $-\dfrac{1}{2}mv_i^2 = F_{net}d$. This means $F_{net}$ is negative, which we expect from the diagram since it's the friction force directed opposite to the direction of the initial velocity. $F_{net} = 0 - F_{fr}$ where I'm taking all the forces to the right (of which there are none, hence the zero), minus the forces to the left (just friction). Therefore $F_{net} = -F_{fr} = - \mu mg$, and then substituting above gives $-\dfrac{1}{2}mv_i^2 = -\mu mg d$, and this leads to the final expression in the video for initial velocity. That's a lot of extra trouble to follow the negative signs, which is why I skipped it, but following the negatives is more rigorous, so it's good stuff.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 6, Problem 41

By j1126990 on Sat, 12/26/2015 - 11:41hello :).., i didnt understand why we take the final position as a reference instead of the initial point ..i thought that we always take initial point as a referance

to keep the numbers positive. positive 15 instead of negative 15

3 years late lol

## Giancoli 7th Edition, Chapter 11, Problem 14

By aheumangutman on Mon, 12/21/2015 - 19:07I apologize but I made a mistake with my original question. I multiplied 1/2 K value (31.3 N m) by the amplitude squared (.65), not by the equilibrium value x=.36 m.

Sorry about that!

## Giancoli 7th Edition, Chapter 11, Problem 14

By aheumangutman on Mon, 12/21/2015 - 18:59Hi Professor Ditchko,

For part c, (solving for the total energy), I set the force of gravity equal to the spring force (with x = 0.36 m) and solved for k (31.3 N m). I then used the total energy equation 1/2(K A^2) and came up with 6.6 J. Why is this wrong? Can I not use the given x=0.36 as the equilibrium position here?

Thanks so much!

## Giancoli 6th Edition, Chapter 10, Problem 11

By cotter.gina on Mon, 12/21/2015 - 15:35Hello, why haven't you drawn the force of gravity going downward? Does the gauge pressure account for the weight of the car as well? Thanks!

## Giancoli 7th Edition, Chapter 9, Problem 19

By moonpen on Sat, 12/19/2015 - 09:05Why wouldn't you use sin37 in Fp and Ft when calculating the Y component of the hinge in part b?

Hi moonpen, thanks for the question. A couple of ways to think about this. First is that the total forces directed down need to equal the total forces up in order for these to be balanced. If there was a 'net' force either up or down then the pole would move up or down. This means the 'y-component' of the hinge upward needs to equal the total weight down of the pole and light.

The second way to think about this is that the resultant hinge force is at some angle to the vertical. The resultant isn't straight up nor straight sideways since it has both 'x' and 'y' components. I think what you're asking about is why not multiply the resultant hinge force by the angle between the pole and the vertical? It's important to notice that the angle of the resultant hinge force is *not along the pole*, so it has a different angle. We don't know what that angle is (although it could be calculated since we know the 'x' and 'y' components). Hope this helps.

All the best,

Mr. Dychko

## Giancoli 6th Edition, Chapter 4, Problem 37

By dsgarrido17 on Thu, 12/17/2015 - 19:05At 3:40 I think it jumps back to A while talking about B. or skips a step

Hi dsgarrido17, I can see what you mean since the video suddenly has a different sound and brightness at that point. It's the result of correcting a typo in the video where I stitched two different 'takes' together. It's still talking about part (a) at 3:40.

Cheers,

Mr. Dychko

## Giancoli 7th Edition, Chapter 11, Problem 34

By aheumangutman on Thu, 12/17/2015 - 11:13Hi Professor Dychko,

You said that theta and X are similar and therefore interchangeable which I understand; but why can we sub out amplitude and replace it with theta max? How are the two related?

Thanks!

Hi aheumangutman, thanks for the question. Seeing as $x$ and $\theta$ are interchangable, let's rewrite the position formula for a simple harmonic oscillator in a way that might answer your question. Normally this position is written as $x = A \cos (2 \pi ft)$, which means the position at a particular time is some fraction (since cos is always less than 1) of the maximum position. So let's rewrite that as $x = x_{max} \cos (2 \pi ft)$. If $x=\theta$ then it stands to reason that $x_{max} = \theta_{max}$ because we're considering $\theta$ to represent a 'position' now, and if the formula is asking for the maximum position (normally called the amplitude), then that's where we write $\theta_{max}$.

Hope that helps,

Mr. Dychko

## Giancoli 6th Edition, Chapter 5, Problem 39

By rbw2306 on Mon, 12/14/2015 - 19:32Why is r4 equal to the square root of .6^2 + .6^2 ?

Forget that comment, pythagorean theorem

## Giancoli 6th Edition, Chapter 4, Problem 53

By SC on Mon, 12/14/2015 - 08:31Isn´t -Fgx = -mgcos22 and not -mgsin22?

Never mind

## Giancoli 6th Edition, Chapter 4, Problem 5

By rinne13 on Sun, 12/13/2015 - 13:39Hello Mr. Giancoli . Could you help me please i have final tomorrow and i cant solve one problem ., i have no one to ask

A 30-g mass moving in positive direction collides head-on with a 10-g mass moving at 60 cm/s in the negative direction. The two masses stick together after the collision, and their velocity is 0.10 m/s. Find the velocity of 30-g mass before collision.

