studiot

Unfortunately I have to agree (and sympathise) with you. +1

With the greatest of respect, old bean, this displays your ignorance. And all ajb was trying to do was explain that relativity is about relative velocity, not 'velocity', funny how that word relative creeps in. FYI, 'linear' and 'vector' are synonymous as they refer to the same thing.

Let us do the second part of the question in two stages and it should become clear why they are working at different rates. Let us first consider the situation where m workers are working and all m workers work at the same rate. So if you like they are all clones or copies of each other. So if one worker is working then the task takes T days. If two workers are working the task takes T/2 days If three workers are working the task takes T/3 days If m workers are working the task takes T/m days. We can consider this as one worker worker working m times as fast. So one worker working m times as fast will complete in T/m days So if one worker works 1.5 times as fast m = 1.5 and will complete in T/1.5 days So days = T/m or m = T/days In our situation, T =50, days =15 so m =50/ 15 = 10/3 times as fast (note this is not a whole number don't worry it will become clear)

More or less exactly what I was considering saying +1

Yes and no. There is not "a (single) state of" time dilation, that implies too much (or perhaps too little?) We call the measurement made by an observer about his own time (his) "proper time". Any other observer in the universe would observe a dilated time if that second observer was in steady motion relative to the first. Since there are an infinite number of possible relative motions (all not greater than c) there are an infinite number of possible "states of dilatation". It does not actually need a real observer for this to be true, just the fact that there could be such an observer.

But A and B work at different rates so each accomplishes a different amount in 6 days or 50 days or 15 days. We have calculated that Mr A accomplishes the job on his own in 50 days. If Mr B worked at the same rate he would also finish in 50 days on his own so together they would finish in half that time ie 25 days. But we are told they finish faster in only 15 days so Mr B must work faster than Mr A. So I ask again how much faster does Mr B work than Mr A?

Yes correct. You are getting there, although I wanted you to say that A working alone takes 30 days to complete 9/15 of the total [math]\left( {\frac{9}{{15}}} \right)a = 30[/math] so [math]a = 50[/math] This was the point of my first question in post#2 We need to do some more work on this part to find b. Please use my 'a' and 'b' notation from post#2 as it will make the next bit easier. Now that we have found 'a', can you answer my second question in post#2?

Well yes that is what it comes down to, but you need to understand the answer to If it takes 15 days to complete the job what fraction of the job is complete after 6 days? and also What fraction of the job remains to be completed?

Unfortunately not. Neither the 1/9 nor the 1/6 is correct (do 1/9 + 1/6 make 1 ?) Edit divide the '1 whole' into days. How many days were there when the job was completed when they were both working together?

Better would be if you told me what fraction of the work remained when B fell ill.

This problem is about ratio, proportion or fractions - whatever you like to call them. You have nearly done the first bit, I will start you off with some notation Let 'a' be the number of days A takes to complete alone Let 'b' be the number of days B takes to complete alone So taking the information you have deduced from the problem can you think of any equation involving a fraction and 'a'? Also can you tell me what is the ratio of speeds at which A and B work?

Yes you could call it that.

That's exactly right! Edit : Did you want to discuss it further?

Hello again, Nicholas. Don't be sorry it is a perfectly good question. I have drawn some very quick pics to show what I mean. It is possible to develop this to a numerical theory that predicts friction coeffiients with reasonable accurcy for many metals. All surfaces are rough at some level of magnification. In (1) we are lowering a block, magnified until a bump shows. In (2) the block touches the table and the bump is the forst point to touch (there must be one) All the weight of the block is now resting on that one point, so it deforms as it cannot support such a load, as in (3) This continues untill a second (third, fourth etc) bump joins and the total surface area of the bumps is just sufficient to support the block as in (4)

After breaking my neck reading the scrolls you posted i thought that they first one gave you the equation and the second one gave you some figures to plug in. So what have you done so far and what do you not understand?

@jajrussel We are all agreed that for a relative speed of 60mph the difference is very slight. I have based my figures on the travellers measuring the distance across america as exactly 3000 miles and therefore measuring their transit time as 180000 seconds and their speed as 60mph. You have mentioned 'invariants'. Invariants means the same for all observers, not just the two in relative motion but others (eg in a helicopter flying above the vehicles) There is only one such invariant that is c the speed of light. However the 60 mph is the same for both the observers in the cars and those at the side of the road. Both observers subject to the same relative motion will observe the same speed (apart from the opposite signs) ie 60mph. So this is not exactly invariant but it is the same for both observers here. So swansont and JohnG will observe the transit time being a tad longer (as noted) and they will measure the distance across America as a tad longer than 3000 miles so that the figure of exactly 60mph is maintained. It is important to realise that both distance and time must change to maintain this equal value of relative velocity.

