Comments

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

Giancoli 6th Edition, Chapter 5, Problem 21

By thesouthportschool on Wed, 03/29/2017 - 22:17

FN = ... +mgcosO

By Mr. Dychko on Mon, 04/17/2017 - 06:45

Hi thesouthportschool, I'm not really sure what the comment is here. I could maybe help if the comment was more specific.

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 19, Problem 19

By brian7989 on Wed, 03/29/2017 - 17:06

From step 2 to step 3, how do you know whether you should combine Req2 with the R on the right of Req2 or R that's below Req 2?

By Mr. Dychko on Mon, 04/17/2017 - 06:41

Hi brian7989, thanks for the question. It's important to combine series resistances before parallel resistances. From step 2 to step 3, Req2 needs to be combined wi the R on the right since it's in series with it. It isn't possible to combine Req2 with the resistor below since it's in parallel with the R below, and it's necessary to know the total resistance of each branch when combining them in parallel. We don't yet know the total resistance of the branch containing Req2 until we add it to the R on the right.

Best wishes,
Mr. Dychko

Giancoli 6th Edition, Chapter 21, Problem 6

By thesouthportschool on Wed, 03/15/2017 - 18:34

I believe you may have gotten the answer wrong and it is 0.036V and not 0.048V.

By Mr. Dychko on Fri, 03/17/2017 - 08:04

Hi thesouthportschool, things look fine after double checking, so please let me know if you're still noticing a discrepancy.

Best wishes with your studies,
Mr. Dychko

Giancoli 7th Edition, Chapter 4, Problem 60

By suriyak786 on Sun, 03/12/2017 - 15:45

If the box is moving up and my positive x component is to the right. Then the Fgx would be going up?

By Mr. Dychko on Fri, 03/17/2017 - 07:58

Hi suriyak786, thanks for this question. The direction of gravity is unaffected by the direction of motion of the box, and it's also unaffected by the coordinate system (whether right is positive or negative). The component of gravity along the ramp will always point down the ramp. Whether that component gets a negative or positive sign is determined by your personal choice of coordinate system (whether "right" is positive or negative, in other words), but the arrow will always point down the ramp.

Hope that helps,
Mr. Dychko

Giancoli 7th Edition, Chapter 3, Problem 37

By suriyak786 on Sat, 03/11/2017 - 20:04

For part b why didn't we also find out Vx. Why did we find Vy

By Mr. Dychko on Sun, 03/12/2017 - 13:17

Hi suriyak786, thank you for your question. At 3:20 we made use of $v_x$ to create an expression for $t$ in terms of things that we know, such as $x$ and $\theta$, and the one thing we don't know, $v$. That was then substituted into the vertical displacement formula, which only has $v_y$ in it since only the vertical component of velocity affects the vertical displacement of the car. Does that help?

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 13, Problem 10

By rdattafl on Sat, 03/11/2017 - 18:00

Mr. Dychko,

The range of temperatures stated in the problem are from -30 degrees C to 50 degrees C. Thus, shouldn't the change in temperature, or delta(T), be 80 degrees C?
With this value of delta(T), the width of the expansion cracks become 12 mm.

By Mr. Dychko on Sun, 03/12/2017 - 13:49

Hi rdattafi, thank you for your question. It turns out that this question calls for a very careful reading since the way it's worded is a bit sneaky. It mentions wanting to know the expansion gap needed at $15^\circ\textrm{ C}$. This expansion gap is needed to deal with the concrete slabs becoming bigger as they get hotter on hot days. The temperatures cooler than $15^\circ\textrm{ C}$ don't concern us since the slabs will only get smaller as they get cooler. $-30^\circ\textrm{ C}$ is a red herring and we can ignore it. The question is not asking "by how much will the concrete expand when changing temperature from $-30^\circ\textrm{ C}$ to $50^\circ\textrm{ C}$". Rather, it's asking for how much of a gap should be left between slabs that are $15^\circ\textrm{ C}$ so that they don't touch and then buckle when they reach $50^\circ\textrm{ C}$.

Best wishes with your studies,
Mr. Dychko

Giancoli 7th Edition, Chapter 2, Problem 36

By suriyak786 on Fri, 03/10/2017 - 21:59

why is there Vmax? Can we just write normal V

By Mr. Dychko on Sun, 03/12/2017 - 13:39

Hi suriyak786, yes, you could write $v$ instead of $v_{max}$, just so long as there's an understanding of what $v$ (or $v_{max})$ is: it's the speed the car reaches after the period of acceleration.

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 3, Problem 37

By idan on Fri, 03/03/2017 - 13:55

I originally tried using the Range formula which gave me a slightly wrong answer. I'm assuming it's not applicable in this situation because of the 1.5 m drop?

