### Giancoli 7th Edition, Chapter 14, Problem 26

By chaegyunkang on Wed, 03/07/2018 - 18:41

I am very confused about how the water can evaporate in room temperature. I know that the water cannot reach 100 celsius from our skin but how can water just evaporate in room temp?

By Mr. Dychko on Sun, 03/11/2018 - 16:30

Hi chaegyunkang, nice question! It turns out that temperature is an average kinetic energy of the particles of the substance. Water at $100^\circ \textrm{ C}$ has molecules with an average kinetic energy sufficient to turn into gas, so many of them do really quickly, and we observe this as boiling. Water at, say, $20^\circ \textrm{ C}$ also has some molecules with enough kinetic energy to turn into gas, but not very many, so we don't see anything special such as bubbles or steam. Nevertheless those few molecules with enough energy to do so turn into gas. This is what's responsible for evaporation, and eventually the luke warm water will all turn to gas for this reason.

Even ice evaporates! A full ice cube tray in your freezer will eventually have much smaller ice cubes after months since some of the molecules have enough kinetic energy to turn into a gas, even though the average kinetic energy is such that the molecules are in the ice phase.

All the best,
Mr. Dychko

### Giancoli 7th Edition, Chapter 8, Problem 58

By aquaoasis14 on Sun, 03/04/2018 - 19:42

why did you multiply it by 4?

By Mr. Dychko on Sun, 03/11/2018 - 16:23

Hi aquaoasis14, thanks for the question. At the 2:00 minute mark in the video, I multiply by 4 since 4 is the lowest common denominator of the fractions in the equation. Doing this gets rid of the fractions since, as a matter of personal opinion, they're annoying! It would be possible to solve the equation just fine while keeping the fractions, if you prefer. The last term in the equation has 1/2 multiplied by 1/2 (making 1/4), keep in mind, so that's why I needed to multiply by 4 instead of 2.

Hope that helps,
Mr. Dychko

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

By chaegyunkang on Thu, 03/01/2018 - 16:50

If the question asked for amount of water, is it okay to end with kg of water? Thank You.

By Mr. Dychko on Sun, 03/11/2018 - 16:19

Hi chaegyunkang, it looks to me like the question is asking for mass per time, which means there needs to be units of mass divided by time, which could be kg / hr, but not just kg.

All the best,
Mr. Dychko

### Giancoli 6th Edition, Chapter 21, Problem 16

By mbnoroozi on Sun, 02/25/2018 - 14:51

Hi Dr. Dychko - I might be missing something obvious but can you explain why the final answer is negative?

Thanks!

By Mr. Dychko on Mon, 02/26/2018 - 20:54

Hi mbnoroozi, thanks for the question. The final answer I have is positive, but I think you're referring to the negative in the formula for the induced EMF? That negative is there to remind us that the induced EMF causes a current directed such that it produces a magnetic flux in the coil that opposes the change in flux that caused the EMF in the first place. Negative in this case can be thought of as "opposite to", in the same way as a car going in the opposite direction to a car with velocity 30 m/s will have a velocity -30 m/s.

Hope this helps,
Mr. Dychko

### Giancoli 7th Edition, Chapter 4, Problem 51

By theovilous on Fri, 02/16/2018 - 07:47

In this problem, can I use the formula, Ffr = μk*Fn? I calculated μk to be .2. So the answer should be .2*60kg*9.8m/s2 = 117.6N. Is this just another way of solving the problem or did I get it wrong? Thank you!!

By Mr. Dychko on Fri, 02/16/2018 - 13:59

Hi theovilous, ah, now I understand your approach. Yes, you can do what you're suggesting. I suspect that if you keep lots of digits in your answer for $\mu_k$ (to avoid intermediate rounding error), that your answer will match the answer here exactly to three significant figures. The approach you're suggesting takes more steps, but it's perfectly valid.

I'm glad you're enjoying the solutions!

