Giancoli Solutions on Video

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  • 1,930 video solutions for all regular problems in Giancoli's 7th Edition and 1,681 solutions for most regular problems in the 6th Edition.
  • Final answer provided in text form for quick reference above each video, and formatted nicely as an equation, like $E=mc^2$. This is useful if you are in the library or have a slow internet connection.

  • Pen colors make the step-by-step solutions clear. Red is used to illustrate algebra steps, and to substitute numeric values in the final step of a solution. When a solution switches to a new train of thought a different pen color emphasizes the switch, so that solutions are very methodical and organized.
  • Solutions are classroom tested, and created by an experienced physics teacher.
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Sample solution

Giancoli 7th Edition, Chapter 5, Problem 12

(4:49)

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Recent questions and answers

Giancoli 7th Edition, Chapter 4, Problem 47

By nategrana on Mon, 10/02/2017 - 18:31

I don't understand, conceptually, how a 2kg box can not move pulling a 5kg box but move it at constant speed pulling a heavier box. Can you please clarify this? Thank you.

By Mr. Dychko on Mon, 10/02/2017 - 20:46

Hi Nategrana, thanks for the question. The difference between the scenarios has to do with static friction vs kinetic friction. The main idea to consider in both parts is that the more massive the box on the table, the harder it is to pull it since it will be pressed up upon more strongly by the surface (aka. the Normal force) and thereby experience more friction. Our goal is to find the maximum mass possible before that friction prevents acceleration. Notice that I mention "prevents acceleration", since being at rest or moving at constant speed are both examples of not accelerating, and in either case the friction is fully equivalent in magnitude to the tension resulting from the hanging mass.

The mass of the box on the table is only one part of the issue, and the other part is the coefficient of friction. The higher the coefficient of friction, the more friction force results from a given mass. So, in part a) we're dealing with the box on the table at rest, and this means the coefficient of friction to use is the coefficient of static friction. In part b) since the box is sliding, we use the coefficient of kinetic friction. The coefficient of kinetic friction is always lower than the coefficient of static friction. This means that for the force of friction to equal the tension force, the case b) with a smaller coefficient of friction calls for greater mass than compared to the case of part a) where a higher coefficient of static friction means a lower mass is needed to equal that same tension force.

Hope that helps,
Mr. Dychko

Giancoli 7th Edition, Chapter 1, Problem 24

By lsugden on Sun, 10/01/2017 - 11:43

I think there is an error in the final exponents. I have all the same values as you do for the density of the proton, however my final answer keeps coming out to 3.9E14 kg. I am getting a value of 2.055E-4 for the volume of the baseball.

By Mr. Dychko on Mon, 10/02/2017 - 20:06

Thank you so much for spotting this lsugden, you're quite correct that I made an error plugging numbers in for the final calculation. I have updated the quick answer.

Thanks again, and best wishes with your studies,
Mr. Dychko

By lsugden on Sun, 10/01/2017 - 11:44

m^3 for the V_baseball

Giancoli 6th Edition, Chapter 2, Problem 5

By tahell2000 on Wed, 09/20/2017 - 16:04

The answer in top left is labeled m/s . Should be cm/s.

By Mr. Dychko on Thu, 09/21/2017 - 09:38

Fixed! Thank you very much for noticing.

Giancoli 7th Edition, Chapter 17, Problem 22

By nategrana on Thu, 09/14/2017 - 16:08

Why do we do scalar addition and not vector addition here? Is it because when we think of potential, we think of a magnitude of available potential occurring as a sort of pool versus, for example, in electric fields, which have an amount and direction its pushing or pulling a charge with a strength?

By Mr. Dychko on Thu, 09/21/2017 - 09:57

Well, scalar addition is used here because we're adding potential energies, and potential energy doesn't have direction. I think it might be helpful to consider what "potential" is, in order to make sense of the fact that it's a scalar. The first thing I always like to remind myself of is that when you hear of "potential", it actually is a lazy way of shortening the more complete term, which is "potential difference". Potential difference is the only thing that can ever be measured. With that said, the next obvious question is "difference" compared to what? In the case of a circuit with a battery, the potential difference is between the positive and negative terminals of the battery. In this question with point charges, the difference is between a point infinitely far away, and the given location of the charge. In other words we have a formula $PE = \dfrac{kq_1q_2}{r}$ that tells us the amount of work an external force (like a hand pushing) would need to do in order to move the charge $q_1$ from infinitely far away to it's given location, which requires work since there's a force of repulsion as a result of $q_2$ (assuming the charges are of the same sign). When there are more charges in addition to $q_2$, as there are on the other two corners of the square in this question, we use the formula a total of 3 times and add the results to find the total. The addition is as scalars since we're adding energies, which don't have direction.

Hope that helps!

Giancoli 6th Edition, Chapter 2, Problem 52

By antoinec467 on Mon, 09/04/2017 - 10:00

Hi how to find the answers for problem 62 and 71?

By Mr. Dychko on Mon, 09/04/2017 - 21:00

Hi antoinec467, 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 Edition, Chapter 16, Problem 15

By nategrana on Thu, 08/31/2017 - 12:09

Why are we looking for d^2 when finding the diagonal distance? The distance of the hypotenuse of a 3,4,5 triangle is not 25 its square root of it which is 5..... A bit confused:(

By Mr. Dychko on Thu, 08/31/2017 - 12:24

Hi nategrana, thanks for the question. To determine the 'x' component of the force on the '2Q' charge, we need the 'x' component of the force due to the '4Q' charge diagonally opposite. The force formula has the distance between the '2Q' and '4Q' charge in the denominator, but it is squared. We're looking for the square of the distance since that's the quantity to plug into the force formula.

Hope that helps,
Mr. Dychko

By nategrana on Thu, 08/31/2017 - 21:00

WOA! Totally missed that. Thank you Mr. Dychko. Great site!

Giancoli 7th Edition, Chapter 2, Problem 22

By mcgracia2008 on Thu, 08/24/2017 - 18:50

I think the result is incorrect. Instead ot multiply 2(88), the tutor divided 2/88.

By Mr. Dychko on Thu, 08/24/2017 - 21:53

Hi mcgracia2008, thanks for looking at the working. I think what you're commenting on is the calculator display in which you see $-28^2/2/88$, but this is the same as $-28^2/(2*88)$. I just like the division signs since it's one less character to press instead of using parentheses for the multiplication in the denominator, but that's just a personal preference.

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 5, Problem 9

By idan on Thu, 08/03/2017 - 15:20

I see how the formula no longer has m in it. However, doesn't friction increase with greater mass which then changes the maximum velocity?

By Mr. Dychko on Thu, 08/31/2017 - 12:31

It seems strange, but true, that mass doesn't affect the maximum velocity. Keep in mind that the friction we're speaking about here is normal to the velocity. In other words, the static friction is perpendicular to the direction of motion. We're speaking only about the friction that changes the direction of the car, so this friction has no effect on the speed.

On the other hand, you might be thinking about "rolling friction", or perhaps "air friction", or in other words the frictions that cause things with wheels to slow down and eventually stop. These types of friction are directed opposite to the direction of motion, and so they indeed do slow the car down. These types of friction are not considered in this question. Rolling friction, indeed, would increase with mass.

All the best,
Mr. Dychko


7th Edition Solutions 6th Edition Solutions Global Edition Solutions