By Mr. Dychko on Sun, 03/12/2017 - 12:53

Hi idan, yes you're exactly right. The range formula was derived using the assumption that the final and initial heights are the same. Since that's not the case here, as you say, the range formula doesn't apply.

Cheers,
Mr. Dychko

Giancoli 7th Edition, Chapter 3, Problem 27

By shichunye on Tue, 02/28/2017 - 13:37

1. why is it 36.6m/s(sin42)-(9.80)(1.50), doesnt the equation say V0sin+ayt?

2. how did you obtain 27.113m/s^2 for Vx?

Giancoli 7th Edition, Chapter 19, Problem 10

By elisabeth.burnor on Tue, 02/28/2017 - 10:24

During my free trial, I had no trouble viewing these videos, but now I cannot get any of the explanation videos to load on my PC Chrome Browser.

By Mr. Dychko on Tue, 02/28/2017 - 11:31

Hi elisabeth.burnor, thank you very much for reporting this. There is a temporary issue currently with the system, called Amazon S3, that hosts the videos and thumbnail images. I'm keeping an eye on https://status.aws.amazon.com/, and I would imagine it won't take them too long to sort thing out. When they do, Giancoli Answers will be back to normal.

Best wishes,
Mr. Dychko

Giancoli 7th Edition, Chapter 2, Problem 42

By idan on Sun, 02/26/2017 - 14:42

By dividing by "t" instead of factoring aren't we losing a physically meaningful answer t=0 (the time the ball was hit)?

By Mr. Dychko on Mon, 02/27/2017 - 00:40

Hi idan, you are making a valid point. You have a sharp eye! Mathematically, it would be more correct to factor and then find the roots of the resulting equation, thereby discovering the answer $t=0$. However, since this question is asking for the "time in the air", $t=0$, as you know, is an extraneous solution, so the effort of finding the solution $t=0$ and then discarding it as extraneous isn't worth the effort. The technique in the video is perfectly fine for a physics class, since math is just a tool for finding solutions to the physical problem, which in this case is "how much time is the ball in the air", not "at what times is the ball at height zero". For the latter, one must follow the technique you suggest by factoring since $t=0$ would be a non-trivial solution, but as it is, $t=0$ is not a solution for determining the "time in the air" problem. In a math class there's no question that to answer "solve this equation", one must factor and include $t=0$ as a solution.

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 2, Problem 26

By idan on Sun, 02/26/2017 - 12:28

I noticed you called "d" distance, but aren't all the kinematic equations actually only referring to displacement?

By Mr. Dychko on Mon, 02/27/2017 - 00:30

Hi idan, you're quite right that the kinematics equations contain displacement, rather than distance. Often times, such as in this question, the two quantities have the same magnitude. The distance and the magnitude of the displacement are the same, in other words, but your comment and awareness that d is in fact always displacement is definitely important in questions for which that's not the case. Technically it's a mistake for me to refer to the factor d as a distance, but I fear that I probably do it quite often, making the distinction mostly just where it's necessary to do so. Hopefully this isn't too confusing!

All the best,
Mr. Dychko

Giancoli 6th Edition, Chapter 8, Problem 64

By thesouthportschool on Sat, 02/18/2017 - 15:37

Is the answer not 453.6926 J or 5*10^2 J, I copied the exact calculations from your final answer of part A (3:49) and did not get 5*10^16 J.

By thesouthportschool on Sat, 02/18/2017 - 15:43

And I also you changed the angular velocity from 3.333*10^-4 to 3.333*10^-11 when doing the calculations to work out the KE

By Mr. Dychko on Wed, 02/22/2017 - 00:37

Hello student from thesoutportschool, thank you for spotting the error. At 3:23 I mention correctly that the angular velocity is $3.333 \times 10^{-4}$, whereas at 3:46 I incorrectly wrote $3.333 \times 10^{-11}$. My calculations were nevertheless done using $3.333 \times 10^{-4}$, so the results are correct. I'll put a note above the video to use $3.333 \times 10^{-4}$ for the angular velocity.

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 12, Problem 15

By margolinw on Fri, 02/17/2017 - 08:04

What is your rationale for deciding to multiply by (r^2/10^beta/10) in part B?

By Mr. Dychko on Wed, 02/22/2017 - 00:14

Hi margolinw, I think you're referring to the algebra step at 3:36, right? Multiplying by $\dfrac{r^2}{10^{\beta/10}}$ is a way to isolate the unknown $r$ by itself on one side of the equation. Multiplying by $r^2$ cancels it on the left side, whereas dividing by $10^{\beta/10}$ cancels that term on the right side, and this results in $r^2$ by itself on the right side. Mission accomplished! Then I switched the sides around and took the square root of $r^2$.