All the best,
Mr. Dychko

By Mr. Dychko on Fri, 02/16/2018 - 09:00

Hi theovilous, thanks for the question. While $F_{fr} = \mu_k F_n$ is a correct formula, I'm not sure how you calculated $\mu_k$? It doesn't look possible with the information given, so I think using acceleration is the only method available for solve this one.

All the best,
Mr. Dychko

By theovilous on Fri, 02/16/2018 - 11:18

I considered retarding force as friction. So I set the formula Ffr = ma --> μkmg=ma --> μk = a/g --> μk = 2/9.8 = .2.

So based on the formula, Ffr=μkFn, Ffr=μkmg = .2 * 60kg * 9.8m/s2 = 117.6N.

Does this approach work as well? my result is very close to your answer which is 120N.

Thank you for your quick response! I am learning a lot from you!

### Giancoli 7th Edition, Chapter 8, Problem 59

By choianchoi on Sun, 01/21/2018 - 17:02

Why is kinetic energy not accounted just before it hits the ground? Also, is the velocity of the center of mass the same with the end of the pole?

By Mr. Dychko on Wed, 01/31/2018 - 07:45

Hi choianchoi, thanks for the question. Let's keep in mind that one end of the pole stays in contact with the ground. It's a pole falling over, in other words. The end touching the ground has velocity zero in that case, at all times. The other end of the pole, on the other hand, has the maximum velocity of any point on the pole. The end initially in the air has a higher linear velocity than the center of mass. Kinetic energy is accounted for, and it's the strategy for calculating the velocity of the end of the pole, but the kinetic energy formula doesn't look like what you're used to from linear problems. We don't use $\dfrac{1}{2}mv^2$. Instead, we use rotational kinetic energy $\dfrac{1}{2}I\omega^2$ instead, and use that to figure out the rotational velocity, which is then used to find the linear velocity of the end of the pole.

All the best,
Mr. Dychko

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

By samkvertus on Thu, 01/18/2018 - 20:54

amazing

By Mr. Dychko on Fri, 01/19/2018 - 07:00

Thanks!

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

By samkvertus on Fri, 01/12/2018 - 11:20

what about the last part of the problem that asked to find the magnitude and direction ?

By Mr. Dychko on Fri, 01/12/2018 - 11:26

Hi samkvertus, thanks for the question. Both the magnitude and directly are actually given here. Direction in this case is "to the left" when the resultant is negative, whereas it's "to the right" when positive. The magnitude is the number when ignoring the negative sign. This is explained more in the video, so consider giving it a second view.

All the best,
Mr. Dychko

### Giancoli 7th Edition, Chapter 3, Problem 44

By chaegyunkang on Wed, 01/10/2018 - 08:58

Dear Mr. Dychoko,

For part (a), I understand how you used the cosine rule to find the unknown vector C. However, I tried to solve the problem similar to what you did in problem 42 and divided the wind velocity into X and Y components. I used to pythagorean theorem to find the resultant vector but was not able to get the right answer. Can you please tell me why it is wrong to use this method.

Thank You

By Mr. Dychko on Wed, 01/10/2018 - 09:20

Hi chaegyunkang, thanks for the question. The method you describe works fine, so your sleuthing should look at how you implemented it. Pay careful attention to whether you subtracted the y-component of the velocity of the air with respect to the ground from the velocity of the plane with respect to the air (instead of adding it). Knowing that $\cos(45) = \dfrac{1}{\sqrt{2}}$, here's what the work will look like in your calculator:
$\Big(688 - \dfrac{90}{\sqrt{2}} \Big)^2 + \Big(\dfrac{90}{\sqrt{2}}\Big)^2 = \textrm{ some number}$
$\sqrt{ \textrm{ some number}} = 627.5953...$ which is the correct magnitude, and I'll leave the angle for you.

Hope this helps,
Mr. Dychko

By chaegyunkang on Wed, 01/10/2018 - 09:20

My mistake was adding the vectors. Thank you so much.

