Comments

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

Giancoli 7th Edition, Chapter 11, Problem 21

By chaegyunkang on Tue, 08/01/2017 - 04:47

This question is the answer for question 20 not 21.

By Mr. Dychko on Thu, 08/31/2017 - 12:34

Hi chaegyunkang, it looks like you're using the Global 7th Edition. Here's the identical solution for problem 20, as you say, in the Global edition: https://www.giancolianswers.com/giancoli-physics-7th-global-edition-solu...

Please navigate for solutions starting with the Global edition: https://www.giancolianswers.com/giancoli-physics-7th-global-edition-chap...

Best wishes,
Mr. Dychko

Giancoli 7th Edition, Chapter 11, Problem 20

By chaegyunkang on Tue, 08/01/2017 - 04:46

This question is the answer for question 19 not 20.

By Mr. Dychko on Thu, 08/31/2017 - 12:36

It's listed as #19 in the Global Edition: https://www.giancolianswers.com/giancoli-physics-7th-global-edition-solu..., the navigation for which begins here: https://www.giancolianswers.com/giancoli-physics-7th-global-edition-chap...

Cheers,
Mr. Dychko

Giancoli 7th Edition, Chapter 11, Problem 19

By chaegyunkang on Tue, 08/01/2017 - 04:43

For part a) it is asking for the amplitude of the motion. The video solved for the wrong question and the answer is also wrong.

By Mr. Dychko on Thu, 08/31/2017 - 12:52

Hi chaegyunkang, from your other questions it's clear that you're using the Global 7th Edition. I think this particular question was removed from the Global Edition.

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 20, Problem 28

By eptrottier on Tue, 07/25/2017 - 12:38

where do you get the formula in the blue?

By Mr. Dychko on Thu, 08/31/2017 - 12:54

Hi eptrottier, that formula is on page 571, labelled 20-7. It's the force between two parallel current carrying wires.

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 3, Problem 30

By choianchoi on Sun, 07/23/2017 - 20:44

Hello, why is the acceleration -9.8 m/s^2 when after the ball reaches the top, the acceleration is +9.8m/s^2?

By Mr. Dychko on Thu, 08/31/2017 - 12:59

Hi choianchoi, I've writen the acceleration as -9.8m/s^2 in the formula, and it's important to stay consistent once you've chosen a coordinate system. That is to say that if down is chosen to be negative, as is often the case, then it has to be considered negative for the entire question. Since we've chosen down as negative, the acceleration due to gravity can never be positive, regardless of where the ball is in it's trajectory.

Best wishes,
Mr. Dychko

Giancoli 7th Edition, Chapter 3, Problem 18

By choianchoi on Sun, 07/23/2017 - 02:44

Hi, as the question asked how far from its base shouldn't the answer be square root (x ^2+y^2)? (displacement between landing point and the jump point)

By Mr. Dychko on Thu, 08/31/2017 - 13:01

Hi choianchoi, actually, since the question uses the word base, it's referring to the bottom of the cliff. The bottom of the cliff is the base of the cliff, so the question is asking only for the horizontal component of the displacement from the jump point to the landing point.

Hope that helps,
Mr. Dychko

Giancoli 7th Edition, Chapter 2, Problem 59

By choianchoi on Sat, 07/22/2017 - 00:57

From 30 s to 50 s why is it curved and not straight?

Thanks

Giancoli 7th Edition, Chapter 4, Problem 8

By chris010727 on Sun, 07/09/2017 - 07:53

Why did you set the initial velocity as 0. Isn't it supposed to be 13 instead because the shot will eventually come to a rest.

Giancoli 7th Edition, Chapter 23, Problem 51

By arielle.lab on Wed, 07/05/2017 - 06:01

Hi! How do you know that do is positive?

Giancoli 7th Edition, Chapter 4, Problem 57

By idan on Thu, 06/22/2017 - 14:40

I found a good explanation online so I'm all set!

