studiot

That was the purpose of this thread - Spreading the Good News. Malicious code needs to spread to be worthwhile to its originators. We can all collectively fight this by broadcasting the antidotes.

This thread was meant as a warning. Even the most competent can be caught out. Yes you can do some of these things if you feel competent, but there is code associated with the link in some cases and as Sensei +1 points out simply arriiving at the link address from your own IP is useful information to a hacker. If you feel competent and want to investigate, you would be safer taking the address to a public net service, they are free in the EU and UK, and protected by a reset protocol such as Deep Freeze. We should all work together to combat those who subvert innocent folks.

Received this Email, this evening Please be aware of this scam as it is highly realistic. Except for the fact that PP have never had the particular email address they sent it to.

I hope you are right, but fear otherwise for the underlined bit. http://www.scienceforums.net/topic/107448-viruses-ransoms-trojans-spys-and-other-malware-into-corporations/

What is clear from this quote is that r x p is not a fundamental definition of angular momentum in the eyes of the author, since it is specified as being only applicable to a point particle, and nothing else. So there must be a more comprehensive definition which includes the angular momentum of thigns which are not point particles. I also note the linked Wikipedia of point particles which clearly indicates that r = 0 for a point particle and you so rudely denied. You were offered, and declined, convivial rational discussion at this more fundamental level.

Thank you Several of my former clients were retired university lecturers who undertook A level marking for some extra income. Most have since quit in disgust in disputes over unsatisfactory marking schemes and model answers, not to mention the online antics with computer based inputting. (Did you see that recent question here as to why everything needs to be online?)

I looked in the back of my cupboard and found one remaining olive branch I had overlooked. As I understand the situation You think you have discovered some circumstances where conservation of angular momentum is not true. I agreed with you there are such circumstances. So I tried to explore these circumstances with you. In fact I put substantial effort into this. But as far as I can tell from your minimalist responses you were not interested in such a project. I will make one comment on your assertion that the equation L = r x p is a definition of angular momentum. It is not a definition, it is is a very simplified formula for calculating the angular momentum in some circumstances. So I will leave you to consider my morning cup of tea that I have just poured. In order to stir the tea into the milk I pour the stream of tea into the cup along the inside of the cup. Thus it has linear momentum as it leaves the spout and enters the cup. But it has zero angular momentum. As the tea enters the cup it swirls round and ceases translation (the cup does not go anywhere) But in swirling round it gains angular momentum but looses linear momentum. So we have an everyday system that starts with all linear momentum and ends with all angular momentum. So neither are conserved since linear momentum is destroyed and angular momentum is created. Do you wish to discuss the analysis of this real world situation or do you wish to close the thread?

Yes I saw this morning when I opened your spoiler that you had reached the 3D conclusion before I did. I don't know if this was obvious in the textbook (for instance a chapter on 3D trig). Also this question demonstrates the need for proper labelling of diagrams. If the pretty picture was a scan from the textbook I would take proper labelling over beauty any time. I think the authors were unduly lazy.

The only way I can make this work is to assume the pretty picture with the shading and no identifying letters is not a plane figure but a 3D picture of an internal corner. Labelling points A by angle alpha, B by angle beta, C at the top and O at the centre/ origin / internal corner. OC forms a vertical axis cut by plane CAB at C. Triangle CAO is a right angled triangle (at O) as is CBO, but triangle OAB is not. Thus triangles COA and COB are at right angles to the plane of AOB. Thus in triangle CBO y = h/tan(theta) This give three known angles and one side in triangle AOB, which can now be solved by the sine rule for d.

I suggest you ignore the right angles. So you can easily establish alpha and beta, giving the angle opposite d (call it gamma). This will give you d by the sine rule as asked for if you know y. To get y by the sine rule you need to solve the upper right hand triangle. Can you do this? I made alpha = 23.57818 degrees beta = 29.55192 degrees gamma = 126.86990 degrees

Exactly. So something is wrong with the problem, or with your reading of it.

So what does that make the angle opposite the side you have labelled as d in the bottom triangle? In other words what does that make angles alpha and beta?

Fair comment. Thank you.