### Giancoli 7th Edition, Chapter 3, Problem 4

By chaegyunkang on Sun, 12/31/2017 - 10:07

Hello, I got a different answer for this question. I divided all the vectors into x and y components and used pythagorean theorem at the end to find the resultant vector. My answer is 22.53 km ( 22.53 degrees North of East ).

By Mr. Dychko on Wed, 01/03/2018 - 21:08

All the best,
Mr. Dychko

### Giancoli 7th Edition, Chapter 6, Problem 3

By missedinger on Sat, 12/02/2017 - 21:05

wondering why you stop @ question 71 for solutions, when the text books has 94 questions?

By Mr. Dychko on Sun, 12/03/2017 - 15:51

Hi missedinger, that's a totally fair question, which I've addressed in the 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 Edition, Chapter 10, Problem 9

By jr59j on Fri, 12/01/2017 - 11:55

Why is it 1/2 of m*g

By Mr. Dychko on Sun, 12/03/2017 - 15:52

Hello, thanks for the question. The lady has two heels, so the weight supported by one heel is half the lady's weight, which works out to $\dfrac{1}{2}mg$.

Cheers,
Mr. Dychko

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

By cm2hn on Mon, 11/20/2017 - 20:20

Why the speed of astronaut is negative? Would I get the same result if I use the capsule as the negative value?

By Mr. Dychko on Tue, 11/21/2017 - 13:20

The only thing that matters is that you're consistent with your coordinate system. This is to say that everything pointing in a particular direction needs the same sign (be it positive or negative doesn't matter, just that they're all the same), and things pointing in the opposite direction have the opposite sign. So, yes, you could make the capsule velocity negative, keeping in mind that this now means the force exerted on the capsule by the astronaut would also be negative in that coordinate system, since it would be directed to the right just the same as the now negative capsule velocity.

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

By cm2hn on Mon, 11/20/2017 - 20:06

Why average force is based on ma*vfa and not ms*vfs?

By Mr. Dychko on Tue, 11/21/2017 - 13:17

Hi cm2hn, we could have used the space capsule final speed in part b), as you suggest, but that's taking a bit of a risk since we calculated the space capsule speed ourselves in part a), and we might have made an error in that calculation. Using the astronaut speed provided by the question is a safer approach, but outside of that, either speed is fine.

### Giancoli 7th Edition, Chapter 6, Problem 25

By idan on Wed, 11/15/2017 - 17:49

When a question doesn't specify an initial speed we can assume it's 0? I'm used to seeing them specify "starting from rest". This is only true with an initial speed of 0.

By Mr. Dychko on Sat, 11/18/2017 - 11:53

Hi idan, thank you for the question. When a question doesn't specify an initial speed, and doesn't imply it by saying something like "lifted off the ground" or saying something else which could be interpreted as an initial speed, then you really don't know what the initial speed is. You can't assume it is zero. In this question, the initial speed doesn't actually matter until you get to part e) where it's asking for the final speed. In part e) it finally says "it started from rest". For parts a) through d), the force needed to impart an acceleration of $0.160 \textrm{ g}$ is the same regardless of what speed it's going initially. Only the acceleration is relevant. For example, consider that the force needed to keep the load at rest would be the same as the force needed to make the load rise at a constant speed. The constant speed doesn't matter so far as the force is concerned.

Hope this helps,
Mr. Dychko

### Giancoli 6th Edition, Chapter 26, Problem 13

By phamk on Fri, 11/10/2017 - 08:57

I don't understand part d. Why aren't we solving for t instead, to get an answer of 26.6 seconds? Shouldn't the time elapsed for the friend always be smaller than the time elapsed for us (the observer)?

By Mr. Dychko on Sat, 11/18/2017 - 12:16

Hi phamk, thanks for your question. Part d) could be phrased more explicity. The friend is now watching the Earth based observer zoom past, and is taking note of the time on the Earth observer's watch, which is at different locations between the start and end time. This means the friend is not measuring proper time since the starting event (the Earth base observer's watch at the initial time) is not in the same place as the ending event (the Earth based observer's watch at the end time).

