For part b), since the electric field is a vector and they're not pointing along the same direction (they're not parallel in other words), it's necessary to use vector addition. Since the angle between them is 90 degrees, using Pythagoras is the most convenient method. This is unlike adding potentials, which are scalars (not vectors), and can be added with regular arithmetic.
Hope this helps,
Mr. Dychko

Just wanted to let you know that there is a typo. For the value of magnification, the quick answer is "-2.9X" but the answer from the video is "-29X." Thank you for your awesome video!

Hi Theovilous, thank you very much for spotting that! I've corrected it. Thanks also for letting me know the videos are helping with your studies. Keep it up!

There is a typo in the answer. In the video, the height of the image is 0.226cm but the answer on top is written 0.266cm. By the way I have been enjoying your physics videos. they are very helpful!!!

I have updated the quick answer. Thank you very much for spotting that, and thanks also for the great feedback. I'm thrilled the solutions are helping with your studies!

Are you referring to the substitution at 3:25? "Substitution" is a strategy for solving sets of related equations (technically called "simultaneous" equations. These are equations that all contain the same variables for the same physical situation). Substitution replaces an unknown variable with an expression containing a different variable, and this process goes on until we're left with an equation containing only one variable, which we can then solve for. Chapter 4 of https://openstax.org/details/books/intermediate-algebra might be helpful.

Thanks for the question. If you want to express the angle in "standard position", which seems to be what you're after with adding something to 270 degrees, you would need to add (90-47.4) to 270. 90-47.4 gives the bit of angle between the terminal arm and the negative y-axis. This standard position angle (312.6 degrees) would be a fine answer as well.
Keep in mind that calculators usually give the "reference angle" (see https://brilliant.org/wiki/reference-angle/) which, in this case, is 47.4 degrees, and because of the way the angle is drawn, this angle is below the negative x-axis.

Hello marthahjalmarsdottir, thanks for the comment. I like getting feedback, and this is a pretty common one, but it's covered in the FAQ: https://www.giancolianswers.com/faq: "It would be enormously time consuming to also answer all the "General Problems" so I'm limiting coverage only to the regular "Problems", of which there are still more than 1700. If you need help with a "General Problem", my suggestion would be to try and find a regular "Problem" that is similar to the "General Problem", and see if you can apply the same problem solving technique."
All the best,
Mr. Dychko

wait but isn't the arclength not the diameter itself though? wouldn't you have to set it to 2pi (because you found the arclength of the crater it would be 180 degrees) and then solve for r, and follow up with multiplying it by 2?

I have some arguments from trigonometry students and their teacher reasoning that for #24 (chapter 3) the answer would be 0. I know this text states on page 63 that Sin2theta is 1 but their reasoning being supported by the calculator is convincing. Would you be able to give more insight to this question?
Thank you.

Hi cmorales, thank you for your question. The textbook question asks "What is the maximum horizontal range of your gun?". The gun is able to shoot, since we're given data about how long a shot straight up takes to return to the gun. This data is useful to determine the initial velocity that's possible for the gun. With that initial velocity in mind, we can calculate the maximum possible range, and it will certainly be more than zero. The angle at which the range is maximum is $45^\circ$, and taking the sine of two times that gives "one", which is the greatest possible sine value of any angle. Keep in mind that we're maximizing horizontal range, not vertical height... I'm just guessing at where the misunderstanding might be...

My homework wanted to see the answer for part B in N*s. My homework is through Pearson Mastering Physics so, I'm not sure if this would apply to everyone.

Hi ragnarlaki, thanks for the comment. This is #13 in my copy of the 7th Edition. Are you using the "Global Edition", perhaps? Solutions for that are here: https://www.giancolianswers.com/giancoli-physics-7th-global-edition-chap.... If not, I'd be curious as to what the ISBN is for your copy of the textbook.

For part c, how can you be certain to use 233.33 m/s as your velocity if this is the velocity give by the problem for the lowest point of the loop? I solved the problem by using 6.0g as my radial acceleration (i got the correct answer but I'm not sure why). Thank You.