How many right angles in a straight line?

Ask Mr Microsoft why Office has to be online these days. Another scenario is the overpowerful IT department that insists everything is run from a central corporate server (which doesn't have to be on the web) but then the bean counters get offered an 'irresistable' deal on online (cloud) storage so buy that instead of paying to upgrade their own server.

I presume that those right angles at the centre are a mistake? This problem is known to surveyors and navigators as the resection problem. that is they determine their position by taking bearing sights on three objects of known position. One solution method is due to Tienstra https://en.wikipedia.org/wiki/Tienstra_formula Here is an online solver to play with http://mesamike.org/geocache/GC1B0Q9/tienstra/

Hello, Tar The situation of reaching orbit is unlikely as no normal gun can send a bullet at anything approaching escape velocity. Without reaching escape velocity or having the benefit of a sustained propulsion behind it any projectile will eventually return to Earth

This is homework help so you need to post both the actual problem and at least one of your failed attempts (perhaps in outline). Your jpg has not taken so I can't see the problem. There is a moderator online right now (swansont) so ask him for help getting your question posted.

You really should have provided more information. So I have offered an amendment to your OP and title, hope you don't mind. A bullet fired horizontally from a gun and a bullet dropped from the same height as the muzzle will reach the ground after the same journey time, ignoring the curvature of the Earth and wind resistance. This is because both bullets start with zero vertical velocity and are only subject to the same (vertical) distance to ground and (vertical) acceleration.

Hello pavel, good to see you back again, you often add something useful to threads. With regard to part (b) of your question the answer is not so simple. It depends upon what you are sampling. Here are two examples. 1) You have divided a cornfield into squares for testing yield. Placing all the tests sequentially in one corner may hit fertile or stony ground. 2) You are a buyer checking a box of apples for % rotten apples. A rotten apple is much more likely to infect neighbours than remote apples. In both of these sampleing sequentially may well produce different resutls from a proper spread of testing. Sampling theory developed from real life situations like these and it is obviously important to ensure that the sample is as representative as possible of the whole sampled population. This is why sampling theory is a huge area in its own right.

Thank you both for this entertaining game of ping-pong and also for introducing me to Bell's paradox, which I had not heard of before. +1 each I don't actually see the issue with Bells. Even in a purely Newtonian universe if you accelerate a compound object hard enough and long enough you will break it. The pair of spaceships and the gossamer thread constitute a compound object. Just as a bag of blood in a centrifuge does. What happens to the bag of blood if you overspin? It demonstrates the Principle of Equivalence very nicely.

In order to have a loop you require a continuum that can contain such a loop. Closed timelike curves can occur in the geometry of spacetime, which is the continuum in which timelike curves of any description occur. You do not therefore need extra dimensions. For a continuum with one totally independent time dimension, there are not enough dimensions to accomodate loops. So you would have to postulate a continuum with additional dimension(s). This is just the geometry. In order for an inhabitant of either conintuum to traverse the loop you would also need an equation of motion that included the loop in its solutions.

In my first post and several times subsequently I asked you very politely to define radius. I expect you have seen the type of fairground ride where a seating capsule or capsules is rotated on the end of an metal arm or arms of fixed length L about a central hub. So here is my question to you about your claim: What is the radius of rotation of this mechanism ? I am not seeking numbers, symbols in a formula will do. Please note this forum requires you to answer this question fully in a way that I can use to calculate the radius. "If the radius is zero then the angular momentum is also zero by definition." There is where you are showing lack of understanding. Consider a rotating sphere with non zero radius. Every point in that sphere has zero radius but is a mass point. The overall sphere has an angular momentum which is the sum of all the individual angular momenta of all the mass points about the central axis of that sphere. The overall mass of that sphere is the sum of the masses of all the mass points. If, as you say, all the mass points on the axis of that sphere have zero angular momentum they contribute nothing to the angular momentum of the sphere, but they do contribute mass to the mass of the sphere. How can this paradox be resolved?

The radius of a point particle is zero, by definition of a point particle. Point particles may possess angular momentum. You didn't address my second point. What is the radius of a twisted space curve?