Post by: Cort on December 25, 2013, 06:32:03 pm
Nonetheless, I am one stubborn bastard and I'm eager to learn. If I'm looking at things the wrong way, or you've got extra ideas/information that you'll want me to be aware of, please do tell. Tell me your secrets. Oh, good buttocks, tell me all your secrets.
Working through holiday homework, this was one of the 'common' exam questions that we were given. I'm trying my best to understand the question, but it's these type of questions that makes me go blank. How would you work this out? By 'describe', in physics exam, do you always must show working out if need be? Or is it all written?
Post by: Aurelian on December 25, 2013, 07:08:39 pm
This in turn implies that the car will decelerate at the same rate in each case.
So, to start off - what is the definition of acceleration? And, given this, what graphical feature will acceleration correspond to on velocity vs. time graphs, as we have here? (NB: We actually have a speed vs. time graph here, but I won't distinguish between velocity and speed for the time being).
EDIT: Also, "describe" in the context of VCE physics generally means words, but often it will be helpful/necessary to include a little bit of mathematics/working (that's not really the case here though).
Post by: Cort on December 25, 2013, 08:47:43 pm
Okay, so, to begin, we're told that in each of the two situations the same car is involved and the same breaking force is used.
This in turn implies that the car will decelerate at the same rate in each case.
So, to start off - what is the definition of acceleration? And, given this, what graphical feature will acceleration correspond to on velocity vs. time graphs, as we have here? (NB: We actually have a speed vs. time graph here, but I won't distinguish between velocity and speed for the time being).
EDIT: Also, "describe" in the context of VCE physics generally means words, but often it will be helpful/necessary to include a little bit of mathematics/working (that's not really the case here though).
Right, the gradient of the breaking force would be ~-33. So if that's the same as the first, does that mean the line would be just on top of the other? Or something like that.
Post by: Aurelian on December 25, 2013, 09:01:14 pm
Right, the gradient of the breaking force would be ~-33. So if that's the same as the first, does that mean the line would be just on top of the other? Or something like that.
On the right track! =)
To be clear, it's the gradient of the sloped part of the graph we're considering, rather than the gradient "of the breaking force", and the magnitude of this gradient represents the magnitude of the deceleration of the car in the first situation.
As discussed above, this deceleration is constant in both situations. That means that the gradient of the sloped part of the speed vs. time graph in each situation will also be the same.
This in turn means that we expect the sloped part of the new graph (which again corresponds to the deceleration of the car) to either be on top of that of the first graph or otherwise parallel.
Now, in the situation for which we're asked to draw the speed vs. time graph, we start at a slightly higher speed than in the situation whose graph we're already given - 65 km/h compared to 60 km/h. If we assume that deceleration begins at the same time in both situations, what will the new graph look like?
(Note, by the way, that we don't actually have to calculate the gradient to answer this question).
Post by: Cort on December 25, 2013, 09:56:10 pm
On the right track! =)
To be clear, it's the gradient of the sloped part of the graph we're considering, rather than the gradient "of the breaking force", and the magnitude of this gradient represents the magnitude of the deceleration of the car in the first situation.
As discussed above, this deceleration is constant in both situations. That means that the gradient of the sloped part of the speed vs. time graph in each situation will also be the same.
This in turn means that we expect the sloped part of the new graph (which again corresponds to the deceleration of the car) to either be on top of that of the first graph or otherwise parallel.
Now, in the situation for which we're asked to draw the speed vs. time graph, we start at a slightly higher speed than in the situation whose graph we're already given - 65 km/h compared to 60 km/h. If we assume that deceleration begins at the same time in both situations, what will the new graph look like?
(Note, by the way, that we don't actually have to calculate the gradient to answer this question).
Lovely. Now I understand. Thanks boss.
Post by: Cort on January 02, 2014, 12:22:23 am
Edit: Another fact that comes into my struggle is my inability to properly remember events/reasons and little things that I should be aware of. Quite amusing how I survived History : Revolutions last year.
Post by: Kanon on January 02, 2014, 03:21:59 am
Rather than just saying wow that's cool, we can see that the numerator is often constant within any system, allowing us to derive
Which is an inverse proportionality relationship, stating that as the radius increases the less force we get. Hopefully this can help you answer/understand the concepts behind if something happens within the system
I hope this answers your questions :)
Post by: Cort on January 08, 2014, 12:09:19 am
Post by: Rod on January 08, 2014, 12:19:37 am
Never got to thank you for that, but thank you. Now there's a little less than 2 weeks to go, but my time wasted attempting to finish my speech has taken up my studying time. I'm worried... should I be doing any form of questions at all, or should I continue learning conceptually? I'm pretty buggered up in balancing enough questions done to understand to wasting time struggling on one.Less than two weeks? What date do you guys start school?
You don't need to understand everything at the moment, so don't stress. Try and go ahead as much as you can, even if you understand 50% of something, it will help because when you learn the topic at school it will come to you immediately. As for questions, I've been doing heaps, if you get stuck you should look for worked solutions on the internet or ask someone on here.
Just keep in mind that most people doing physics haven't even opened the text book, so as much as you read ahead, and have a basic understanding of what's to come, will be really beneficial for you.
Hope that helps,
Rod
Post by: PB on January 08, 2014, 11:16:46 am
My apologies if I have confused you with my advice. I am simply paraphrasing the very same tip that has been given to me by Michael Li, Istafa and Nliu (lxnl) - all of whom have gotten absolutely ridiculous study scores and 99.95s.
Ok, so you DON'T want to be FULLY understanding everything at the moment. That would be impossible unless you are some kind of savant! What you want to be doing is understanding the main concepts of the course. So, do not go into the little tidbits and minute intrinsicalities of the subject, but merely grasp the basic stuff of each topic yeah? For example - momentum and collisions, just recognise that in every isolated system, momentum is NEVER lost. Thats it. Just the barest summaries of each concept.
I highly recommend this StudyOn website thingo or the Studyon booklet which my friend lent me, which provides amazingly succinct and straight-to-the point summaries about certain concepts. For me, reading the textbook can sometimes be confusing and extremely time-consuming, having to critically sift through a whole lot of irrelevent garbage.
And by all means, if doing a few questions here and there helps crystallise a certain physics idea in your head then go ahead and do them! Don't let me or anyone else tell you otherwise.
When I said not to do questions - I simply meant not to start diving into every imaginable question to fully understand everything about the subject. You simply won't be able to get through the whole course (my mistake when I was where you are right now).
So my take home message is this - by simply confining yourself to understanding the surface concepts of the subject, you should be able 1. actually get through the whole course 2. get a big picture of the subject. Which is ideally what you want for each Math/Science subject entering into VCE.
