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Giancoli 7th Edition, Chapter 5, Problem 31

By chaegyunkang on Tue, 09/18/2018 - 16:35

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?

Giancoli 7th Edition, Chapter 11, Problem 21

By kevin.rosario on Sat, 09/15/2018 - 23:12

I 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 - 16:18

How do we know that the initial velocity is zero? If it was given a push I thought it would be a non-zero velocity.

By Mr. Dychko on Sat, 09/15/2018 - 20:52

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 - 16:03

If the car was skidding what difference would it make to the problem? Does it change to kinetic friction when it starts to skid?

By Mr. Dychko on Sat, 09/15/2018 - 20:46

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.

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 4, Problem 55

By chaegyunkang on Sat, 09/15/2018 - 14:39

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?

By Mr. Dychko on Sat, 09/15/2018 - 20:41

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 - 18:50

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.

By Mr. Dychko on Wed, 09/05/2018 - 21:00

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 19, Problem 32

By thesouthportschool on Mon, 08/27/2018 - 18:10

this is epic

Giancoli 7th "Global" Edition, Chapter 3, Problem 16

By hugokatich on Mon, 08/20/2018 - 19:43

Also, what rule did you use for the triangle in part b? Which rule indicates that Vrx = V1x + V2x?

By Mr. Dychko on Fri, 08/24/2018 - 22:51

Thank you for the question. Adding vectors using components is one of the methods for adding vectors. The component method, which is probably the most popular method to use, involves finding the sum of the component of each vector along an axis, and then taking the total as the component of the resultant along that axis. Vrx = V1x + V2x is just an algebraic way of saying that.

All the best,
Mr. Dychko

Giancoli 7th "Global" Edition, Chapter 3, Problem 16

By hugokatich on Mon, 08/20/2018 - 19:39

Where did the answer of 62.2 units come from? Given that the square root of 90^2 - (-65^2) is equal to 111.02?

By Mr. Dychko on Fri, 08/24/2018 - 21:40

Hi hugokatich, thank you for the question. It's important to notice the minus sign in the expression there, which doesn't go away after squaring -65. Yes, -65 squared becomes positive, but the expression is for finding a leg of a right triangle, not for finding the hypotenuse, so the expression has a minus between the squared terms. $\sqrt{90^2-(-65)^2} = 62.2$.

Best wishes,
Mr. Dychko

Giancoli 7th Edition, Chapter 2, Problem 23

By umajacobshome on Thu, 08/02/2018 - 20:28

hi why in the bottom equation can you just say that average velocity is the two terms summed and divided by two if average velocity was previously defined as displacement over time? This seems like the equation for speed

By Mr. Dychko on Fri, 08/03/2018 - 11:23

Hi umajacobshome, thanks for the question. Average velocity is indeed defined as displacement over time, as you say. There is a special case though when acceleration is constant, in that the average velocity in this case can also be calculated as the sum of the initial and final velocities divided by 2. Written as an equation, that is to say $v_{avg} = \dfrac{v_i + v_f}{2}$ in the special case when acceleration doesn't change.

Hope this helps,
Mr. Dychko

Giancoli 7th Edition, Chapter 24, Problem 29

By justin.sunny.bains on Mon, 07/30/2018 - 16:06

Why is m 1.5 and not just 1?

By Mr. Dychko on Mon, 07/30/2018 - 17:58

Hi justin.sunny.bains, thanks for the questions. We're looking at a diffraction pattern here, and I would suggest referring to Figure 24-21 from the textbook to see the pattern of minima in the diffraction pattern. They occur at integer multiples of the wavelength. The twist in this question is that it tells us the position of the maxima, for which we need to deduce a formula from Figure 24.21. The maxima occur half way between the minima, so we need to add 1/2, as in $D \sin{\theta} = (m + \dfrac{1}{2})\lambda$.

Best wishes with your studies,
Mr. Dychko

Giancoli 7th Edition, Chapter 21, Problem 12

By justin.sunny.bains on Tue, 07/24/2018 - 17:23

Should the answer be negative?

By Mr. Dychko on Tue, 07/24/2018 - 20:27

Hi justin.sunny.bains, thanks for your question. I suppose the better answer would have a negative sign, but I didn't feel it was that important in this case. The negative sign is meant to indicate that the induced EMF creates an induced magnetic flux that opposes the change in magnetic flux in the solenoid. This means the induced EMF will have a direction opposite to the EMF that increased to the 5.0 A current. Outside of that, we don't really have a circuit diagram with a specific direction specified, so it's not really clear otherwise what direction a negative sign would mean. Nevertheless, given a choice between the two answers, including a negative would be a marginally better answer.

Best wishes with your studies,
Mr. Dychko

Giancoli 7th Edition, Chapter 4, Problem 22

By chaegyunkang on Sun, 07/01/2018 - 15:55

where is the normal force for part a) ?

By chaegyunkang on Sun, 07/01/2018 - 18:57

Oh, I realized that free body diagram only shows the forces acting on the object, not on other objects....

