physics questions

last SG · 2012-01-29T22:13:55+00:00

StudentQns posted this question at physicsforum.comhere's the illustrationhttp://postimage.org/image/4ure9j3sr/

StudentQns SG

Oh ok thanks. I was revising on practical electricity and I thought of this. they say as resistance increases more heat is produced as P=I^{2R but then power is equal to V}2/R too. So as R increases P increases but also as R increases P decreases. so j don’t really understand this.

thanks for the help and happy Chinese new year

eagle SG

P is not directly proportional to R.

As R increases, V and I may change accordingly. It then depends on how much they change before you can decide whether power will change.

StudentQns SG

oh ok i understand. um, i have another question now its related to dynamics. I’m not very sure if its part of my O level syllabus but I’m quite confused about this, here it is:

For example, when I punch a wall, what determines the amount of “impact” on the wall? Is it force (or something else, like momentum P=mv?!)? Since F=ma, the force generated would be proportional to how much mass my hand has as well as its acceleration. However, it seems counterintuitive to say that my punch would not produce a force if my hand moves at a constant velocity (a=0). Certainly, a car moving at constant velocity would produce an “impact” when it hits a passer-by.

Is the acceleration in F=ma then referring to the decceleration upon my hand hitting the wall? From e.g. 2m/s to 0m/s. If this is so, then whether or not previously my hand was accelerating or not doesn’t matter, only the ‘final velocity’ immediately before impact. However, since a = (final – initial velocity)/time taken, what is the time taken for the decceleration then? Is it instantaneous? How could it be, since that would entail the denominator of the equation being 0? Even if it’s a minute fraction of a second, doesn’t this mean that the acceleration would be quite huge (e.g. -2/0.001) and so will the resulting force? This doesn’t seem right since a punching hand shouldn’t cause so much force?

Thanks so much for helping!

eagle SG

This is under the concept of impulse, which is about the time interval in which a force is delivered.

It is not in O levels syllabus, but in A levels syllabus.

Yes, the deceleration can be quite large. The shorter the time, the larger the force. This is the basis for impulse of a force.

StudentQns SG

Oh ok thanks for the help eagle! I have some magnetism questions hope u dont mind

StudentQns SG

i think the post messed up

original post:
Oh ok thanks for the help eagle! I have some magnetism questions hope u dont mind

how will the magnetic field look like when a bar magnet is placed on a iron wall? The north and south pole are in contact with the iron wall. I know that that bar magnet will be attracted so I know that a north and south pole will form. However, I’m not sure how it will look like. I would greatly appreciate it if u could explain this to me with a diagram.

Also, I’m not very sure because the north and south pole are made up and since they have the same properties, the founder of magnets can name the poles anyway they want (meaning a north can be a south and vice versa). I know that due to permeability, the magnetic field will tend into the material with a higher permeability. but as mentioned the north can be a south and vice versa, so im very confused on this part.

thanks so much for the help!

StudentQns SG

This is under the concept of impulse, which is about the time interval in which a force is delivered.

It is not in O levels syllabus, but in A levels syllabus.

Yes, the deceleration can be quite large. The shorter the time, the larger the force. This is the basis for impulse of a force.

Oh ok thanks for the help eagle! I have some magnetism questions hope u dont mind

1) how will the magnetic field look like when a bar magnet is placed on a iron wall? The north and south pole are in contact with the iron wall. I know that that bar magnet will be attracted so I know that a north and south pole will form. However, I'm not sure how it will look like. I would greatly appreciate it if u could explain this to me with a diagram.

Also, I'm not very sure because the north and south pole are chosen to be that way and since they have the same properties, the founder of magnets can name the poles any way (north can be a south and south a north). So I'm not sure how to draw it.i know that magnetic field will rather go through a material with a higher magnetic permeability, but then as previously mentioned a south can be a north and so on. so when i place a magnet on a iron wall, im not sure if the north pole's magnetic field will pass through the iron first. so im very very confused here.

2) why when magnetic shielding occurs, the iron bar (for eg) is also being induced so eg a north pole is placed in front of the iron bar which is perpenticular to the bar magnet's north pole. when this happens two south poles will be induced in the middle and 2 north poles at the two end. but when we draw this out, we only draw the magnetic field by the magnet. but wouldnt there be the magnetic field of the iron bar itself when it is induced? so why wont anything be attracted at the bottom (opposite side of the iron bar) since i think that there will be its own magnetic field? thanks so much for the help!

eagle SG

Been too busy to really reply... I think I sleep an average of 3.5 hrs a day for the past week to complete my stuff :x

You mentioned about definitions. That is right. For magnetic field lines, it is also defined as field lines exiting from a north pole and moving towards a south pole. By convention, magnetic field lines are always in closed loops. As you move towards uni studies, you will see closed loop integrals.

I can't draw out the diagram now, but when magnets touch iron bars, what will happen is that the iron bar becomes a temporary or induced magnet. Basically, the field lines of the magnet will be concentrated and aligned along the iron bar all the way to it's end poles. The field lines will then exit the bar and move all the way back to the magnet.

A website with some graphics can be found here, but it might be too advanced for O levels.

http://www.coolmagnetman.com/magshield.htm

I don't understand your 2nd question. Would be good if you have a diagram. Basically, what you are describing is similar to a magnetic monopole, or so I think. Tradtionally, there is no conclusive evidence of the existence of magnetic monopole

http://en.wikipedia.org/wiki/Magnetic_monopole

However, in recent years, researchers have observed the existence of such monopoles

http://www.sciencedaily.com/releases/2009/09/090903163725.htm

http://www.sciencedaily.com/releases/2010/10/101018074540.htm

Magnetic monopoles are however out of O level, A level and first year uni physics syllabus.