Giancoli 7th Edition, Chapter 4, Problem 57

By idan on Thu, 06/22/2017 - 14:16

How do you know the angle theta used to resolve the components is the same as the angle the incline makes with the horizontal? Please refresh my geometry!

Giancoli 7th Edition, Chapter 16, Problem 30

By aheumangutman on Mon, 06/05/2017 - 19:27

Hi Professor Dychko,
I'm confused on why you would not solve for the x and y components of both electric fields and then take the magnitude of the net electric field.
Thanks!

Giancoli 7th Edition, Chapter 16, Problem 30

By aheumangutman on Mon, 06/05/2017 - 19:27

Hi Professor Dychko,
I'm confused on why you would not solve for the x and y components of both electric fields and then take the magnitude of the net electric field.
Thanks!

Giancoli 7th Edition, Chapter 28, Problem 2

By kbick on Sun, 05/07/2017 - 11:33

Hi! Just wondering when we can use the small angle approximation - you say here that it is when theta is less than zero, but how can that be?

By Mr. Dychko on Sat, 05/13/2017 - 13:03

Hi kbick, this is a good question since the small angle approximation is often a very useful trick to make the algebra much easier. Yes, I know it seems unbelievable that you can just replace $sin(x)$ with simply $x$, so what you should do to convince yourself that it's acceptable when $x$ is small (I'm saying $x$, but $\theta$, or whatever variable represents the angle in your equation) is this: plot the graph of $y=x$ and $y=sin(x)$ on the same graph, but there's a catch. Zoom in on the graph so that you can only see $x$ between, say, -1 and 1. You'll notice the graphs look the same! This is the reason the approximation works, is that within this restricted domain of $-1 \leq x \leq 1$, both functions $y=x$ and $y=sin(x)$ give the same result. When you zoom out and look at larger values of $x$, then of course the graphs do not look the same, and you can no longer claim that $x \approx sin(x)$, like you can for small $x$.

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 31, Problem 14

By fridley.26 on Wed, 04/26/2017 - 11:46

Your video says released and the answer says required. Just wanted to point it out was confused at first but then after watching the video I understood what happened.

By Mr. Dychko on Fri, 04/28/2017 - 12:24

Ah, thanks fridley.26! Sorry for the confusion. I've updated the quick answer to say released.

Cheers,
Mr. Dychko

Giancoli 7th "Global" Edition, Chapter 8, Problem 19

By lina09037788 on Tue, 04/25/2017 - 06:04

angular position should be in radians, so it's 23*2*3.14 not 23 revolutions.

By Mr. Dychko on Fri, 04/28/2017 - 12:22

Hi flsktkd, thanks for the comment. Putting angular position in radians is often the better choice, and it's the official best unit for expressing angular position, that's true. However, revolutions are still used frequently since it's a unit that people can better understand. When looking at the tachometer in a car, it shows the engine speed in revolutions per minute, for example. Since this question asks for the final angular speed in rpm, it makes sense to use revolutions throughout, although it doesn't specify units for part (a) so I suppose it would be fine to use radians per second squared there.

Best wishes,
Mr. Dychko

Giancoli 7th Edition, Chapter 6, Problem 43

By suriyak786 on Sat, 04/22/2017 - 21:22

if the elevator is falling, wouldn't H be negative?

By Mr. Dychko on Fri, 04/28/2017 - 12:14

Hi suriyak786, thanks for the question. Yes, since we haven't specified the coordinate system, it's true that up is the positive direction, and down is negative. The elevator would have a negative velocity while it's falling. However, it's initial displacement is positive since we've done one unconventional thing here: the origin (in other words the location where the vertical position is zero) is not where the elevator starts. Instead we've chosen the origin, also called the reference level when talking about vertical position, to be at the height of the fully compressed spring, which is below where the elevator starts. This choice of reference level simplified our algebra a bit. $h$ is the displacement of the elevator compared to the reference level, and it is positive since it's above the reference level, and we have the conventional coordinate system with up being positive.