By Mr. Dychko on Mon, 07/02/2018 - 15:21

Yep, you got it! :)

Giancoli 7th Edition, Chapter 19, Problem 20

By cm2hn on Sat, 06/30/2018 - 21:13

Hello, I was trying yo watch the video, but it is unable to play. Thanks.

By Mr. Dychko on Mon, 07/02/2018 - 15:20

Hello cm2hn, have you tried this video again? In case the Javascript enabling the PDF embedded display was interfering with viewing the video, I have made the PDF into a clickable link instead. I havn't heard reports from other students about difficult viewing videos, so I suspect it's an isolated issue. Please try the video again and let me know if there are still any problems.

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 3, Problem 16

By chaegyunkang on Wed, 06/27/2018 - 20:21

I think the answer to this question is wrong. The coordinate of the resultant vector is (-142.2 , 65) which means that it is located in quadrant 2. Since the value of tangent is negative in quadrant 2 it makes sense as well. The angle is 24.56 degrees above the negative x - axis, not positive x-axis.

By Mr. Dychko on Fri, 06/29/2018 - 18:19

Hi chaegyunkang, thank you for spotting that, and for the thoughtful analysis. You're quite right! I'll change the written answer, and made a note about the video.
All the best with your studies,
Mr. Dychko

Giancoli 7th Edition, Chapter 3, Problem 48

By brysongrondel on Sat, 06/02/2018 - 20:49

If I am not familiar on the sine law and I am not being instructed in the sine/cosine law, what formulas from the book can be used to arrive at the same solutions?

By Mr. Dychko on Sat, 06/02/2018 - 21:17

So... the alternative to using the sine law here (or cosine law in other questions) is to do a lot of work resolving vectors into components, adding those components, then using the Pythagorean Theorem to turn the resultant components into the final resultant vector. The sine and cosine laws are designed to make the solution much, much quicker, and I'm quite sure the time you put into learning them will pay off by avoiding the much larger amount of time dealing with components. Places to start would be https://en.wikipedia.org/wiki/Law_of_sines and https://en.wikipedia.org/wiki/Law_of_cosines. Short term pain for long term gain!
Good luck,
Mr. Dychko

Giancoli 7th Edition, Chapter 3, Problem 47

By brysongrondel on Sat, 06/02/2018 - 19:58

I'm confused on why the last problem did not utilize the Pythagorean theorem to find the actual speed traveled by the swimmer and the actual distance traveled, but for this problem, we did both... I solved the last problem (#46) using the a^2 + b^2 approach and found the angle using the inverse tangent, but my answer was wildly different. Any insight into how to reconcile this in my brain?

By Mr. Dychko on Sat, 06/02/2018 - 21:12

Hi brysongrondel, thanks for the question. I understand that something here is confusing, but it isn't clear to me which question/solution you're comparing with, since #46 isn't really comparable to this one... I notice you asked another question, so I'll follow up there.

Giancoli 7th Edition, Chapter 2, Problem 12

By brysongrondel on Mon, 05/28/2018 - 16:30

So, first off, I have to say that I love this site so far. I am kind of struggling with the issues that arise from rounding errors. I am familiar with significant figures, but I am wondering if there is a better rule of thumb. Normally I don't round until I get the final answer, but I got problem #7 wrong. On this problem, I did all of the steps correctly, but I rounded up the times t1 and t2 to 2 sig figs and ended up getting 62 km/h. Is there anything I should know that might be helpful as I prepare for tests and future problem sets?
Thanks!

By Mr. Dychko on Wed, 05/30/2018 - 10:21

Hi brysongrondel, thank you very much for the nice feedback, and your question. The general idea is to avoid rounding until the final answer. Rounding values before a final answer is called intermediate rounding error, and causes the final answer to differ as a result of the rounding. So, generally, avoid rounding until the end. It's a topic that calls for a bit of patience since most numbers need rounding eventually, and the answer you're comparing to may have chosen a different number of digits than you, resulting in a different answer than what you obtained despite your following the best practice of keeping "lots" of digits until the end. Different people could have different opinions on what "lots" means. I would say that two additional digits, beyond those that are significant , should be kept with intermediate numbers.
Hope that helps,
Mr. Dychko

Giancoli 7th Edition, Chapter 16, Problem 29

By cm2hn on Mon, 05/28/2018 - 12:12

Hello, in the part to replace cos45 if I write sq root 2 / 2 (that is what the calculator gives me) instead of 1 / sq root of 2, at the end I guess it will give me a different answer. Right?