I converted : 30 g = 0.03 kg ; 10 g= 0.01kg ; V2 = - 0.6 m/s ; V prime = 0.1 m/s

I found P = 0.004, i have problem to find V1 of mass 0.03 kg

thank you so much in advance

yana

## Giancoli 7th Edition, Chapter 8, Problem 1

By moonpen on Fri, 12/11/2015 - 18:39I thought I would let you know, for some reason you wrote the answer for b) is 1.95 rad, however in the textbook and in your video it is stated the answer is 1.05 rad.

Hi moopen, thanks for letting me know about this typo. The '9' is right beside the '0' so that must be why I accidentally entered 1.95 instead of 1.05.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 13, Problem 53

By Mr. Dychko on Wed, 12/09/2015 - 21:50Hi merkinthedark, thank you for spotting this error. To get the final answer I should have solved for $\dfrac{v_{rms_1}}{v_{rms_2}}$ which would give $\sqrt{\dfrac{m_2}{m_1}}$, and then plugged the numbers as shown in the video. The final answer would be correct for $\dfrac{v_{rms_1}}{v_{rms_2}}$. If we pause to apply the "reality check" to the answer (by asking yourself "self, does this number make sense?") we would expect the isotope with less mass to be faster than the heavier isotope since the speed depends only on temperature (which is the same for each in this question), and inversely proportional to mass (meaning smaller mass gives higher speed) according to $v_{rms}=\sqrt{\dfrac{3kT}{m}}$. This means, with the numbers plugged in the way they are in the video, this should be solving for $\dfrac{v_{rms_1}}{v_{rms_2}}$, not $\dfrac{v_{rms_2}}{v_{rms_1}}$. I've flagged this video for a retake one day and made a note in the quick answer section.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 10, Problem 24

By andtorres on Tue, 12/08/2015 - 18:30Giancoli,

The correct answer on your drawing shows to the 4th power but the correct answer is to the third power. It shows this on your calculator too.

Hi andtorres, thanks for spotting that. Yes indeed, the final answer should be times ten to the third power. I've updated the quick answer and made a note for students.

Best wishes,

Mr. Dychko

## Giancoli 7th Edition, Chapter 13, Problem 53

By merkinthedark on Tue, 12/08/2015 - 18:29in the blue equation it shows the square of m1/m2 but when you solve it you put the square of m2/m1...can you explain why you did not flip Vrms2/Vrms1 but did so for the left side of the equation?

## Giancoli 7th Edition, Chapter 13, Problem 3

By shine07 on Mon, 12/07/2015 - 19:33Never mind. I read the question wrong.

## Giancoli 7th Edition, Chapter 13, Problem 3

By shine07 on Mon, 12/07/2015 - 19:26Can you please include the formulas first rather than just plugging in the numbers?

## Giancoli 7th Edition, Chapter 8, Problem 67

By jessicaannehoch on Mon, 12/07/2015 - 16:58Hello Mr. Dychko,

I just would like to point out that while you entered the correct formula into the calculator, in the written equation you forgot to square 3.0m in the last part of part "B".

Thank you for your work!

## Giancoli 7th Edition, Chapter 5, Problem 23

By 1609702237 on Sat, 12/05/2015 - 06:45Good job

## Giancoli 7th Edition, Chapter 12, Problem 37

By merkinthedark on Thu, 12/03/2015 - 17:46I believe that part B ask for the new fundamental frequency having the pipe filled with Helium, would you mind double checking.

Hi markinthedark, yes, nice catch. I recalculated the length with helium rather than recalculating the fundamental frequency. To find the fundamental frequency of an open ended tube, with the length calculated in part (a) as $0.58537 \textrm{ m}$, and the speed of sound in helium as $1005 \textrm{ m/s}$ (this is a number I just had to look up), the fundamental frequency is $f_1 = \dfrac{v}{2l} $$= \dfrac{1005 \textrm{ m/s}}{2 * 0.58537 \textrm{ m}}$$=858.43 \textrm{ Hz} = 858 \textrm{ Hz}$. I've flagged the video to update it one day, and I'll post a note in the quick answer section. Let me know if you still have any questions about this problem.

Hope this helps,

Mr. Dychko

## Giancoli 6th Edition, Chapter 5, Problem 47

By devon.gibson2016 on Thu, 12/03/2015 - 01:54Where did you get 6.38 x 10^6 since the radius of the Earth is 6.38 x 10^3?

Hi devon, the difference is in the units. The Earth's radius is $6.38 \times 10^6 \textrm{ m}$, or $6.38 \times 10^3 \textrm{ km}$.

Cheers,

Mr. Dychko

## Giancoli 7th Edition, Chapter 13, Problem 8

By Mr. Dychko on Tue, 12/01/2015 - 20:58There isn't really an explicit equation mentioned in the textbook for this. Notice that the equation resembles an equation for a straight line $y = mx + b$. This means if you know the "y" intercept (zero in this case), then you can add that to the quantity that varies (the length of the alcohol column in this case) times the amount by which the temperature changes per change in length. That's how I think of it to myself, but that might not make a lot of sense as an explanation... Here's another take:

Consider the first line where we're calculating $\dfrac{\Delta T}{\Delta l}$. I wrote this with numbers, but it could have been written with variables. It would look like this: $\dfrac{\Delta T}{\Delta l} = \dfrac{T - T_o}{l - l_o}$. Multiply both sides by $l-l_o$ gives $\dfrac{\Delta T}{\Delta l}(l-l_o)$. Then add $T_o$ to both sides, and there you have it: the equation of interest. Hopefully one of those explanations helps.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 13, Problem 8

By Mr. Dychko on Tue, 12/01/2015 - 20:51A student wrote: I am confused as to how you derived the equation after finding the ratio. Thank you very much.