The result of each person reporting the same time elapsed on the other person's watch when their own watch shows 20.0 s passed has some symmetry to it. We expect similar results from each perspective since no one perspective is special compared to the other.

Relativity is definitely confusing since it contradicts our slow moving day-to-day experience!
Hope this helps,
Mr. Dychko

### Giancoli 7th Edition, Chapter 6, Problem 65

By Mr. Dychko on Tue, 11/07/2017 - 10:45

### Giancoli 7th Edition, Chapter 6, Problem 20

By fortunado09 on Sun, 11/05/2017 - 17:34

The force of the ball on the glove is in joules? The answer is in Newtons, though. What am I missing?

By Mr. Dychko on Tue, 11/07/2017 - 10:48

Hi fortunado09, sorry about this one. I misspoke in the video. Force is always in Newtons, of course! I'll make a note above the video for other students.

### Giancoli 7th Edition, Chapter 6, Problem 65

By fortunado09 on Sun, 11/05/2017 - 16:07

Caught this one too, where are you?

### Giancoli 7th Edition, Chapter 6, Problem 63

By fortunado09 on Sun, 11/05/2017 - 15:54

You’ve got 70km/h in your work, but it’s 80km/h in the problem and your calculator!

By Mr. Dychko on Tue, 11/07/2017 - 10:42

Nice catch! I'll mention this error in the final answer. Thanks again for reporting it.

### Giancoli 7th Edition, Chapter 6, Problem 42

By fortunado09 on Sun, 11/05/2017 - 13:11

How do you get the units N/m? I’m getting kg/s^2 which is not a thing I’m familiar with.

How do you know to use 9.8m/s^2 per g? Is that just gravity when dealing with g’s?

By Mr. Dychko on Tue, 11/07/2017 - 10:34

Ah, and yes, "g's" are assumed to be the acceleration of gravity near the surface of Earth, so it can always be substituted with $9.8 \textrm{ m / s}^s$ unless the question explicitly says something else (such as "on Mars" or "10 000 m above the Earth").

By Mr. Dychko on Tue, 11/07/2017 - 10:32

Hi fortunado09, N/m is another way of saying $\textrm{kg / s}^2$. Using one or the other is just a matter of personal preference, and I prefer $\textrm{ N/m}$ since I think it better conveys the idea of a spring constant being the number of Newtons of force the spring will exert for each meter that it's compressed. If you expand the Newtons in $\textrm{ N / m}$ to base units, you'll find am "m" cancels, leaving you with your correct alternative units of $\textrm{ kg / s}^s$.

Cheers,
Mr. Dychko

### Giancoli 7th Edition, Chapter 6, Problem 65

By kfcheung1016 on Mon, 10/30/2017 - 06:31

there is a typing error
the ans is 5.4×104 W in (a.
Thank you

By kfcheung1016 on Mon, 10/30/2017 - 06:32

5.3×10^4 *

### Giancoli 7th Edition, Chapter 4, Problem 42

By tnreifsnyder on Wed, 10/25/2017 - 21:55

If the box is pushed, why is there no force in the positive direction on the x-axis? Also, why is there no net force?

By Mr. Dychko on Tue, 11/07/2017 - 10:56

Hi tnreifsnyder,

There is no force in the positive x-axis since the shove that made the box initially start moving has already finished. We begin the question at the moment when the shove is finished and the box has an initial speed of $3.5 \textrm {m/s}$, and we want to know how far it will go. If there was no friction it would move forever (that's Newton's first law). However, there is in fact a net force along the x-axis: it's friction since that's the only x-axis force! We calculate how far the box will go given it's initial speed imparted by the shove, and the acceleration (or deceleration if you prefer that term) due to the friction force.

Hope that helps,
Mr. Dychko

### 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