I hope that clears things up :P If you have any other questions please feel free to ask, I absolutely hate confusing people!
Kind Regards,
PB
P.S. I know the idea of understanding only the surface of your subjects is a little wishy-washy and undefined. However, it really is up to your disgression to know how deeply you should to dig into each physics topic at this point of time :P
Post by: Cort on January 09, 2014, 05:52:17 pm
Post by: PB on January 10, 2014, 12:28:58 am
Post by: Cort on January 15, 2014, 12:07:53 am
Post by: lzxnl on January 15, 2014, 12:27:47 am
Second law isn't needed for VCE purposes at all, but one consequence is that objects move faster when closer to the sun
Third law is
I trust you know what each variable is.
Post by: Cort on January 25, 2014, 03:04:08 pm
Thanks,
Cort.
Post by: Rod on January 28, 2014, 03:39:33 am
Thank you for that Izxnl. Since I'm nearly finished doing some scouting for photonics (before I dive into a real understanding), I just wanted to ask: What is the mentality that you have/use/adopt when you're imagining and thinking about frequency waves/signals? It's a concept that I really cannot apply into any situation as of yet, because it feels so paper thin. That kinda sucks because this makes it harder for me to grasp modulation.,
Thanks,
Cort.
Think of you computer! Go to a speed check and check out your bandwith. Go to phone and see what kind of wire it is, is it copper, coaxial or optical ??? Use a shitty phone to prank call someone in Ireland, how attenuated does your signal get?
Is that better? :)
Sorry for not replying sooner, I read this but I skimmed it any didn't know you were asking a question
Rod
Post by: Cort on January 31, 2014, 12:13:13 am
I've been attempting to answer some motion basics again. Frankly, it's quite an embarrassment. I even struggle with Checkpoints motion basic. This is not good at all.
A car travelling with a constant speed of 80kmh^-1 passes a stationary motorcycle policeman. The policeman sets off in pursuit, accelerating uniformly to 80kmh^-1 in 10.0s and reaching a constant speed of 100 kmh^-1 after a further 5s. At what time will the policeman catch up with the car?
This was from Heineman Physics 12.
On another note, I think the biggest problem for me is that, I have no idea what to do with the numbers/equations after I've been given. Even though I'm aware of the problem, and what's its asking me, I can't seem to make the connection from thinking outside the box. I think an equation is an equation, and that's it. Plug and play. Perhaps I need to just 'practice more'? Any tips to break this restraint?
Post by: lzxnl on January 31, 2014, 12:45:21 am
So for your equation, for instance, the policeman will catch up with the car when it travels the same distance as the car.
The question you gave is slightly complicated. Let us break it down.
After a time t hours, the car will travel 80 km/h * t hours = 80t km.
The policeman's movement falls into three stages. When accelerating to 80 km/h in 10s or 1/360 hours, we're given an initial speed (0), final speed (80 km/h) and time (10 s). To find the displacement, we use x=t(u+v)/2 = 1/360 hours*(0+80)/2 km/hr = 1/9 km.
Then, in the second stage, the policeman goes from 80 to 100 km/h in 5 s. Using the above formula again, we have x = t/2*(u+v) = 1/2*5/3600 hours*(80+100)/2 km/hr = 450/7200 km = 45/720 km = 9/144 km = 1/16 km. (as you can see I'm doing this as I go LOL)
Note how I've written in all of the units. You can cancel them like they were regular pronumerals and this shows you the importance of converting units; if I had 1 km/h * 1 second = 1 km s/h, I would have to use the fact that 1 second is 1/3600 of an hour to get 1/3600 km.
In the third stage, the policeman travels for a time t hours - 15 seconds = (t-15/3600) hours = (t-1/240) hours. As it moves at a constant 100 km/hr speed, the distance travelled in this third stage is speed*time = (t-1/240)*100 km. In total, for a time t, the policeman travels 1/16 + 1/9 + 100(t-1/240) km. This is equal to the distance travelled by the car, so set this equal to 80t. Solve for t and you'll have the time in hours.
I get 20t = 100/240 - 25/144 = 50/120 - 25/144 = 60/144 - 25/144 = 35/144
t = 7/576 hours. Someone please check my working. In an exam, please use your calculator :P
Post by: Cort on February 01, 2014, 02:32:32 pm
First of all, learn what all of the symbols in each equation mean. For instance, in v^2 = u^2 + 2ax, v is the final velocity, u is the initial velocity, a is the acceleration and x is the displacement. This equation is only valid for constant acceleration, so a is constant. Also, x is the difference in position of the particle from when the velocity is u to when to the velocity is v. Don't just rote learn and formula bash.
So for your equation, for instance, the policeman will catch up with the car when it travels the same distance as the car.
The question you gave is slightly complicated. Let us break it down.
After a time t hours, the car will travel 80 km/h * t hours = 80t km.
The policeman's movement falls into three stages. When accelerating to 80 km/h in 10s or 1/360 hours, we're given an initial speed (0), final speed (80 km/h) and time (10 s). To find the displacement, we use x=t(u+v)/2 = 1/360 hours*(0+80)/2 km/hr = 1/9 km.
Then, in the second stage, the policeman goes from 80 to 100 km/h in 5 s. Using the above formula again, we have x = t/2*(u+v) = 1/2*5/3600 hours*(80+100)/2 km/hr = 450/7200 km = 45/720 km = 9/144 km = 1/16 km. (as you can see I'm doing this as I go LOL)
Note how I've written in all of the units. You can cancel them like they were regular pronumerals and this shows you the importance of converting units; if I had 1 km/h * 1 second = 1 km s/h, I would have to use the fact that 1 second is 1/3600 of an hour to get 1/3600 km.
In the third stage, the policeman travels for a time t hours - 15 seconds = (t-15/3600) hours = (t-1/240) hours. As it moves at a constant 100 km/hr speed, the distance travelled in this third stage is speed*time = (t-1/240)*100 km. In total, for a time t, the policeman travels 1/16 + 1/9 + 100(t-1/240) km. This is equal to the distance travelled by the car, so set this equal to 80t. Solve for t and you'll have the time in hours.
I get 20t = 100/240 - 25/144 = 50/120 - 25/144 = 60/144 - 25/144 = 35/144
t = 7/576 hours. Someone please check my working. In an exam, please use your calculator :P
I appreciate the help, but I'm afraid you'll have to drop a view IQ points whilst I'm here. I can understand the first stage and how you made the situation -- but what boggles me is why we have to find the displacement, when it just asks at the time that it's catching up to the car? Is it because we have to find time, we normally find the distance..or? Also, I lobbed "180 *(5=3600) /2" and it instead gave me 1/8 instead of 1/16km. But I apologise, beyond that I cannot understand it further, (that's my vault on my end), especially when you're talking about the third stage.
