Comments

Giancoli 7th Edition, Chapter 8, Problem 67

By jessicaannehoch on Mon, 12/07/2015 - 16:58

Hello Mr. Dychko,
I just would like to point out that while you entered the correct formula into the calculator, in the written equation you forgot to square 3.0m in the last part of part "B".

Thank you for your work!

Giancoli 7th Edition, Chapter 5, Problem 23

By 1609702237 on Sat, 12/05/2015 - 06:45

Good job

Giancoli 7th Edition, Chapter 12, Problem 37

By merkinthedark on Thu, 12/03/2015 - 17:46

I believe that part B ask for the new fundamental frequency having the pipe filled with Helium, would you mind double checking.

By Mr. Dychko on Sun, 12/06/2015 - 21:38

Hi markinthedark, yes, nice catch. I recalculated the length with helium rather than recalculating the fundamental frequency. To find the fundamental frequency of an open ended tube, with the length calculated in part (a) as $0.58537 \textrm{ m}$, and the speed of sound in helium as $1005 \textrm{ m/s}$ (this is a number I just had to look up), the fundamental frequency is $f_1 = \dfrac{v}{2l} $$= \dfrac{1005 \textrm{ m/s}}{2 * 0.58537 \textrm{ m}}$$=858.43 \textrm{ Hz} = 858 \textrm{ Hz}$. I've flagged the video to update it one day, and I'll post a note in the quick answer section. Let me know if you still have any questions about this problem.

Hope this helps,
Mr. Dychko

Giancoli 6th Edition, Chapter 5, Problem 47

By devon.gibson2016 on Thu, 12/03/2015 - 01:54

Where did you get 6.38 x 10^6 since the radius of the Earth is 6.38 x 10^3?

By Mr. Dychko on Sun, 12/06/2015 - 21:23

Hi devon, the difference is in the units. The Earth's radius is $6.38 \times 10^6 \textrm{ m}$, or $6.38 \times 10^3 \textrm{ km}$.

Cheers,
Mr. Dychko

Giancoli 7th Edition, Chapter 13, Problem 8

By Mr. Dychko on Tue, 12/01/2015 - 20:58

There isn't really an explicit equation mentioned in the textbook for this. Notice that the equation resembles an equation for a straight line $y = mx + b$. This means if you know the "y" intercept (zero in this case), then you can add that to the quantity that varies (the length of the alcohol column in this case) times the amount by which the temperature changes per change in length. That's how I think of it to myself, but that might not make a lot of sense as an explanation... Here's another take:
Consider the first line where we're calculating $\dfrac{\Delta T}{\Delta l}$. I wrote this with numbers, but it could have been written with variables. It would look like this: $\dfrac{\Delta T}{\Delta l} = \dfrac{T - T_o}{l - l_o}$. Multiply both sides by $l-l_o$ gives $\dfrac{\Delta T}{\Delta l}(l-l_o)$. Then add $T_o$ to both sides, and there you have it: the equation of interest. Hopefully one of those explanations helps.

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 13, Problem 8

By Mr. Dychko on Tue, 12/01/2015 - 20:51

A student wrote: I am confused as to how you derived the equation after finding the ratio. Thank you very much.

Giancoli 7th Edition, Chapter 8, Problem 13

By Mr. Dychko on Tue, 12/01/2015 - 20:37

$a_R = \omega^2 r$ comes from equation 8-6 on page 202. It's a different way of writing centripetal acceleration. The other way of writing it is $a_R = \dfrac{v^2}{r}$, but with a substitution for $v$ as $v = \dfrac{\omega}{r}$, you end up with $a_R = \omega^2 r$.

Hope that helps,
Mr. Dychko

Giancoli 7th Edition, Chapter 8, Problem 13

By Mr. Dychko on Tue, 12/01/2015 - 20:32

A student wrote: "I was looking at your solution for problem 8 in chapter 13. I am confused as
to how you derived the equation that you used. I could find no reference of
it in the book or other resources. Thank you for your help."

Giancoli 7th Edition, Chapter 10, Problem 14

By hannahsmidnight on Tue, 12/01/2015 - 15:43

Where did the pi come from?

By Mr. Dychko on Tue, 12/01/2015 - 20:29

Hi hannahsmidnight, we need to calculate the area of the circular piston, so that's where $\pi$ comes from: the area of a circle is $\pi (\dfrac{d}{2})^2$.

