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Sample solution
Giancoli 7th Edition, Chapter 5, Problem 12
(4:49)
Giancoli 7th Edition, Chapter 2, Problem 41
By sheumangutman on Tue, 09/27/2016 - 06:57Why cant the formula v1=vo+At be used for this problem assuming v1=0?
Giancoli 7th Edition, Chapter 3, Problem 42
By julia.wolfe on Mon, 09/26/2016 - 12:54At 1:45 why is the x component 4.24 + 1.7 shown as equaling 2.124? Shouldn't it be 5.94 since the stair climb and boat direction are the same?
Oops nevermind! I had copied some numbers wrong!
Giancoli 7th Edition, Chapter 17, Problem 39
By sanghoonwilliamk on Sat, 09/24/2016 - 21:45i think this should be .63 microfarads
Thanks sanghoonwilliamk, you're totally right. $0.63 \textrm{ }\mu\textrm{F}$ is correct. While the working is correct in the video, I made a careless error turning the scientific notation into a decimal, and I put a note about this in the quick answer.
Thanks for the sharp eye!
Mr. Dychko
Okay gotcha~ thanks for the clarification.
Giancoli 7th Edition, Chapter 3, Problem 18
By sheumangutman on Sat, 09/24/2016 - 19:00Can you please explain why the acceleration in this question is assumed to be positive while it was negative in question #17? Thanks.
Hi sheumangutman,
Thanks for the good question. Choosing whether the upward or downward direction is positive is called "choosing a coordinate system". Choosing a coordinate system is something you do for every solution, and the choice is a matter of personal preference. The only rule is that you're consistent, which means if you choose the downward acceleration to be positive, then it means downward velocity also must be positive, and the same goes for downward displacement. I choose down to be the positive direction since it meant there were no negative signs in my work, which I think looks just a little cleaner. The question is asking only for distances anyhow, which are positive, so I might as well make my calculations result in the final answer, rather than finding a negative downward displacement (which would have happened if I made the more conventional choice of upward being positive for the coordinate system) and then taking it's "magnitude" to get the final positive answer for height of the cliff.
Hope that helps,
Mr. Dychko
Giancoli 7th Edition, Chapter 2, Problem 42
By lcbishop on Sat, 09/24/2016 - 12:13For part c) What specific factors make this an estimate?
Hi Icbishop, did you post this question on the wrong video? I don't see a part c) for this problem.
Cheers,
Mr. Dychko
Well I double checked and I do indeed have a part c on #42 chapter 2. It is the 7th Edition as well. The question also matches up with the solution provided.
Giancoli 7th Edition, Chapter 2, Problem 42
By sheumangutman on Sat, 09/24/2016 - 10:44For the 2nd part of the question, why can't you use the formula Vf=Vo+2at?
Thanks.
Hi sheumangutman,
Thanks for the question. I think you mean to suggest $v_f = v_o + at$, correct? That approach would be equally fine, provided you establish that $v_f = -v_o$, which is true since it returns back to the original launch height. This means your suggested formula, with a substitution for $v_f$, would become $-v_o = v_o + at$ which rearranges to $t = \dfrac{-2v_o}{a}$ which is the same formula shown at 1:20 in the video. You have to be a little cautious with your suggested formula, however, since it works only when you can tell how the final velocity compares with the initial velocity. $v_f = -v_o$ only when something returns to it's original height. If something fell into a hole after being launched upwards, meaning it returns to a different height, you would be better off using the $d=v_ot + \dfrac{1}{2}at^2$ formula. For this problem, however, the formula you suggest would be just fine.
All the best,
Mr. Dychko
Giancoli 7th Edition, Chapter 4, Problem 32
By zzdawit@yahoo.com on Thu, 09/22/2016 - 22:04but i got a different value when I chose to use y as negative . what can I do about it ? are both yours and mine correct ?
Hi zzdawit, choosing y as negative is a valid thing to do, but it should lead to the same answer. You have to take care that you distinguish between the acceleration of the two blocks. The accelerations are not the same anymore. They have opposite signs, and I would guess that you might consider carefully examining your work to see that you took their opposite signs into account. My comment above explains a bit more about this, but let me know if you have more questions.
All the best,
Mr. Dychko
Giancoli 7th Edition, Chapter 4, Problem 13
By zzdawit@yahoo.com on Wed, 09/21/2016 - 07:55what is 9.8N/kg
Hi zzdawit, and thanks for the question. $9.8 \textrm{ N/kg}$ is an alternative way of writing what you might be used to seeing as $9.8 \textrm{ m/s}^2$. The two forms are equivalent, and the one you choose to use is just personal preference. The two forms have different names. $9.8 \textrm{ N/kg}$ is called the Gravitational field strength, whereas $9.8 \textrm{ m/s}^2$ is named the acceleration due to gravity. When solving problems that require the force due to gravity I prefer to write it as $9.8 \textrm{ N/kg}$ since it's easy for me to see the $\textrm{kg}$ cancel when multiplying it by $\textrm{kg}$, leaving $\textrm{N}$, whereas for problems involving the speed or acceleration of something falling I prefer to write $9.8 \textrm{ m/s}^2$.
To illustrate how the units are the same even though they look different, consider this (writing units here, not variables, for this "dimensional analysis"): $\textrm{N/kg} = \textrm{kg} \times \textrm{m/s}^2 / \textrm{kg} = \textrm{m/s}^2$ where the Newtons were expanded into kilograms times meters per second squared since $F = ma$.
The short answer: $9.8 \textrm{ N/kg} = 9.8 \textrm{ m/s}^2$. Use them interchangeably as you prefer.
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