Post by: hongkyho on February 01, 2014, 03:20:01 pm
Also, I lobbed "180 *(5=3600) /2" and it instead gave me 1/8 instead of 1/16km.
You are right :)
why we have to find the displacement
Think of this situation like you would for two simultaneous questions in maths. Common knowledge in maths says that for two simultaneous equation, we need to find two variables that are present in both equations; in this case, displacement and time.
Or, just look at the question. It states: At what time will the policeman catch up with the car? The policeman will only catch up to the car once it has traveled the same distance from the starting point. Therefore, the time element we are trying to look for is specific to this certain condition.
The answer you should get is 65/2 seconds or 13/1440 hours.
Hope this helps :)
Post by: lzxnl on February 01, 2014, 04:52:01 pm
I appreciate the help, but I'm afraid you'll have to drop a view IQ points whilst I'm here. I can understand the first stage and how you made the situation -- but what boggles me is why we have to find the displacement, when it just asks at the time that it's catching up to the car? Is it because we have to find time, we normally find the distance..or? Also, I lobbed "180 *(5=3600) /2" and it instead gave me 1/8 instead of 1/16km. But I apologise, beyond that I cannot understand it further, (that's my vault on my end), especially when you're talking about the third stage.
I plugged my working into a calculator and I still seem to get 1/16. Hmm.
Note that in my working, it's 1/2*5/3600 hours*(80+100)/2 km/hr. On the numerator we have 1*5*180. On the denominator we have 2*3600*2. Dividing these through seems to get 1/16 for me.
As hongkyho said, the question is fundamentally asking for when the policeman's displacement is equal to that of the motorist's.
This question works in three stages, as there are three stages of motion. The policeman travels for 10 and 5 seconds in the first and second stages respectively. The distance travelled by the cyclist is given by 80t, where t is measured from the beginning. At the end, the policeman travels at a constant 100 km/h, so his distance travelled is 100T, where T is the time he actually travels at 100 km/h.
As the time for the motorcyclist and the policeman move the same, the t for the cyclist is the same t for the policeman. However, the policeman has spent 15 seconds in the first two stages of motion, so he really only spends t-15 seconds in the third stage of constant speed. That's why I've put in 100(t-15) is the distance travelled in the third stage.
Post by: hongkyho on February 01, 2014, 05:09:48 pm
Note that in my working, it's 1/2*5/3600 hours*(80+100)/2 km/hr.
I think you are halving it twice.
Post by: lzxnl on February 01, 2014, 05:27:42 pm
The flaws of working things out on a computer screen :P
Post by: Cort on February 09, 2014, 01:23:32 pm
I normally understand what to do if the question is the same for at least 3/4 times...however this doesn't happen. So, suggestions?
Post by: PB on February 09, 2014, 02:14:50 pm
What I find really useful is to note down a general warning in a box in your notes of that topic. Say for motion, if you keep forgetting to convert kmh-1 to ms-1. Put a WARNING sign in your motion notes and note down that you should remember to convert etc. This is so that when you are reading through notes prior to SACs, these little notices helps you cover each weakness you might have for each topic.
Post by: Rod on February 12, 2014, 03:55:45 pm
Where are you up to Cort? I'm only up to Q 21, but I've been doing more chem and health checkpoints.
We should keep each other update to keep up motivated !!
PS - PB, Lxnl - did you freaks end up completing all 783 qs? When do you think me and Cort should complete them by?
Cheers :)
Post by: lzxnl on February 12, 2014, 06:05:02 pm
There are 783 questions in the 2014 checkpoints D::::
Where are you up to Cort? I'm only up to Q 21, but I've been doing more chem and health checkpoints.
We should keep each other update to keep up motivated !!
PS - PB, Lxnl - did you freaks end up completing all 783 qs? When do you think me and Cort should complete them by?
Cheers :)
Some of the questions in Checkpoints are for a different detailed study; I didn't do those.
But yes, I did all the other questions in the physics 2013 Checkpoints that were relevant.
I think I completed the questions in SWOTVAC LOL
Post by: Cort on February 12, 2014, 06:12:41 pm
Now, I've got some questions concerning tension itself. I understand that it acts as the 'driving force' for the object that it is being pulled (eg. trailer). My general conceptual question is: How do I exactly know which equation do I exactly use? Or, simply, how should I be aware of what the question is actually asking me to find? I'll demonstrate. I'll compare the questions, what I did and what the answers was.
ex.1: A car of mass 1000kg is being towed on a level road by a van of mass 2000kg. There is a constant retarding force due to air resistance and friction, of 500N on the van, and 300N on the car. The vehicles are travelling at a constant speed.
Q: What is the value of the tension, T, in the towbar?
My answer and reasoning: Since tension is being pulled BY the van, that means that tension is only being acted on the car. Hence:
T-Fr= Fnet. Since fnet=0 from constant speed, then
T-Fr=0.
T-500N (from the van)=0.
Thus T=500N. This question was wrong, because T=300N from the car itself, not that van that was towing it. Why is this so? I thought tension was dependent on the object that is pulling it..hence the FR from the van. An object that is being pulled (the car) provides no form of any tension.
ex2. A car of mass 1300kg has a caravan of mass 900kg attached to it. The car and caravan move off from rest. They have an initial acceleration of 1.25ms^2.
Q: What is the tension in the coupling between the car and the caravan as they start to accelerate?
Again, my reasoning was this: Since tension occurs from the car pulling from the car, the only orce acting on the trailer is tension. Thus to find the tension, I did this.
T-Fr=Fnet. Since Fr is not provided, I assumed it was zero.
T=2750N.
This is again wrong; as the answer states that Fd is being provided solely by the coupling for the trailer. Hence, T=m*a
T= 900kg*1.25ms^2.
So, besides wondering why I got those two questions wrong, my third question is from these two examples: How do I know which type of 'tension' am I meant to exactly find? I cannot seem and put my tongue on it, but evidently there's a fault in my logical thinking that I'll link to amend. Do you might know the reason why I am thinking like this?
Thanks,
Cort.
Post by: Cort on February 12, 2014, 06:13:35 pm
Some of the questions in Checkpoints are for a different detailed study; I didn't do those.
But yes, I did all the other questions in the physics 2013 Checkpoints that were relevant.
I think I completed the questions in SWOTVAC LOL
Yes, but given your scores, I think you're a freak of nature (No intent to offend). But really, you are.