StudentQns SG

P is not directly proportional to R.

As R increases, V and I may change accordingly. It then depends on how much they change before you can decide whether power will change.

Hi sorry for jumping back, when incorrect wire is used, then R will increase. But then since v=ri, then won't I decrease. So when p=vi or I^2r then won't they be the same? Thankd

eagle SG

If V is a constant, i.e. from a battery, I will decrease when r increase

the changes of I^2 R is hard to determine because there are 2 variables changing, unless you know what you are doing with the mathematics

StudentQns SG

So what does it mean when I anyhow use a wire at home, what effects will there be? Since the voltage in at home will be constant so if R increases what will be the effect? As V should be constant then won’t there be no change?

eagle SG

yes, current will change.

Under practical electricity topic, I usually teach students the danger of connecting too many appliances in parallel. Resistance will drop greatly, resulting in an over current which can cause a short circuit.

Usually can link to how a circuit breaker works. :)

Actually your tutor should have taught you all these. :x

StudentQns SG

Then how does the thickness of a wire cause more heat to be produced? Thanks for the help! Since R increases, but V=RI and I will reduce, so P=VI won’t P be reduced n P=I2R and P=V2/R.

I’m not sure about this affecting it. Thanks for thr help! Smile

UltimaOnline SG

Actually your tutor should have taught you all these. :x

He said, "My physics tutor is injured (internal injury, vomitting blood, may never recover) so he’s unable to teach me now. N my school teacher is very bad to me (always fleeing from me in terror whenever I want to ask her my incessant questions)".

It's good to be a deep, critical thinker and have the passion to go beyond the syllabus. But trying to resolve all your incessant questions on an non-personal online forum, is an inefficient process and quite simply impossible to satisfy you (and frustrating to those who would try to help you out by replying your questions).

What you need (in addition to doing self-research and self-learning through Wikipedia and other reliable websites, for 'A' level Chemistry use Rod Beavon, Jim Clarke and Doc Brown), you probably need to change your private tutors to new ones who can better satisfy your academic needs, than your current tutors (which you've personal problems with, as you indicated in your private message).

Anyway, for the record, I won't be replying to any of StudentQns' O level Chem questions (I don't wish to enable such an unfeasible learning approach); and when he gets to A levels in 2 years time, I still may not reply his A level Chem questions, depending on whether I observe any changes in his approach to this matter.

Edit :
Just in case of misunderstanding, StudentQns, this is not a personal attack on you (the 1st paragraph above is just for kidding). Feel free to continue posting all your Chem and Physics and Math questions here. Even if I don't reply you, lots of other folks will. All the best with finding better private tutors for yourself.

StudentQns SG

Thanks ultimaonline, that’s why I think that its a curse to be so curious. As forg eagle, thanks for all the help and patience with me, I know that I’m a real bother.

As for my physics question, I’m unsure how a thinner or thicker sure will affect the amount of heat being produced. Since V=RI, and V is constant. Since R increase, so I should decrease so that they will multiply to get same value of V. So when I use the power formula P=VI, I don’t see a increase in power. If 100V=100ohmx1A, then power is 100W. But when I increase R to 200ohm I becomes 0.5A so power becomes 50W. So does this mean with higher resistance (thinner wires) less heat is produced?

Thanks for the help. I feel that this is a part of the syllabus so I’m asking you this question. Thanks a lot for all the help!

Darkness_hacker99 SG

@StudentQns

In the past, when my teachers see me, the first thing they do is to turn back and walk fast fast to avoid me. Just like what UltimaOnline said, I would acutally do self study and research on my own and try to digest it myself. The process of self-learning is kinda enjoyable for me. Over time, I develop this habit to only approach my teachers for help when there is a very challenging question that I'm dying to know the solution or a particular concept which are fit for serious discussion.

When you're doing your work, it's good to take out a piece of paper and list down

I remember once I was told by my facilitator that "whatever questions that you will be asking, most likely this question has been answered by somebody else. It's the matter whether if you could find the answer on the internet." It's true, I always remember this and hence the quote "Google is your best friend". I say "Youtube" is my buddy

eagle SG

Thanks ultimaonline, that’s why I think that its a curse to be so curious. As forg eagle, thanks for all the help and patience with me, I know that I’m a real bother.

Thanks for the help. I feel that this is a part of the syllabus so I’m asking you this question. Thanks a lot for all the help!

Hi, let me show you some values

Let

V = 100V
I = 1 A
Hence, R = 100 ohms

P = VI = 100W
P = V^2 / R = 100^2 / 100 = 100 ohms
P = I^2 R = 1^2 * 100 = 100 ohms

If R increases to 200 ohms, V = 100V and I = 0.5A,

P = VI = 50W
P = V^2 / R = 100^2 / 200 = 50 ohms
P = I^2 R = 0.5^2 * 200 = 50 ohms

So yes, for a constant voltage, higher resistance will lead to lesser power being dissipated.

eagle SG

@StudentQns

When you're doing your work, it's good to take out a piece of paper and list down

I always tell people there was once in JC, there was this maths question I think for 3 days before finally solving it while I was brushing teeth in the morning.

Maybe because my ego was too high, I always believed that if I couldn't solve, my friends couldn't too, so there was no point asking them

StudentQns SG

Oh so when they say more resistance, the current must be constant and voltage can change. Thanks for the help eagle and patience with me. But in a home circuit, isn’t voltage fixed? Thanks for the help!

Darkness hacker, I see u r something like me. I’ll pm u my situation.

Jacky Woo SG

why dun u google, you best friend in internet.

utilise and make good use of it.

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