Best wishes,
Mr. Dychko

Giancoli 7th Edition, Chapter 24, Problem 53

By spanda2u on Sun, 11/29/2015 - 15:25

Hello,
there is a second question in this problem: "What if the glass is to appear dark?"
I am curious to see that solution.
thanks,
Spanda

By Mr. Dychko on Sun, 11/29/2015 - 20:40

Hi spanda2u, well, in a pinch I can give a written solution: the minimum separation would be zero, or if you want a non-zero answer then the first separation for destructive interference would be $225 \textrm{ nm}$ ($230 \textrm{ nm}$ with sig. figs. considered). It's important to know why "zero" creates destructive interference ("zero" isn't really zero, but rather a microscopic separation in which a very tiny air gap still exists despite the two pieces of glass touching). At "zero" separation, the ray reflecting from the bottom piece of glass undergoes a $\dfrac{1}{2}\lambda$ phase shift since it's initially in a medium of low index of refraction (a microscopic layer of air) and then reflecting off a medium of high index of refraction (bottom piece of glass). This business about a ray undergoing $\dfrac{1}{2}\lambda$ phase shift when reflecting from a medium of higher index of refraction that the medium in which it is initially is just a rule that has to be memorized. Since the ray reflected from the bottom piece of glass is shifted $\dfrac{1}{2}\lambda$, whereas the ray reflected from the bottom of the top piece of glass is not phase shifted at all (since it's initially in a high index medium, glass, reflecting off a low index medium of air), and there is no additional path length travelled by the bottom ray (or a negligible path length to be precise), these two rays are $\dfrac{1}{2}\lambda$ out of phase with respect to each other and so they destructively interfere (dark!). The next dark separation would occur when the separation between the pieces of glass results in an additional path length travelled by the bottom ray to be $\lambda$. The bottom ray travels the separation twice (once down, then again back up), so the separation that causes an additional path length of $\lambda$ is $\dfrac{1}{2}\lambda$. This makes the first non-zero dark separation equal to $\dfrac{1}{2}450 \textrm{ nm} = 225 \textrm{ nm} \simeq 230 \textrm{ nm}$.

Hope that helps,
Mr. Dychko

By spanda2u on Wed, 12/02/2015 - 20:47

Thank you.

Giancoli 7th Edition, Chapter 8, Problem 70

By aheumangutman on Sat, 11/28/2015 - 09:53

Hi Mr. Dychko,
For solving for the inertia of the horizontal rod, why do we use the I = (1/12 M L^2) equation and not the I = (1/3 M L^2) equation?
Thanks!

By Mr. Dychko on Sun, 11/29/2015 - 20:08

Hi aheumangutman, the position of the axis of rotation affects which equation to use. The textbook has some good illustrations on pg. 210 in Figure 8-20. This particular question says the axis of rotation is positioned at the center of the rod, which is what makes the equation $I=\dfrac{1}{12}ML^2$.

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 8, Problem 26

By brf15 on Sat, 11/28/2015 - 00:15

ur the best

Giancoli 7th Edition, Chapter 8, Problem 1

By brf15 on Sat, 11/28/2015 - 00:00

GOOD JOB MR DITCHKO!!!!

Giancoli 7th Edition, Chapter 6, Problem 42

By tmesser on Tue, 11/24/2015 - 12:32

I found the distance using d = vf^2-vi^2/2a and I got 8.9 m. When I substitute that into the 1/2kx^2=1/2mv^2 equation I get a k equal to 10500 N/m. What am I doing wrong?

By Mr. Dychko on Tue, 11/24/2015 - 20:17

Hi tmesser, good question. The tricky thing here is that acceleration is not constant. All of the kinematics formulas, like $d = \dfrac{v_f^2-v_i^2}{2a}$, assume acceleration is constant, so it can't be used in this question. Only conservation of energy can be used, plus the Hook's law for a spring, and $F_{net}=ma$, all three of which don't assume constant acceleration. The acceleration isn't constant since the force applied by the spring on the car changes depending on how much the spring is compressed. In the beginning the spring pushes only lightly, so the car's deceleration is small, but then as the spring is compressed it pushes harder on the car thereby increasing the deceleration.

Hope that helps,
Mr. Dychko

Giancoli 7th Edition, Chapter 4, Problem 51

By sueqrahn on Tue, 11/24/2015 - 04:09

Hey Mr. Giancoli,

Thank you so much for your videos! They are really helpful! Just so you know, this problem asked for the retarding force so the units should be Newtons, not kilograms.

By Mr. Dychko on Tue, 11/24/2015 - 20:10

Hi sueqrahn, thanks for noticing that typo. You're quite right that it should be written as Newtons and not kilograms.

Keep up the good work,
Mr. Dychko

Giancoli 6th Edition, Chapter 4, Problem 37

By rinne13 on Sun, 11/22/2015 - 20:30

Dear Mr.Dychko thank you so much for your help :)
Best regards ,
Yana

By Mr. Dychko on Mon, 11/23/2015 - 09:36

You're welcome! =)

Giancoli 7th Edition, Chapter 24, Problem 18

By spanda2u on Sat, 11/21/2015 - 19:33

oh, i thought red light has longer wavelengths and is slower than the shorter wavelengths of blue!