Post by: lzxnl on February 12, 2014, 06:48:50 pm
Heya Rod. I'm about up to 40-something. But I'm being butt-blasted in every question it's not funny at all. And thanks again PB, great help, haha.
Now, I've got some questions concerning tension itself. I understand that it acts as the 'driving force' for the object that it is being pulled (eg. trailer). My general conceptual question is: How do I exactly know which equation do I exactly use? Or, simply, how should I be aware of what the question is actually asking me to find? I'll demonstrate. I'll compare the questions, what I did and what the answers was.
ex.1: A car of mass 1000kg is being towed on a level road by a van of mass 2000kg. There is a constant retarding force due to air resistance and friction, of 500N on the van, and 300N on the car. The vehicles are travelling at a constant speed.
Q: What is the value of the tension, T, in the towbar?
My answer and reasoning: Since tension is being pulled BY the van, that means that tension is only being acted on the car. Hence:
T-Fr= Fnet. Since fnet=0 from constant speed, then
T-Fr=0.
T-500N (from the van)=0.
Thus T=500N. This question was wrong, because T=300N from the car itself, not that van that was towing it. Why is this so? I thought tension was dependent on the object that is pulling it..hence the FR from the van. An object that is being pulled (the car) provides no form of any tension.
ex2. A car of mass 1300kg has a caravan of mass 900kg attached to it. The car and caravan move off from rest. They have an initial acceleration of 1.25ms^2.
Q: What is the tension in the coupling between the car and the caravan as they start to accelerate?
Again, my reasoning was this: Since tension occurs from the car pulling from the car, the only orce acting on the trailer is tension. Thus to find the tension, I did this.
T-Fr=Fnet. Since Fr is not provided, I assumed it was zero.
T=2750N.
This is again wrong; as the answer states that Fd is being provided solely by the coupling for the trailer. Hence, T=m*a
T= 900kg*1.25ms^2.
So, besides wondering why I got those two questions wrong, my third question is from these two examples: How do I know which type of 'tension' am I meant to exactly find? I cannot seem and put my tongue on it, but evidently there's a fault in my logical thinking that I'll link to amend. Do you might know the reason why I am thinking like this?
Thanks,
Cort.
Let's look at the forces on the car in the first question. There are only two forces on that: tension, and the resistive force of 300 N. As the car is moving at constant speed (presumably in a straight line), the net force is zero. Thus, the tension must equal the resistive force. Here, the resistive force on the van only affects the driving force necessary to maintain a constant speed, not the tension force itself.
Just consider tension as a regular force with a direction. Then, consider objects separately.
If you apply this logic to the second question, you see that the tension is the only thing accelerating the caravan (the tension slows the car down; a larger driving force is thus required for the car to tow a caravan, but the driving force isn't mentioned in these questions; perhaps that is confusing you). Now, your mistake here was assuming that the mass was the mass of the combined car-caravan system. If you just consider the caravan separately and replace the tension force with any other pulling force, you see that the pulling force is the only force responsible for the acceleration on the caravan. Also, as the car is now not considered (this pulling force could be any pulling force), the mass in question is the mass of the caravan only. Therefore, we have T=ma=900*1.25=1125 N.
In any question with systems of objects, consider each object separately and find out what forces are actually acting on each object. The car only provides tension to the caravan and trailer, nothing more.
Yes, but given your scores, I think you're a freak of nature (No intent to offend). But really, you are.
Me? Freak of nature? I'll take that as a compliment :P
Post by: PB on February 13, 2014, 01:29:34 am
PS - PB, Lxnl - did you freaks end up completing all 783 qs? When do you think me and Cort should complete them by?
Cheers :)
haha I am definitely not smart enough to be a freak, lxnl yes, but not me XD. How on earth can you find the time to finish all the questions?!?! I think it would have been impossible for me to have finished even half the questions due to other subjects and other time consuming responsibilities. I think I probably did about 1/4 or less of Checkpoints unfortunately :P
Now, I don't mean to demotivate you when I say this, I am merely telling you what I did :S
I mean, definitely try to finish all the questions if you can, but I honestly think it is quite an ambitious task :PP
My recommendation for you is to simply do all the relevant questions relating to your SACs' topics. Do only minimal questions on the topics that are not tested on your SACs just so you know about them. But once it comes to the end-of-year exam, start to learn the insides and out of the whole course by doing many exams.
That is exactly what I did for Physics and it seemed to work out fine for me :P
If you can finish all the questions, by all means, go for it. But if you are struggling for time, which most of you probably will be throughout the year - maybe try to give my method a go. It is quite efficient.
The weird thing is that I did at least 70% of Checkpoints questions for Chemistry (a whole heap more than for physics) and I happened to do worse lol :P ah well.
Post by: Rod on February 13, 2014, 08:40:58 pm
haha I am definitely not smart enough to be a freak, lxnl yes, but not me XD. How on earth can you find the time to finish all the questions?!?! I think it would have been impossible for me to have finished even half the questions due to other subjects and other time consuming responsibilities. I think I probably did about 1/4 or less of Checkpoints unfortunately :PSome excellent tips here thank you so much.
Now, I don't mean to demotivate you when I say this, I am merely telling you what I did :S
I mean, definitely try to finish all the questions if you can, but I honestly think it is quite an ambitious task :PP
My recommendation for you is to simply do all the relevant questions relating to your SACs' topics. Do only minimal questions on the topics that are not tested on your SACs just so you know about them. But once it comes to the end-of-year exam, start to learn the insides and out of the whole course by doing many exams.
That is exactly what I did for Physics and it seemed to work out fine for me :P
If you can finish all the questions, by all means, go for it. But if you are struggling for time, which most of you probably will be throughout the year - maybe try to give my method a go. It is quite efficient.
The weird thing is that I did at least 70% of Checkpoints questions for Chemistry (a whole heap more than for physics) and I happened to do worse lol :P ah well.
And you are a 'freak' PB :)
Post by: Cort on February 16, 2014, 03:59:04 pm
Or is it because energy stored equation (1/2 * k * delta x^2) does have a force constant (which is the spring constant, k), and there is also a change in displacement as well? Hence, one can assume that:
WD = F * x = 1/2*m*v^2 = 1/2 * k* delta x ^2?
Post by: Cort on February 16, 2014, 04:12:37 pm
I think I'm starting to get the hang of this.
Lastly, is the change in displacement in hooks law/Epe the same as "compression"? Since change in displacement occurs when it is compressed and/or stretched?
Post by: lzxnl on February 16, 2014, 09:33:22 pm
In regards to energy, how on earth is work done on the spring the same as energy stored in the spring? I thought that for work done to occur, it conceptually requires some kind of force constant and a sense of displacement? That is, WD = F * x. This is perhaps why WD = F*x = 1/2 * M *v^2, right?