Hope that helps,
Mr. Dychko

Giancoli 7th Edition, Chapter 16, Problem 5

By julia.wolfe on Sun, 02/12/2017 - 12:41

How are you supposed to know that the comb gained (rather than lost) electrons? I know that the chapter mentions that plastic items develop negative charges when rubbed, but I thought the problem would specify that the charge acquired was -3 microcoulombs if the comb had gained (vs. lost) electrons?
Thanks,
Julia

By Mr. Dychko on Wed, 02/22/2017 - 00:24

Hi julia.wolfe,

I think the question would be worded better if it said $-3.0 \textrm{ } \mu \textrm{C}$. I don't think you're meant to know that plastic gains electrons, although in this case, without the question specifying a negative charge, knowing the properties of plastic is the only way to know. I would presume that in any real test question that the negative sign would be given.

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 11, Problem 28

By donnarrnc on Fri, 02/10/2017 - 18:10

Is this calculation correct? It shows 4/pi^2 on the calculator display.

By Mr. Dychko on Fri, 02/10/2017 - 23:42

Hi donnarrnc,

Thanks for the question. Yes, things are all good here. The calculator display shows /4/pi^2, which is another way of saying what you might have been expecting: /(4pi^2).

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 18, Problem 36

By tuh20232 on Thu, 02/09/2017 - 15:50

I got confused with the P= 75 w and the P = the power consumed would you please explain the difference? Thank you

By Mr. Dychko on Fri, 02/10/2017 - 23:48

Hi tuh20232, thanks for the question. P = the power consumed, always. On the packaging for a bulb, the manufacturer will give the answer to the power calculation, assuming a particular voltage. For a bulb sold in the US, the manufacturer will assume 120 V in the calculation for the bulb power, where for a bulb sold in the EU, the manufacturer will assume 240 V.

P = the power consumed, whereas the P = ## rating for a bulb depends on where the bulb is sold.

Hope that helps,
Mr. Dychko

Giancoli 7th Edition, Chapter 19, Problem 18

By saie.joshi on Sun, 02/05/2017 - 11:21

do the nodes initially not mean anything then? I thought the nodes would make the resistors not in parallel or series.

By Mr. Dychko on Sun, 02/05/2017 - 12:24

Hi saie.joshi,

Thanks for the question. This diagram is a bit tricky since some resistors are in series, whereas others are in parallel. We break it down step by step. If current reaching a node has a choice between multiple paths, then the resistors along the two (or more) paths are in parallel. If current reaching a node has only one option, and that is to proceed through another resistor, then that next resistor is in series. Does that help?

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 10, Problem 22

By rdattafl on Thu, 02/02/2017 - 11:00

In the video, you wrote 1.29 * 10^5 Pa, but in the main answer section, you wrote 1.20 * 10^5 Pa. Merely for clarity, can you correct the second answer?

By Mr. Dychko on Sun, 02/05/2017 - 12:25

Thank you very much for spotting that rdattafl. I've updated the quick answer.

Giancoli 7th Edition, Chapter 2, Problem 30

By m_iqbal on Tue, 01/31/2017 - 19:20

Hi,
I have a question that if the object has constant negative velocity, will the object speed up or slow down?

By Mr. Dychko on Sun, 02/05/2017 - 12:35

Hi m_iqbal, thanks for the question. The negative sign for velocity indicates direction only. It says nothing about whether the object is speeding up or slowing down. A negative velocity means only that the object is moving in the negative direction, which is typically to the left, although a person solving a problem is free to redefine the coordinate system by saying "I feel like making the negative direction to the right". It's a bit unusual to say "right is negative", but you do need to be prepared for this idea since occasionally I do say "let down be the positive direction", rather than the traditional "up is positive", when doing so simplifies the mathematics by avoiding so many negative signs in a formula.

A change in speed is given by acceleration, so your next question might be: does a negative acceleration mean the object is speeding up or slowing down. The answer here depends on whether the object currently has a negative velocity, or not. If the acceleration has the same sign as the velocity, it's speeding up. An acceleration with an opposite sign to the velocity means the object is slowing down. So, running through the list of possibilities: +a & +v => speeding up, -a & +v => slowing down, +a & -v => slowing down, -a & -v => speeding up. The last scenario is probably the one you'll want to take note of, where a negative acceleration means the object is speeding up when it already has negative velocity.

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 9, Problem 48

By jsh2672 on Tue, 01/31/2017 - 15:56

Why are you multiplying by 1/100? Shouldn't you be multiplying by 100 to get the percentage?

By Mr. Dychko on Sun, 02/05/2017 - 12:43

Thank you very much jsh2672 for noticing that! I've corrected the quick answer to show the result after multiplying by 100, and put a note about the error in the video.

Best wishes,
Mr. Dychko