By Mr. Dychko on Mon, 11/23/2015 - 09:43

Hi spanda2u, you're quite right that red has a longer wavelength, but indeed, as shown in the video, "chromatic dispersion" for most materials usually means the long wavelength colors travel faster. Here's the Wikipedia article (it's rather densely written unfortunately).

Giancoli 7th Edition, Chapter 8, Problem 6

By rafih6 on Wed, 11/18/2015 - 16:24

Nice dude

Giancoli 7th Edition, Chapter 2, Problem 15

By rafih6 on Tue, 11/17/2015 - 01:23

I dont understand what you did in this video can you please explain it?

By Mr. Dychko on Tue, 11/17/2015 - 20:56

Hi rafih6, I'll give it a try, but if this doesn't help then try again with a more specific question. The general idea in this question is that you know the total time for two different things to finish travelling the same distance. The distance is the separation between the bowler and the pin. The first thing travelling is the ball towards the pin, and the second thing travelling is the sound of the impact which takes some time to travel from the point of contact with the pin back to the bowler. I set up three equations to summarize these points, then do some algebra by doing substitution between equations to combine them and eliminate unknowns.

Maybe that helps?
Mr. Dychko

Giancoli 7th Edition, Chapter 10, Problem 7

By merkinthedark on Mon, 11/16/2015 - 13:36

can you explain why we cube 1220x10^3m when in the chart given its already in kilometers

By Mr. Dychko on Tue, 11/17/2015 - 20:50

Hi merkinthedark, thanks for the question. The inner core radius needs to be cubed in order to find the volume of the inner core. The volume formula is $V = \dfrac{4}{3}\pi r^3$, with the radius cubed. This might not be your question however.... maybe you're wondering if we could use the radius in kilometers instead of meters, and that when using kilometers there's no need to cube the radius? It doesn't work like that since the formula still requires the radius to be cubed regardless of what the units are. Cubing kilometers will give a different answer than cubing meters, so there is a difference, and as usual formulas expect "meters, kilograms, seconds" most of the time.

Hope that helps,
Mr. Dychko

Giancoli 7th Edition, Chapter 17, Problem 67

By bonafideluy on Thu, 11/12/2015 - 07:53

is there any solutions to misconception questions?

By Mr. Dychko on Thu, 11/12/2015 - 22:12

Hi bonafideluy, it would be enormously time consuming to also answer all the other categories of problems, like "General Problems", and "Misconceptual Questions" so I'm limiting coverage only to the regular "Problems", of which there are still more than 1,700 solutions, and hopefully that's helpful enough for most students.

All the best,
Mr. Dychko

By bonafideluy on Thu, 11/12/2015 - 07:54

Are they listed in the back of the test book or are they just not as important?

Giancoli 6th Edition, Chapter 4, Problem 44

By devon.gibson2016 on Wed, 11/11/2015 - 10:33

Why is the friction force not in the negative direction?

By Mr. Dychko on Wed, 11/11/2015 - 20:14

Hi devon.gibson2016, yes, it's typical that friction is often in the negative direction since kinetic friction, such as air friction, is often what's slows things down. In this case air friction would be in the opposite direction to the direction of motion, and positive would be taken as the direction of motion, making friction pointing in the negative direction. In this particular question, however, the friction we're speaking of is causing the motion. In this question we're dealing with static friction, which is the friction force exerted on the drag racer tires by the road. The drag racer tires push backwards on the road, and the road in turn pushes forward on the tires. This forward static friction force is the Newton's 3rd law counterpart to the force exerted backward on the road by the tires. This static friction force is in the same direction as the positive direction of motion, so it's positive.

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 5, Problem 25

By tmesser on Fri, 11/06/2015 - 07:35

Why can you not get the velocity by saying it is 1/2 of the final velocity (i.e., 37.5 m/sec)? Why is this different from the velocity found using the equation?

By Mr. Dychko on Fri, 11/06/2015 - 09:00

Hi tmesser, thanks for the question. First, let's see if I understand why you're considering $v = \dfrac{v_f}{2}$. Does what you're thinking come from the equation $v = at$? If so, it's correct to say that after half the time, you would expect to reach half the velocity. However, that's after half the time has passed, whereas this question is asking for the velocity after half the distance has been covered. Since the car is accelerating, it will reach it's "half way velocity" at a time before the "half way distance", which you noticed with your calculation giving $37.5 \textrm{ m/s}$, whereas the velocity at the half-way position is $53 \textrm{ m/s}$.

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 4, Problem 26

By msmgl on Tue, 11/03/2015 - 17:26

I still don't understand why FxA = FxB in this problem. With similar problems in chapter 3 the vector components were never equal unless the angles were equal. Please explain what in this problem specifically leads to knowing that FxA = FxB. Thank you!