Or is it because energy stored equation (1/2 * k * delta x^2) does have a force constant (which is the spring constant, k), and there is also a change in displacement as well? Hence, one can assume that:
WD = F * x = 1/2*m*v^2 = 1/2 * k* delta x ^2?
Work = force by distance only when the force is constant. For a spring, it's not constant.
If we consider stretching the spring by a tiny displacement dx such that the force is effectively constant over the entire displacement, the differential work done dW is given by force*distance = F dx. Now, finding the total work done, we integrate over whatever x values the spring moves over. Now, F = -kx and the change in potential energy is equal to the negative of the work done (work = change in kinetic energy, sum of changes in kinetic and potential energies is zero by conservation of energy), so the potential energy is the integral of kx dx, or just 1/2 kx^2.
As for how kinetic energy is 1/2 mv^2, the easiest way of thinking about this is accelerating an object of mass m from rest to a speed v at a constant rate. Remember v^2 = u^2 + 2ax? Well, if accelerated from rest, u=0, so v^2/2 = ax. For a constant force, work = Fx = max = m(ax) = mv^2/2 which is the familiar formula for energy.
Also: work done on a spring is NOT 1/2*k(change in x)^2. It is 1/2*k(final x^2 - initial x^2).
Also, can one also assume that in terms of energy transfers, it's rather obvious by what direction it's travelling to give you an idea of how to set up the equation? That is, if it states that it is travelling horizontally, that means that in terms of energy, it will only be kinetic as work done to some other form; eg. spring p.e. However, the question shows that it has travelled in some form of an angle (like down a hill), it has BOTH Gpe + Ke in the system of Total energy?
I think I'm starting to get the hang of this.
Lastly, is the change in displacement in hooks law/Epe the same as "compression"? Since change in displacement occurs when it is compressed and/or stretched?
Yes, down the hill the object has both KE and GPE. However the height isn't the distance travelled on the hill; it's just the vertical displacement.
The change in displacement in Hooke's law is just how much the spring is stretched from the point of no force. A compression generally increases the change in displacement, yes.
Post by: Cort on February 17, 2014, 07:28:08 pm
Work = force by distance only when the force is constant. For a spring, it's not constant.
If we consider stretching the spring by a tiny displacement dx such that the force is effectively constant over the entire displacement, the differential work done dW is given by force*distance = F dx. Now, finding the total work done, we integrate over whatever x values the spring moves over. Now, F = -kx and the change in potential energy is equal to the negative of the work done (work = change in kinetic energy, sum of changes in kinetic and potential energies is zero by conservation of energy), so the potential energy is the integral of kx dx, or just 1/2 kx^2.
Wait, I never knew this. So because Work = Change in Ke, Work also = 0 due to the conversation of energy by Change in Ke + change in Pe? Hence that is why change in pe = negative work (aka change in Ke)? I'm just a tad confused because I need some sort of concrete formula/picture to help me out there. Sorry!
Edit: DON'T SAY THIS IS BECAUSE TOTAL ENERGY = Pe (either Gpe or Epe) + Ke = CONSTANT BECAUSE OF CONVERSION OF ENERGY RIGHT?
ALSO, if that means that it is the negative of work done = the energy would still be 'postive' in the end because you cannot have negative energy?
Post by: lzxnl on February 17, 2014, 08:14:44 pm
Wait, I never knew this. So because Work = Change in Ke, Work also = 0 due to the conversation of energy by Change in Ke + change in Pe? Hence that is why change in pe = negative work (aka change in Ke)? I'm just a tad confused because I need some sort of concrete formula/picture to help me out there. Sorry!
Edit: DON'T SAY THIS IS BECAUSE TOTAL ENERGY = Pe (either Gpe or Epe) + Ke = CONSTANT BECAUSE OF CONVERSION OF ENERGY RIGHT?
ALSO, if that means that it is the negative of work done = the energy would still be 'postive' in the end because you cannot have negative energy?
Conservation of energy states that
You can't have negative kinetic energy, that is correct. If something has only 2 J of kinetic energy, you cannot possibly do -3 J of work on it. As soon as the object loses all kinetic energy, it's at rest and can only gain kinetic energy, so kinetic energy never becomes negative.
Potential energy is a different story. These can be negative. Gravitational potential energy (in space, not the mgh one on Earth) is always negative, for instance. I don't quite understand your query here though. I've only said that the signs of the changes in energy may be negative.
Post by: Cort on February 19, 2014, 12:17:54 am
Conservation of energy states thatso
so
You can't have negative kinetic energy, that is correct. If something has only 2 J of kinetic energy, you cannot possibly do -3 J of work on it. As soon as the object loses all kinetic energy, it's at rest and can only gain kinetic energy, so kinetic energy never becomes negative.
Potential energy is a different story. These can be negative. Gravitational potential energy (in space, not the mgh one on Earth) is always negative, for instance. I don't quite understand your query here though. I've only said that the signs of the changes in energy may be negative.
Ah, bloody hell. Lacking logical reasoning sucks.
Another note: I'm having trouble understanding why an object will not go faster if it reduced its mass. Here's the question:
"Jack and Jill are racing their toboggans down an icy hill. Jack and Jill are of similar mass are using the same type of tobogann.When Jack is a certain distance from the end of the race they are travelling with the same velocity. Jack is behind Jill and decides that if he is going to win the race he must lighten his tobogann so he pushes a box containing their ice-skating gear off the side of his toboggan."
Q: Explaining whehter this will be a successful way for Jack to catch up to Jill
I stated:
- Jack will not catch up to Jill because pushing sideways box means he gets sideways force in the opposite direction.
- Jack will only catch up if he pushed it directly behind him because he gains forward momentum.
My troubles:
I cannot fathom why (as the answer states) "From a conversation of momentum, there will be no change in the forward momentum of the box or tobogan" and "Lightening his toboggan will not cause Jack to gain or lose downhill momentum" (Why so? Is it because he pushed it sideways, and not backwards instead? And how does not lightening his toboggan not allow Jack to gain momentum because of less mass?
Post by: Zealous on February 19, 2014, 05:59:15 pm
"Jack and Jill are racing their toboggans down an icy hill. Jack and Jill are of similar mass are using the same type of tobogann.When Jack is a certain distance from the end of the race they are travelling with the same velocity. Jack is behind Jill and decides that if he is going to win the race he must lighten his tobogann so he pushes a box containing their ice-skating gear off the side of his toboggan."For an inclined plane:
Q: Explaining whehter this will be a successful way for Jack to catch up to Jill
There is no mass component in this equation. If he reduces his weight, he will still accelerate at the same rate as before (and Jill will also accelerate at that same rate).