By Mr. Dychko on Mon, 05/28/2018 - 13:23

Hello cm2hn, thanks for your question. Well, the ultimate proof is always in trying your variation and see the effect, but I can say that $\dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}$. If you multiply $\dfrac{1}{\sqrt{2}}$ by "1", you won't change it's value, but if you make "1" look funny as, say, $\dfrac{\sqrt{2}}{\sqrt{2}}$, you'll find that it turns $\dfrac{1}{\sqrt{2}} $ into $\dfrac{\sqrt{2}}{2}$. We know that they're equivalent since you can turn one into the other by multiplying by 1.
All the best,
Mr. Dychko

Giancoli 6th Edition, Chapter 7, Problem 34

By rodri100172 on Thu, 05/17/2018 - 18:51

Thanks bro! you're a lifesaver! Say hi to Brian for me.

Giancoli 7th Edition, Chapter 3, Problem 17

By nguyen112212 on Tue, 05/15/2018 - 19:51

Hi Sr. I am wondering what happened to the other "Y" when you were solving for time. It's the part where t = square root of 2Y/AY.

By Mr. Dychko on Tue, 05/15/2018 - 21:23

Hello, thanks for the question. The "y" in the denominator is just a subscript for the "a" (acceleration), to label the acceleration as "the acceleration in the y direction". It is not a separate factor, but maybe I wrote it a bit too big. The equation is $t = \sqrt{\dfrac{2Y}{a_y}}$.

All the best,
Mr. Dychko

Giancoli 6th Edition, Chapter 5, Problem 43

By pathak_s on Sun, 05/06/2018 - 17:44

Hey why doesn't the equation for velocity (2*pi*r)/T work for this question as well?

Giancoli 7th Edition, Chapter 19, Problem 24

By aquaoasis14 on Wed, 04/18/2018 - 18:01

why did the R disappear finding the currents to the after problem?(the last problem)

Giancoli 7th Edition, Chapter 22, Problem 14

By william.g.parker on Wed, 04/04/2018 - 15:46

You wrote 0.314 as the answer for B instead of 0.341

By Mr. Dychko on Wed, 04/04/2018 - 15:52

Fixed! Thank you!

Giancoli 7th Edition, Chapter 1, Problem 28

By abgoooor201596 on Tue, 04/03/2018 - 02:10

Why your not estimating the Numbers 362 and 76 to one SIGFIG. ?

By Mr. Dychko on Tue, 04/03/2018 - 09:57

Hello, these are good questions. 365 days / year is a definition, so it's precise to three significant figures. 78 years / lifetime is an estimate, and personal opinion says that being precise to the "ones" place is appropriate, but it's just an opinion. Keep in mind that you never round numbers until the final answer. If you round before, even if a number such as 78 should be 80 since it has one significant figure, suppose, you would still nevertheless do calculations with 78, not 80. Only final answers get rounded, otherwise your calculation has an intermediate rounding error, which is to say that the calculation will be dramatically different only on account of rounding, not because any quantities have actually changed.

Hope that helps,
Mr. Dychko

By Mr. Dychko on Fri, 04/06/2018 - 19:47

Hmmm, that's a thought provoking example. The goal is always to make your calculation as precise as appropriate. Avoiding intermediate rounding error is normally part of that. However, example 1-6 illustrates that if the surrounding values have only one significant figure, as is the case with educated guesses, having only one significant figure in pi is acceptable since, in this context, greater precision isn't called for on account of the other imprecise quantities in the calculation. Think of these significant figure rules as guidelines which are open to interpretation, not as firm rules.

All the best,
Mr. Dychko

By abgoooor201596 on Fri, 04/06/2018 - 00:50

Ok That is good idea

but remember the example (Example 1–6) (How much water is in this lake?) in the book . he is rounded the value of pi to 3
is he rounded for to be 1 SIGFIG or other reason since pi is also in term of (definition) as you mentioned above ?

Thanks

Giancoli 7th Edition, Chapter 1, Problem 25

By abgoooor201596 on Tue, 04/03/2018 - 01:28

hi I've a question
How we gona apply this idea with this Number 3.0*10^0
Thank You

By Mr. Dychko on Tue, 04/03/2018 - 09:52

Hi abgoooor201596, $3.0 \times 10^0$ has 2 significant figures.

Cheers,
Mr. Dychko

Giancoli 7th Edition, Chapter 8, Problem 41

By theovilous on Sat, 03/31/2018 - 10:31

The actual marry-go-round disk is the total weight - two persons. Which is 560-50 = 510 kg. So, Wouldn't the moment of inertia have to be (2.5)^2(510/2 + 2(25)? I don't understand why you used 560 when calculating the moment of inertia.

By Mr. Dychko on Sat, 03/31/2018 - 22:06

Hi theovilous, thanks for the question. What I'm seeing is that the text says

Assume the merry-go-round is a uniform disk of radius 2.5 m and has a mass of 560 kg, and two children (each with a mass of 25 kg) sit opposite each other on the edge.

which gives 560kg as the mass of the merry-go-round disk, not including the children. Perhaps you read that differently and interpreted that the children were included in the 560kg?

Best wishes,
Mr. Dychko

By theovilous on Sun, 04/01/2018 - 10:44

I read the question wrong! thanks for pointing that out! Thank you so much as always!!

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