By Mr. Dychko on Tue, 11/03/2015 - 20:14

Hi msmgel, thanks for the question. This problem says the sum of the forces is straight up. Put in other words, this means the net force is straight up. The only way for that to be possible is if the horizontal components of $F_A$ and $F_B$ are of equal magnitude in opposite directions.

Hope that helps,
Mr. Dychko

Giancoli 7th Edition, Chapter 6, Problem 39

By nschidlowsky on Sat, 10/31/2015 - 15:39

I understand now. Thank you for your help!

Giancoli 7th Edition, Chapter 6, Problem 39

By nschidlowsky on Fri, 10/30/2015 - 14:51

Why is there no gravitational potential energy when the spring is compressed?

By Mr. Dychko on Fri, 10/30/2015 - 20:06

Hi nschidlowsky, thanks for the question. You mention referring to part b), but the concept is also applies to part a) as well. Let's clarify one thing about gravitational potential energy: an absolute value never exists, but rather you can only ever calculate a difference in gravitational potential energy between two heights. Instead of saying "Gravitational Potential Energy", we should always be saying "Difference in Gravitational Potential Energy between the final position and the reference level", but that's too much to say, so we say "Gravitational Potential Energy" for short. When using the typical formula $E_p = mgh$, where $h$ is the height above the ground, this is actually a calculation of the difference in potential energy between the ground (the ground is called the reference level in this case) and the position $h$ meters above the ground. In this particular video, instead of comparing to the ground, I calculated the gravitational potential between the final height and the initial height of the ball on the compressed spring. The initial position of the ball on the compressed spring is the reference level. That means the height of the ball initially is zero meters above the reference level, so $h$ is zero, and so there's no "gravitational potential energy".

Taking the reference level to be at the initial height of the compressed spring is strategic since it eliminates some variables: it means there is no gravitational potential energy term on the "initial" side of the energy conservation equation. There only exists elastic potential energy.

Hope this helps,
Mr. Dychko

By nschidlowsky on Fri, 10/30/2015 - 14:52

Just to clarify, I'm referring to part (b). Thanks!

Giancoli 7th Edition, Chapter 4, Problem 59

By gctso on Sat, 10/17/2015 - 19:17

Oh, I'm sorry. I misread the question as the height of the crate before release is 8.15m.

Giancoli 7th Edition, Chapter 4, Problem 59

By gctso on Sat, 10/17/2015 - 19:14

Why don't you use the distance of the inclined plane (hypotenuse) in part b? Shouldn't we use acceleration (which I think we made it to a horizontal component) and the distance of the same plane?

Giancoli 7th Edition, Chapter 6, Problem 54

By p.zeppelin on Wed, 10/14/2015 - 22:41

Why did you multiply everything by 2?

By Mr. Dychko on Thu, 10/15/2015 - 09:32

@p.zepplin, multiplying by "2" is an algebra step to remove the $\dfrac{1}{2}$ from the left side in order to isolate $v_f$.

Cheers,
Mr. Dychko

Giancoli 7th Edition, Chapter 6, Problem 40

By p.zeppelin on Wed, 10/14/2015 - 17:02

I'm confused. I did the problem, but for each point my initial velocity was obtained from the Final velocity of the previous point. For example, the final velocity I obtained for point 2 was 25m/s, which is correct. But for point 3 I assumed the initial velocity would be the final velocity of point 2 (25m/s) and assumed the initial height = 0 while the final height = 26m. Did the same process for the next point where my Vi=2.4m/s (obtained from Vf of point 3) and my Initial height=12 (subtracted 26 from 14). I thought your point of reference was the lowest point, which I though in point 4 would be 0 and the highest 26-14? Sorry I'm just really confused. Cab you tell me why my process is wrong? Thank you!

By Mr. Dychko on Thu, 10/15/2015 - 09:30

Hi p.zepplin, the important thing to be mindful of with conservation of energy questions is to choose one point of reference for the entire question, and don't change that reference point. You can choose any reference point you want, but after making the choice you can't change it.

Just to make this point clear, consider a book on a table top 3m above the floor. If you move it horizontally, the potential energy with respect to the floor doesn't change because its height hasn't changed. However, if you change your reference point to be the table top after moving the book horizontally then it would suggest the potential energy became zero after moving the book horizontally, which is not a correct conclusion and illustrates the problem of changing your reference point during a question.

I usually choose a reference point which is at the lowest since otherwise you'll get a negative potential energy, which I find confusing. The lowest point in this question is point "2". All heights are measured with respect the the reference level at point "2". Take another stab at this problem now that you've read this and let me know how it goes.

Cheers,
Mr. Dychko