Why? If Jack reduces his weight, it will reduce his weight force (N) due to gravity, leading to a decrease the normal force and also the force pushing him down the slope will also be reduced.
We've got less mass, but we've also got less force (down the hill), so acceleration stays the same.
Hopefully that makes sense.
Post by: Cort on February 19, 2014, 11:25:29 pm
For an inclined plane:
There is no mass component in this equation. If he reduces his weight, he will still accelerate at the same rate as before (and Jill will also accelerate at that same rate).
Why? If Jack reduces his weight, it will reduce his weight force (N) due to gravity, leading to a decrease the normal force and also the force pushing him down the slope will also be reduced.
We've got less mass, but we've also got less force (down the hill), so acceleration stays the same.
Hopefully that makes sense.
It does. Thank ye
Post by: Cort on March 02, 2014, 04:17:58 pm
If initial speed (Vo) is given, does that mean initial velocities for the vertical component exist, even if it is launched at a cliff? Because I think that would make some sense. Conversely, if it is projected horizontally from a cliff, and no angle/initial velocity is given, then Uv = 0?
Post by: Aurelian on March 02, 2014, 04:40:18 pm
I tend to confuse myself when it comes to solving projectile motions..the thing is, for all types of those projectile motion situations ( such as horizontal launch at a cliff; launch from an angle) I do no know whether my initial velocities (in the vertical component) is correct or not.
If initial speed (Vo) is given, does that mean initial velocities for the vertical component exist, even if it is launched at a cliff? Because I think that would make some sense. Conversely, if it is projected horizontally from a cliff, and no angle/initial velocity is given, then Uv = 0?
Just use your common sense and think only about how the object you're dealing with is moving through space. If an object is being launched "horizontally", then that means its initial motion is entirely entirely left or right and that the object is not moving up or down at all to start with. Since the object isn't moving up or down in any way, then clearly that object cannot have an initial (non-zero) vertical velocity; the object is stationary with respect to vertical motion.
An object launched with an initial (non-zero) speed at some angle to the horizontal, however, will be moving up or down as well as left or right. That is, we will be able to represent this angled-speed as some vector combination of a left/right velocity and an up/down velocity. Since the object is initially moving in a vertical direction as well as a horizontal direction, clearly it will have an initial (non-zero) vertical velocity.
Whether a projectile is launched from a cliff or from the ground doesn't change anything concerning the above reasoning; all we need to know about when finding initial velocities is how the object is moving through space at the start of its flight. Surrounding scenery only becomes relevant later on in the problem solving.
Does that help? =)
Post by: Cort on March 02, 2014, 04:53:09 pm
Just use your common sense
But I have none D:
and think only about how the object you're dealing with is moving through space. If an object is being launched "horizontally", then that means its initial motion is entirely entirely left or right and that the object is not moving up or down at all to start with. Since the object isn't moving up or down in any way, then clearly that object cannot have an initial (non-zero) vertical velocity; the object is stationary with respect to vertical motion.
An object launched with an initial (non-zero) speed at some angle to the horizontal, however, will be moving up or down as well as left or right. That is, we will be able to represent this angled-speed as some vector combination of a left/right velocity and an up/down velocity. Since the object is initially moving in a vertical direction as well as a horizontal direction, clearly it will have an initial (non-zero) vertical velocity.
Whether a projectile is launched from a cliff or from the ground doesn't change anything concerning the above reasoning; all we need to know about when finding initial velocities is how the object is moving through space at the start of its flight. Surrounding scenery only becomes relevant later on in the problem solving.
Does that help? =)
That makes total sense, since I never thought about it that way. I'm still trying to break away my bonds of "sub this crap in and get outta here" mentality..but it's slow progress I guess. Thanks.
Post by: Cort on March 02, 2014, 05:33:37 pm
The ball leaves the tee horizontally and hits the hole. The horizontal distance between the tee and the hole is 155m. The vertical distance is 45m. There is a tailwind that effectively cancels any air resistance.
a) Calculate the speed of the ball off the tee:
I've worked out the angled speed to be ~60ms^-1. Apparently the answer is ~52 -- which is the horizontal speed. Did I read the question wrong? Or am I working it out wrong?
Post by: lzxnl on March 02, 2014, 05:45:02 pm
Ah, bloody hell. Lacking logical reasoning sucks.
Another note: I'm having trouble understanding why an object will not go faster if it reduced its mass. Here's the question:
"Jack and Jill are racing their toboggans down an icy hill. Jack and Jill are of similar mass are using the same type of tobogann.When Jack is a certain distance from the end of the race they are travelling with the same velocity. Jack is behind Jill and decides that if he is going to win the race he must lighten his tobogann so he pushes a box containing their ice-skating gear off the side of his toboggan."
Q: Explaining whehter this will be a successful way for Jack to catch up to Jill
I stated:
- Jack will not catch up to Jill because pushing sideways box means he gets sideways force in the opposite direction.
- Jack will only catch up if he pushed it directly behind him because he gains forward momentum.
My troubles:
I cannot fathom why (as the answer states) "From a conversation of momentum, there will be no change in the forward momentum of the box or tobogan" and "Lightening his toboggan will not cause Jack to gain or lose downhill momentum" (Why so? Is it because he pushed it sideways, and not backwards instead? And how does not lightening his toboggan not allow Jack to gain momentum because of less mass?
One of my students asked me this question today and upon consideration, this is my response.
Lightening the toboggan has no effect on Jack's momentum. Why? Well, the argument is that p=mv, and if we reduce m we must increase v. That argument implicitly uses conservation of momentum on the extended toboggan-box system, and the mass of that system is NOT reduced. Therefore, you can't say that lightening the toboggan will immediately speed it up.
Second issue: when Jack throws the box off the toboggan, the box is still moving down the hill in the same direction as the toboggan. If you consider the toboggan-box group as a single system, they form a system only acted upon by gravity. The force on the box, coming from the toboggan, is an internal force and cannot affect the momentum of the system as a whole.
Also, we can split the system up into the toboggan and the box individually. Although initially we think of the box on top of the toboggan, the momentum of the toboggan is really independent of the box; the toboggan, moving at a fixed speed, has the same momentum regardless of if the box is on top of it. Pushing the box off the tobbogan is then akin to pushing sideways on a wall. As the person is pushing sideways, the downhill motion is unaffected and the toboggan moves at the same rate. You're right, it does have to do with the fact that the box is thrown sideways.