Hi maizer01, thanks for the question. Are you referring to the 1/2 in front of each kinetic energy term in part b)? They don't cancel since 1/2 is not a common factor of all terms. There is a potential energy term which is not multiplied by 1/2, which means 1/2 is not a common factor.
All the best,
Mr. Dychko

Hey, would you be able to put up the solution for question 13 of the chapter? It was a question on the number of electrons transferred after charge transfer.

There is a mistake in the video. Although the text is correct when talking about the permeability constant you say 4pi x 10^-4 as the text shows, the correct number is 4pi x 10^-7

For part a, I made the kinetic energy zero thinking that the initial state would be when he was stationary on top of the platform (with only gravitational potential energy). Why is it wrong to use this as my initial state? I do understand how you made your initial state as the moment the person jumps off the platform.

For the two values of m2 that you calculated from the quadratic equation, how do you know that m2 is the larger value and not the other value from the quadratic equation?

Hi chaegyunkang, the presumption here is zero initial velocity, after which a force is applied in order to provide some acceleration. Bobsleds start from rest and the athletes run beside it, pushing it along, before jumping inside (which they do at the 75 meter mark in this case). I suppose some familiarity with bobsledding is called for here, but nevertheless, if an initial speed isn't given in a question, chances are it must be zero. The force applied gives extra acceleration (beyond that provided by gravity), not initial velocity.

Hi chaegyunkang, it's always true that kinetic friction is less than static friction, all else being equal. In this question, however, it doesn't really matter what type of friction is mentioned, since we're not given the coefficient anyway. What matters is the deceleration the car, from which we calculate the coefficient. Be it a static or kinetic coefficient is not important. The important thing is that the type of friction on the incline is the same type of friction on the level surface so that we can be sure that whatever we calculated on the level surface will still apply on the incline.

How do we know that it is a frictionless ramp? Is it because the question does not give us the coefficient of friction so we assume that it is frictionless?

Hi chaegyunkang, thanks for the question. Since the question has a truck driving on the ramp, we assume there is only rolling friction, which we assume is negligibly small. There is no sliding, so there's no kinetic friction. While there is static friction, since we assume the brakes are not being applied (since the ramp is used when the brakes have failed), this means the tire is not applying a horizontal force due to static friction, and so the pavement is not in turn applying a Newton's 3rd law reaction force on the truck. In other words, static friction is not slowing down the truck. There is only rolling friction, which we can ignore.

Question #39 a ball is thrown straight up with a speed of 36 m/s. how long does it take to return to its starting point. Question is my teacher teaches out of 7th edition I have 6th edition but can not find this question in my book or on this website.It says on top of my page ch# 2 and page ref 2-7 but cant find it also how come you don't post question you just go to answers only . need to see question sometimes so I can find which edition it is in.

Hi EddieG, thanks for the question. I would love to post the questions, but I've avoided that since it would be a copyright issue with the publisher since I didn't create the questions. Perhaps you can ask your teacher to photocopy just the problems from their 7th Edition text for your reference?
All the best,
Mr. Dychko

## Giancoli 6th Edition, Chapter 24, Problem 10

By dtearth on Sun, 05/19/2019 - 11:54this is minor but I think the LaTeX on this page's answer might be missing a "}", it's showing up as "1.0 \textrm{ cm"

You're totally right, thank you for spotting that!

## Giancoli 7th Edition, Chapter 17, Problem 27

By joe.robson1234 on Wed, 04/24/2019 - 07:41Why did you use Pythagoras' theorem when you could have just added the electric fields? Any reason?

For part b), since the electric field is a vector and they're not pointing along the same direction (they're not parallel in other words), it's necessary to use vector addition. Since the angle between them is 90 degrees, using Pythagoras is the most convenient method. This is unlike adding potentials, which are scalars (not vectors), and can be added with regular arithmetic.

Hope this helps,

Mr. Dychko

Thank you, this helps a lot.