Well you said is was N13500 If you are asking what does overhead mean, it is the amount of money you must spend to maintain a shop, market stall or whatever the trader has. You said he was a shopkeeper so his shop costs him the same amount , whether he sells anything or not (electricity, water, perhaps rent, insurance etc)

Well that is not the way we learn, teach or practice Science from an early level. Take for instance the typical highschool maths questions 'A mass M is suspended from a light inextensible string...............' A rigid rod of length l supports two blocks....' 'A frictionless pulley.....' Yet these methods yield reproducible results adequate for many purposes. People every day bet their lives on calculations known to be inaccurate or based on known false assumptions because observation (experience) has shown that those calculations are adequate for purpose. Sometimes we have results without explanation and scientific theories are offered a conceivable explanations, such as the reversing magnetic fields of Earth, but we do not rely on them because they also predict effects we do not observe. Sometimes we have theories that offer apparantly good explanations and the authors of better ones struggle against the establishment to promote them, (isotasy v plate techtonics for instance), but usually further observations come to light which weed out the inadequate. These days we seek corroboration or exception more actively than ever, because we are aware of this. But we should always be wary of specifying how any theory 'should be' . I said 'adequate through experience' above this is a process that gives us increasing confidence in a theory as we make more and more observations that agree with its predictions, but fail to find exceptions. This is the case with Quantum Mechanics, And I commend it to the House.

http://www.physics.smu.edu/scalise/www/misc/crackpot/crindex.html

I was asked to explain this comment I made in another thread, but since discussion would be off topic there I have started this one. I'm sure I have missed many obvious thoughts so comments from Engineers would be welcome. Paradoxically engineers use centre lines both because something balances about the centre line and because things are irregular about the centre line and for other practical reasons as well. Working through my examples 1) If we drill a hole or bang a nail in we want to mark the spot where to place the point of our drill or nail. This is at the intersection of two centre lines. 2) So let us say we are going to drill these holes to bore out the cylinders in an engine block. It is mechanically important that these are in a straight line to balance the cranking forces, so we reference them to a single line, the centre line. 3) Another good reason for using a centre line is that the edges of the workpiece (the block will be a rough casting) so the edges do not form good reference lines. Nor will the corresponding edges of any two blocks be the same. 4)Of course the edges may be straight and regular, but they may taper. They are therefore then difficult to measure from. Do you measure at right angles to the edge or what? Measurements are usually made in right angled (rectangular) coordinate systems. 5) Say I have a beam that I place on two supports so that as shown in (5a). If I now place a heavy weight on it so it bends as in (5b), the top surface is now shorter than the bottom surface. 6) So there must be an intermediate point where there is no change of length as in (6). It can be proved that this point is the centreline of such a beam. So it makes s ense to measure from the line of zero change. 7) Because the shortening of the top surface is caused by the ebnding introducing a compressive force and the lengthening of the bottom surface by a corresponding tensile force again there must be a line where there is zero distorting force and this again lies along the centre line. Another name for this is the neutral axis. 8)Returning to fig4, if that is actually part of a tapering block or cut in earthworks or whatever and I need the volume, I can calculate the volume as an area times a length. But what area and what length? Well it turns out that if I take the average of the two end areas and multiply this by (you guessed it) the distance along the centre line the product gives me the volume.

Turnover = sales ie it is just another word for total takings. Annual turnover = annual sales = (average weekly sales ) x 52 Net Profit = Sales - Costs Costs = Cost of goods + overheads

Doesn't that give you a nice warm feeling. +1

This would seem to me to be a correct statement. Let us say you are referring to your 60mph road speed and you three are driving in convoy. The point is that the velocity v in the formulae is not 60mph, it is the relative velocity between your three convoy cars, which is precisely zero. This, of course, leads to zero time dilation between their clocks observed by the three drivers. So they measure their transit time across America as 180000 seconds. However since swansont was standing talking to JonG by the side of the road, their clocks measured a transit time of 180000.00000001 seconds.

Are you sure you posted this in the correct thread? Where did the'rocekt experiment' come into this?

Of course having got values for L and W you can check to see if they correctly match the conditions of the problem.