This is from Checkpoints:
The ball leaves the tee horizontally and hits the hole. The horizontal distance between the tee and the hole is 155m. The vertical distance is 45m. There is a tailwind that effectively cancels any air resistance.
a) Calculate the speed of the ball off the tee:
I've worked out the angled speed to be ~60ms^-1. Apparently the answer is ~52 -- which is the horizontal speed. Did I read the question wrong? Or am I working it out wrong?
So...initial vertical speed is zero. Vertically, we have 45 = 1/2 at^2 = 5t^2. t=3 (zero initial vertical velocity)
Horizontally, as the horizontal acceleration is zero, we have 155m = speed * time = speed * 3s. Speed is around 51.67 m/s.
Post by: Cort on March 10, 2014, 08:15:11 pm
I'm now touching on Circular Motion. I'm somewhat confuzzled when it comes to the Loop-de-loop ones; especially when it asks about the 'minimum' speed for car to go around at the top. Is there a reason why you would cancel out or why the reaction force (N) become zero in this instance? It's drilling my head in. Thanks.
I've did some reading and it's something related to being 'minimum' so that it equals the g = 9.81ms^-2 or something. Something like that, right?
Edit 2: Right, it has to do with the speed being equal or greater than 9.81 itself to keep it from falling down. If it's smaller than that, gravity will cause the car to fall down. The problem is..how does THAT relate to the normal force at all?
Post by: PB on March 10, 2014, 11:16:35 pm
Therefore, the fact that there is zero normal force indicates that the car is not pressing against the rails at all - it is at the verge of falling off. And if it goes any slower, it WILL fall off.
Post by: Cort on March 10, 2014, 11:30:03 pm
The reason why the normal even exists at the top is because the car is pressing against the rails, and hence, the rails presses back (creating that normal force on the car).
Therefore, the fact that there is zero normal force indicates that the car is not pressing against the rails at all - it is at the verge of falling off. And if it goes any slower, it WILL fall off.
Right, I really am stupid. So it's apparent that when it comes to 'minimum/least speed' it means the verge of it falling off as you said. Bah. That means that Fn=0.
Post by: PB on March 11, 2014, 01:05:02 am
Post by: Cort on March 16, 2014, 01:17:51 am
This is from checkpoints
A Small satellite orbits Mars. It has a kinetic energy of 3.0 * 10^10J, and it is at a constant distsancen of 8.0^10^7from the centre of Mars. What is the weight of the satellite at this height?
The furthest I got to was..
W =m*g = m*v^2/r = ??
Post by: lzxnl on March 16, 2014, 10:22:47 am
Gravity confuzzles me.
This is from checkpoints
A Small satellite orbits Mars. It has a kinetic energy of 3.0 * 10^10J, and it is at a constant distsancen of 8.0^10^7from the centre of Mars. What is the weight of the satellite at this height?
The furthest I got to was..
W =m*g = m*v^2/r = ??
mv^2/r = GmM/r^2
1/2 mv^2 = GmM/2r = kinetic energy = 1/2 × weight force × radius
Post by: Cort on March 16, 2014, 03:14:24 pm
mv^2/r = GmM/r^2
1/2 mv^2 = GmM/2r = kinetic energy = 1/2 × weight force × radius
1/2 mv^2 = GmM/2r -- how did you go from r^2 to 2r?
Post by: Cort on March 16, 2014, 03:33:17 pm
Calculate the mass of the Earth from the following data:
radius of moon's orbit= (3.8*10^8); G = 6.67*10^-11, Priod = 28 days
I've done T^2/R^3 = GM/4*pi^2 and it ain't working. Because I assumed that the moon would be a satellite here, hence the m's cancel out.
Similarly,
Calculate the altitude of a satellite in a geo-stationary orbit.
Data: G = 6.67*10^-11, Me = 6.4*10^24, Re = 6.5*10^6m.
Would T^2/R^3 be useful here? Is there any special properties when it comes to satellites in geo-stationary orbit?
Post by: lzxnl on March 16, 2014, 05:13:11 pm
1/2 mv^2 = GmM/2r -- how did you go from r^2 to 2r?
Because the kinetic energy term doesn't have a "r" term in the denom but the force term does?
Alright..This is really confusing.
Calculate the mass of the Earth from the following data:
radius of moon's orbit= (3.8*10^8); G = 6.67*10^-11, Priod = 28 days
I've done T^2/R^3 = GM/4*pi^2 and it ain't working. Because I assumed that the moon would be a satellite here, hence the m's cancel out.
Similarly,
Calculate the altitude of a satellite in a geo-stationary orbit.
Data: G = 6.67*10^-11, Me = 6.4*10^24, Re = 6.5*10^6m.
Would T^2/R^3 be useful here? Is there any special properties when it comes to satellites in geo-stationary orbit?
The moon is a satellite, yes. Therefore the M in your first formula is the mass of the Earth.
Geostationary orbit => period = 86400s. That's it. Then, be careful about the wording "altitude", not radius.
Post by: Cort on March 16, 2014, 06:01:16 pm
Because the kinetic energy term doesn't have a "r" term in the denom but the force term does?
The moon is a satellite, yes. Therefore the M in your first formula is the mass of the Earth.
Geostationary orbit => period = 86400s. That's it. Then, be careful about the wording "altitude", not radius.
Thanks - but I'm afraid I'm still confused. Do you mind if you explained about how Ke is found? 1/2 mv^2 = GmM/2r = kinetic energy = 1/2 × weight force × radius - still don't understand.
For geostationary orbit, do you always assume the period would be 1 day = 86400s? So every time geostationary is mentioned, just whip down T= 1day? Is there a reason why we might use T =1 day = 86400s?
Post by: lzxnl on March 16, 2014, 07:01:24 pm
Thanks - but I'm afraid I'm still confused. Do you mind if you explained about how Ke is found? 1/2 mv^2 = GmM/2r = kinetic energy = 1/2 × weight force × radius - still don't understand.
For geostationary orbit, do you always assume the period would be 1 day = 86400s? So every time geostationary is mentioned, just whip down T= 1day? Is there a reason why we might use T =1 day = 86400s?
I really can't explain it. I've written out all the explaining there is.
mv^2/r = GmM/r^2 => centripetal force requirement
Multiply both sides by r/2 to convert the left hand side to kinetic energy
mv^2/2 = GmM/2r
That's the definition of a geostationary orbit; same period as the Earth's rotation
Post by: rhinwarr on March 16, 2014, 07:06:26 pm
Quote
A Small satellite orbits Mars. It has a kinetic energy of 3.0 * 10^10J, and it is at a constant distsancen of 8.0^10^7from the centre of Mars. What is the weight of the satellite at this height?
w = mg = mv^2 / r (because the centripetal acceleration is being caused by the weight force)
Ek = 1/2 mv^2
2Ek = mv^2
sub 2Ek into the 1st equation
w = 2Ek / r
w = 750N
Quote
For geostationary orbit, do you always assume the period would be 1 day = 86400s?