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 11, Problem 4

By dshadowalker on Thu, 04/20/2017 - 10:14

My answer comes out to 316N/m can you show how to input into the calculator? I like when you show the calculator because this assists me in using the calculator and making sure I have the equation right.

By Mr. Dychko on Fri, 04/28/2017 - 12:04

Hi dshadowalker, thanks for the question. I get 316 N/m too! :) The difference is that I've rounded it to two significant figures, which explains why the answer is $320 \textrm{ N/m}$. Yes, showing calcs. with the on-screen calculator is nice, but it was getting very time consuming doing that so I included it only for a few chapters.

Best wishes with your studies,
Mr. Dychko

Giancoli 7th Edition, Chapter 21, Problem 44

By julia.wolfe on Sun, 04/16/2017 - 09:00

Why don't you subtract the potential due to the inductance of the coil? I thought the minus sign in the E = -L (dI/dt) formula was meant to indicate that the voltage would be in the opposite direction following Lenz's law (current increasing > flux increasing > induced current needs to decrease flux to compensate > current flows in other direction > voltage drop in other direction). What am I missing?

By Mr. Dychko on Mon, 04/17/2017 - 06:10

Hi julia.wolfe, thanks for the question. What's important to notice is that the sign of the potential difference will be the same across the resistor as it is across the inductor. In both cases, the circuit element causes a potential drop. (I say "circuit element" since it's OK to model this circuit as containing a single resistor and a single inductor connected by zero-resistance wires). Recall that when using Kirchoff's Laws to analyze a circuit, you make the potential across a resistor negative (it's a potential drop, in other words) when traversing the resistor in the same direction as the current. Because the inductor opposes the change in current, as you correctly mentioned, this inductor also causes a potential drop to oppose the change (note that, contrary to what you mentioned, the current does not end up flowing in the other direction. Rather, it's increase is opposed, and the inductor would disappear once the current stops changing). It would be fine to call both potential drops negative if you wish, but the question isn't specific about direction, so calling both positive is fine when answering a question asking for a "potential difference".

Hope that helps,
Mr. Dychko

Giancoli 7th Edition, Chapter 13, Problem 47

By elkinsk on Sat, 04/15/2017 - 18:32

i think the temperature is 22° not 20°?

By Mr. Dychko on Mon, 04/17/2017 - 06:18

Hi elkinsk, thank you for noticing that! You're quite correct that the temperature should be $22^\circ\textrm{C}$ instead of $20^\circ\textrm{C}$. It turns out that the final answer is the same in either case since the 22 or 20 get added to 273 in order to convert to Kelvin, and the percent difference between 295 and 293 is small. I'll put a note for other students about this, and thanks again.

Best wishes with your studies,
Mr. Dychko

Giancoli 7th Edition, Chapter 14, Problem 3

By girondebordeaux92 on Sun, 04/09/2017 - 03:19

I think specific heat capacity of 4.186 is missing in the calculation? Can you please clarify

By Mr. Dychko on Mon, 04/17/2017 - 06:33

Bonjour girondebordeaux92, thanks for the question. The 4.186 figure you're quoting is, with units, $4.186 \dfrac{\textrm{kJ}}{\textrm{kg} \cdot \textrm{C}^\circ}$. Understanding units answers your question, since if you consult table 14-1 you'll notice that an alternative way to mention the specific heat of water is to write $1.00 \dfrac{\textrm{kcal}}{\textrm{kg} \cdot \textrm{C}^\circ}$. It turns out that the "kcal" way of writing specific heat, as shown in the video, if just convenient in this case since the energy of the candy bar is giving in kcal.

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 18, Problem 35

By cj_vana on Mon, 04/03/2017 - 01:41

For part b, should the 120V be squared?

By Mr. Dychko on Mon, 04/17/2017 - 06:35

Hi cj_vana, thanks for pointing this out. It turn out that I did, in fact, square the $120 \textrm{ V}$ when doing the calculation, even though I forgot to write that the number is squared in the working. I'll make a note about this for other students.

All the best,
Mr. Dychko