## Giancoli 7th Edition, Chapter 25, Problem 31

By theovilous on Fri, 04/12/2019 - 17:41Just wanted to let you know that there is a typo. For the value of magnification, the quick answer is "-2.9X" but the answer from the video is "-29X." Thank you for your awesome video!

Hi Theovilous, thank you very much for spotting that! I've corrected it. Thanks also for letting me know the videos are helping with your studies. Keep it up!

## Giancoli 7th Edition, Chapter 23, Problem 54

By theovilous on Mon, 04/01/2019 - 18:13There is a typo in the answer. In the video, the height of the image is 0.226cm but the answer on top is written 0.266cm. By the way I have been enjoying your physics videos. they are very helpful!!!

I have updated the quick answer. Thank you very much for spotting that, and thanks also for the great feedback. I'm thrilled the solutions are helping with your studies!

## Giancoli 7th Edition, Chapter 14, Problem 24

By soja78866 on Mon, 03/11/2019 - 08:34How did you get 2.1x10^5 j/kg? That number is not in the problem.

This number is found in a data table 14-3 titled "Latent Heats" on page 397. It's the latent heat of vaporization for oxygen.

## Giancoli 6th Edition, Chapter 9, Problem 12

By kaam.wilson on Sun, 03/10/2019 - 16:02why was the substitution necessary

I don't understand that part

Are you referring to the substitution at 3:25? "Substitution" is a strategy for solving sets of related equations (technically called "simultaneous" equations. These are equations that all contain the same variables for the same physical situation). Substitution replaces an unknown variable with an expression containing a different variable, and this process goes on until we're left with an equation containing only one variable, which we can then solve for. Chapter 4 of https://openstax.org/details/books/intermediate-algebra might be helpful.

## Giancoli 6th Edition, Chapter 4, Problem 10

By kaam.wilson on Sun, 03/10/2019 - 14:16what is normal. I dont quite understand its use in physics diagrams

"Normal" mean 90 degrees. It's a term that's also used in math. It also goes by another name: "Right angle".

## Giancoli 6th Edition, Chapter 3, Problem 4

By kaam.wilson on Sun, 03/10/2019 - 13:25Since the angle is in the fourth quadrant with y being negative and x being positive, shouldn't you add 47.4 to 270 degrees?

Thanks for the question. If you want to express the angle in "standard position", which seems to be what you're after with adding something to 270 degrees, you would need to add (90-47.4) to 270. 90-47.4 gives the bit of angle between the terminal arm and the negative y-axis. This standard position angle (312.6 degrees) would be a fine answer as well.

Keep in mind that calculators usually give the "reference angle" (see https://brilliant.org/wiki/reference-angle/) which, in this case, is 47.4 degrees, and because of the way the angle is drawn, this angle is below the negative x-axis.

## Giancoli 7th "Global" Edition, Chapter 5, Problem 9

By Mr. Dychko on Sun, 01/27/2019 - 10:29Hello marthahjalmarsdottir, thanks for the comment. I like getting feedback, and this is a pretty common one, but it's covered in the FAQ: https://www.giancolianswers.com/faq: "It would be enormously time consuming to also answer all the "General Problems" so I'm limiting coverage only to the regular "Problems", of which there are still more than 1700. If you need help with a "General Problem", my suggestion would be to try and find a regular "Problem" that is similar to the "General Problem", and see if you can apply the same problem solving technique."

All the best,

Mr. Dychko

## Giancoli 7th "Global" Edition, Chapter 5, Problem 9

By marthahjalmarsdottir on Sun, 01/27/2019 - 09:06Could you add general problems? :)

Hehe

## Giancoli 7th Edition, Chapter 9, Problem 36

By allosh12357 on Tue, 12/25/2018 - 18:42why we took the normal force (upward) not the weight (downward)???

## Giancoli 7th Edition, Chapter 28, Problem 8

By ZreLL on Wed, 11/28/2018 - 15:57We are missing a squared sign for C in the equation

## Giancoli 7th Edition, Chapter 8, Problem 3

By maizer01 on Sun, 11/11/2018 - 07:27wait but isn't the arclength not the diameter itself though? wouldn't you have to set it to 2pi (because you found the arclength of the crater it would be 180 degrees) and then solve for r, and follow up with multiplying it by 2?