Geostationary orbit is where the satellite's period of orbit is equal to the period of rotation of the centre of mass. This is so the geostationary satellite stays above the same point on the centre of mass throughout its orbit.
Yes, the period will always be one day IF the centre of mass is the Earth, because the period of rotation of the Earth is one day.
However, if the centre of mass is something else, eg. Mars, then it will be different.
Post by: Cort on March 17, 2014, 11:40:55 pm
Thanks,
Cort.
Post by: lzxnl on March 18, 2014, 07:00:27 am
Out of curiousity and to consolidate - can you assume (although you never should!) that everytime it mentions a 'orbit circular path' you always go Fc = Fg? What about Kepler's Third Law? Does it apply in situations where it is an orbital circular path, although it was intended for ellipitical equations?
Thanks,
Cort.
A circle IS an ellipse, just that both semi-major and semi-minor axes are equal.
And yes for your question on centripetal forces.
Post by: jesterino on March 20, 2014, 04:58:43 pm
"In an adventure park ride, people are strapped in a small cage that is whirled in a vertical circle on cables. The cables make an angle of 30 with the vertical. The radius of the circle is 8.0m. The cage has a total of mass 250kg."
(From the previous question, net force was worked out to be 3125 N.)
Show that the tension in each of the cables at the top of the circle is close to 360 N.
So far I have drawn out a triangle with 30 at the top, T (tension) as the hypotenuse, and mg as the vertical component. I'm quite sure this is incorrect though
This question is killing me, I have no idea on how to reach the end result.
Post by: PB on March 20, 2014, 11:11:13 pm
If the net force in this circular motion is 3125N, then it should stay as 3125N anywhere in the circle- as long as the velocity is the same and the radius is the same. Else, the cage would spiral off (too little net force towards the centre)or spiral towards the centre (too much net force).
So at the top of the circle, the net force will still be 3125N. The only things that could be changing throughout the circular motion are the constituents of the net force - weight force and tension force. We know that weight force will always stay (250kg x9.8 =2450N) - therefore it is the tension force that is changing.
At the top of the circle, you can take the weight force away from the net force (3125-2450N = 675N). Assuming that there are two cables, 675 / 2 = 337.5N, which is kind of close to 360N :P
Edit: Assuming that there are two cables is the only way I can make sense of this question...
Post by: jesterino on March 22, 2014, 08:08:54 pm
3125 = 2Tcos30 + mg, where T is tension i'm guessing.
So if at the top: net force = tension + mg
How did they deduce tension = 2Tcos30? I also attached the diagram of the problem below.
Post by: RKTR on March 22, 2014, 09:55:05 pm
N=3125-mg
there are two cables
divide N by 2
cos30= (N/2) / T
Tcos30=N/2
N=2Tcos30
Post by: souka on March 24, 2014, 09:36:05 pm
Which when solved would equal ~360.8 N
Post by: Cort on April 10, 2014, 11:29:11 pm
Currently I'm assuming that you use r= half the x length when it is used for translational equilibrium/balancing, right? This is for materials & structures.
Thanks,
Cort.
Post by: lzxnl on April 10, 2014, 11:32:56 pm
Reading up/doing some questions regarding torque. I know the rule is t = f*r = f*r*sin theta if the angle is involved. My problem is, for r, when do you know that you have to measure it from the centre of the object, or just use the full radius of the object? Whhich/what keywords should I be aware of that will tell me when I should be using r = half of the x length, or r = use the full length?
Currently I'm assuming that you use r= half the x length when it is used for translational equilibrium/balancing, right? This is for materials & structures.
Thanks,
Cort.
The radius r is measured from the axis of rotation. Identify where that is first.
Post by: Cort on April 10, 2014, 11:39:17 pm
The radius r is measured from the axis of rotation. Identify where that is first.
Ah, thanks for the heads up. The axis of rotation is where..well, where it rotates. r measures from that direction, apparently, but in different situations the set up will be different. Thanks for the thought!
Post by: Cort on April 14, 2014, 11:35:11 pm
"When cantilevers are joined at the centre of the span, there is no reduction in force. These are the same as if the beams were not connected."
- What does it mean when there is no "reduction in force...at the centre of the span" ?
Thanks,
Cort.
Edit: For the caniliever below, why is that the force on the left goes downwards? I assumed that both the left and right pillars will have an upwards force.
Post by: Cort on April 22, 2014, 08:50:06 pm
q738 (Structures and materials)
"Two wires, A and B, are of equal length. A has twice the diameter of B. The Young's modulus of A is twice that of B. if they are subjected to the same load, and A stretches by 10mm, the wire B will stretch by x. Which of the following is closest to the value of x?"
The answers would state, that after writing out the Young's Modulus for Mat A and Mat B, you'll do Mat A/Mat B to eliminate Fa*La/Fa*Lb. I can understand the result of dividing those two together..but its understanding why they would do it so suddenly. is it meant to be some kind of general knowledge I'm supposed to know? If so, would there be another way to actually approach, and break down the ideas contained within the question?
Thanks,
Cort
Post by: Cort on May 11, 2014, 11:58:54 pm
Post by: Thorium on May 12, 2014, 07:31:41 am
When it says RMS voltage -- does that mean voltage = 12v?
Yes in other words. Rms stands for root mean square
Post by: Stevensmay on May 12, 2014, 11:25:12 am
When it says RMS voltage -- does that mean voltage = 12v?
RMS voltage of an AC voltage can be found by dividing it by root(2). It doesn't mean voltage is always 12V however.
Post by: lzxnl on May 12, 2014, 08:26:46 pm
Post by: Cort on May 12, 2014, 09:34:24 pm
Post by: Stevensmay on May 13, 2014, 12:38:23 am
Thank you all. Is it required to know about rms and multiplying it and what not in the exam?
From memory I believe it is.
Post by: Thorium on May 13, 2014, 07:17:53 am
Thank you all. Is it required to know about rms and multiplying it and what not in the exam?
I read somewhere in my notes, it mentioned that this part is embedded in further electronics detailed study not in the normal electronics. So if you are going to be doing further electronics, then it should be good to have a prior knowledge.
Post by: Cort on May 15, 2014, 10:30:36 pm
Furthermore, mind if someone took me through the process of thought? What are the things I should be actively aware of?
Thanks,
Cort