## Giancoli 7th Edition, Chapter 3, Problem 24

By cmorales on Thu, 11/01/2018 - 14:22I have some arguments from trigonometry students and their teacher reasoning that for #24 (chapter 3) the answer would be 0. I know this text states on page 63 that Sin2theta is 1 but their reasoning being supported by the calculator is convincing. Would you be able to give more insight to this question?

Thank you.

Hi cmorales, thank you for your question. The textbook question asks "What is the maximum horizontal range of your gun?". The gun is able to shoot, since we're given data about how long a shot straight up takes to return to the gun. This data is useful to determine the initial velocity that's possible for the gun. With that initial velocity in mind, we can calculate the maximum possible range, and it will certainly be more than zero. The angle at which the range is maximum is $45^\circ$, and taking the sine of two times that gives "one", which is the greatest possible sine value of any angle. Keep in mind that we're maximizing horizontal range, not vertical height... I'm just guessing at where the misunderstanding might be...

## Giancoli 7th Edition, Chapter 26, Problem 26

By ZreLL on Thu, 11/01/2018 - 13:29My homework wanted to see the answer for part B in N*s. My homework is through Pearson Mastering Physics so, I'm not sure if this would apply to everyone.

## Giancoli 7th Edition, Chapter 16, Problem 13

By ragnarlaki on Fri, 10/19/2018 - 06:58Oh I’m sorry, I didn’t realize I had the global edition

No worries, and thanks for letting me know that solved the problem!

## Giancoli 7th Edition, Chapter 16, Problem 13

By ragnarlaki on Thu, 10/18/2018 - 09:18This is not problem 13, 7th edition

Hi ragnarlaki, thanks for the comment. This is #13 in my copy of the 7th Edition. Are you using the "Global Edition", perhaps? Solutions for that are here: https://www.giancolianswers.com/giancoli-physics-7th-global-edition-chap.... If not, I'd be curious as to what the ISBN is for your copy of the textbook.

Best wishes,

Mr. Dychko

## Giancoli 7th Edition, Chapter 5, Problem 12

By chaegyunkang on Wed, 10/17/2018 - 17:00For part c, how can you be certain to use 233.33 m/s as your velocity if this is the velocity give by the problem for the lowest point of the loop? I solved the problem by using 6.0g as my radial acceleration (i got the correct answer but I'm not sure why). Thank You.

Oops I misread the part where it said "same speed".

## Giancoli 7th Edition, Chapter 6, Problem 36

By maizer01 on Sun, 10/14/2018 - 16:23what about the 1/2's? don't they cancel?

Hi maizer01, thanks for the question. Are you referring to the 1/2 in front of each kinetic energy term in part b)? They don't cancel since 1/2 is not a common factor of

allterms. There is a potential energy term which is not multiplied by 1/2, which means 1/2 is not acommonfactor.All the best,

Mr. Dychko

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

By kim.c13 on Wed, 10/10/2018 - 23:19On this problem, the equation for elastic collisions is vA - vB = v'A - v'B, but why is the one on #25 negative for the final velocity (v') side?

Wait! Nevermind - got it.

## Giancoli 7th "Global" Edition, Chapter 16, Problem 10

By hugokatich on Thu, 10/04/2018 - 23:24Hey, would you be able to put up the solution for question 13 of the chapter? It was a question on the number of electrons transferred after charge transfer.

Hi hugokatich, isn't question 13 here: https://www.giancolianswers.com/giancoli-physics-7th-global-edition-solu...?

## Giancoli 7th "Global" Edition, Chapter 18, Problem 33

By ingimar2112 on Thu, 10/04/2018 - 13:20nice

Thanks!

## Giancoli 7th Edition, Chapter 20, Problem 26

By ZreLL on Sun, 09/30/2018 - 09:17There is a mistake in the video. Although the text is correct when talking about the permeability constant you say 4pi x 10^-4 as the text shows, the correct number is 4pi x 10^-7

Thank you very much ZreLL, I've added a note to the quick answer mentioning this error.

All the best,

Mr. Dychko

Your welcome. Thank you for all the help with physics.

## Giancoli 7th Edition, Chapter 6, Problem 38

By chaegyunkang on Sat, 09/22/2018 - 07:03For part a, I made the kinetic energy zero thinking that the initial state would be when he was stationary on top of the platform (with only gravitational potential energy). Why is it wrong to use this as my initial state? I do understand how you made your initial state as the moment the person jumps off the platform.

## Giancoli 7th Edition, Chapter 5, Problem 31

By chaegyunkang on Tue, 09/18/2018 - 16:35For the two values of m2 that you calculated from the quadratic equation, how do you know that m2 is the larger value and not the other value from the quadratic equation?

## Giancoli 7th Edition, Chapter 11, Problem 21

By kevin.rosario on Sat, 09/15/2018 - 23:12I did part A on my calculator and I came up with 22.64942355. How did you come up with 2.29Hz?

Thank you.

## Giancoli 7th Edition, Chapter 4, Problem 63

By chaegyunkang on Sat, 09/15/2018 - 16:18How do we know that the initial velocity is zero? If it was given a push I thought it would be a non-zero velocity.

Hi chaegyunkang, the presumption here is zero initial velocity, after which a force is applied in order to provide some acceleration. Bobsleds start from rest and the athletes run beside it, pushing it along, before jumping inside (which they do at the 75 meter mark in this case). I suppose some familiarity with bobsledding is called for here, but nevertheless, if an initial speed isn't given in a question, chances are it must be zero. The force applied gives extra acceleration (beyond that provided by gravity), not initial velocity.

Best wishes with your studies,

Mr. Dychko

## Giancoli 7th Edition, Chapter 4, Problem 61

By chaegyunkang on Sat, 09/15/2018 - 16:03If the car was skidding what difference would it make to the problem? Does it change to kinetic friction when it starts to skid?

Hi chaegyunkang, it's always true that kinetic friction is less than static friction, all else being equal. In this question, however, it doesn't really matter what type of friction is mentioned, since we're not given the coefficient anyway. What matters is the deceleration the car, from which we calculate the coefficient. Be it a static or kinetic coefficient is not important. The important thing is that the type of friction on the incline is the

same typeof friction on the level surface so that we can be sure that whatever we calculated on the level surface will still apply on the incline.All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 4, Problem 55

By chaegyunkang on Sat, 09/15/2018 - 14:39How do we know that it is a frictionless ramp? Is it because the question does not give us the coefficient of friction so we assume that it is frictionless?

Hi chaegyunkang, thanks for the question. Since the question has a truck driving on the ramp, we assume there is only rolling friction, which we assume is negligibly small. There is no sliding, so there's no kinetic friction. While there is static friction, since we assume the brakes are not being applied (since the ramp is used when the brakes have failed), this means the tire is not applying a horizontal force due to static friction, and so the pavement is not in turn applying a Newton's 3rd law reaction force on the truck. In other words, static friction is not slowing down the truck. There is only rolling friction, which we can ignore.

Hope this helps,

Mr. Dychko

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

By EddieG on Wed, 09/05/2018 - 18:50Question #39 a ball is thrown straight up with a speed of 36 m/s. how long does it take to return to its starting point. Question is my teacher teaches out of 7th edition I have 6th edition but can not find this question in my book or on this website.It says on top of my page ch# 2 and page ref 2-7 but cant find it also how come you don't post question you just go to answers only . need to see question sometimes so I can find which edition it is in.

Hi EddieG, thanks for the question. I would love to post the questions, but I've avoided that since it would be a copyright issue with the publisher since I didn't create the questions. Perhaps you can ask your teacher to photocopy just the problems from their 7th Edition text for your reference?

All the best,

Mr. Dychko