studiot

OK It's good that you only want hints not answers since we don't provide full answers in this forum. Surely you must have had some thoughts? Here are some hints. 1)What is the relationship between work and energy? [math]Work = \int {F(x)dx} [/math] So can you see what to do to obtain the force from the energy and the function which describes x? 2) The total energy = E What forms may the total take and how does that energy budget compare with the values of U on the graph. Have you, for instance, enough energy if E = 5 to traverse the whole graph? Again your graph shows potential energy as negative for certain values of x. Can PE ever be negative?

These books listed are accessible from middle high school through senior high and onto first degree level, except the last which took me right the way through O level, A level, Degree, Professional Institution and Postgrad and is still going strong. Atkins Molecules. (P W Atkins) a really delightful book described by some as the most beautiful chemistry book ever written. The Mathematical Mechanic (Walter Levi) a brilliant melding of maths and physics. Levi sets out to prove mathematical theorems in as many physics ways as possible. Cats Paws and Catapaults (Steven Vogel). Compares and contrasts the approach by Nature asd Man to the solution of important engineering problems. describes materials science v muscle action and much more. From Calculus to Chaos (Acheson.) The development of non linear dynamics brought to life and a uniquely simple way. The Penguin and Dictionary of Curious and Interesting Geometry. (David Wells) Fascinating reading for a dark winter's night. Did you know Napoleon was responsible for a theorem in Maths? The Day the Earth Nearly Died (Michael J Benton) Wonderful balanced roundup geological study of the five main mass extinctions in the Earth's geological history and more. and finally that piece de resistance A Compendium of Mathematics and Physics (Myler and Sutton)

No. The current that charges/discarges a capacitor flows in conductors attached to the capacitor. It is a conduction current. This current is external to the capacitor. The displacement current is the current that 'appears to flow' in the region of non conductance. That is within the capacitor. It is internal to the capacitor.

There is more to it than this. Others have already said that you should be comparing the indefinite integral and the definite integral. The anti-derivative is something else again. The definite integral is a pure number it is not a function. The indefinite integral is a function. I suggest accepting these for now until the significance of this distinction really becomes apparent when you proceed to the applications of calculus. And well done (+1) to endercreeper for being prepared to put up the work.

The title of this thread is permutations and combinations. John and imatfaal have correctly identified that combinations are the appropriate operation here, since order does not matter in this question. John has further identified an ambiguity with the posing of the data. You may wish to know that The number of combinations of n things r at a time in which p things always happen is n-pCr-p The number of combinations of n things r at a time in which p things never happen is n-pCr

Well it's good to experiment as this is the ultimate test of a theory. I think you will be in for a suprise or two when you do your experiment. What do you know about the pressure in a gas (or any other fluid for that matter)?

Perhaps I should expand on this. There is no magnetic field associated with the generation of displacement current. It is a purely electric effect. However a magnetic field appears as a result of the changing electric field. The displacement current may be input to the usual formula [math]B = \mu {i_D}[/math] To calculate the value of the magnetic field Where iD is the displacement currrent.

The op actually specifies better in post#3 However there are many pitfalls in referring to zero volts as earth. Take for instance an integrated (HiFi ?)amplifier. It will have zero volts rails for the preamplifier and zero volts rails for the power amplifier. It may also have separate + and - supply rails. It may have red and black speaker terminals. It may have a terminal on the case marked Earth. These may all be on the same circuit board, as in the Marshall 8000 series. If multichannel it may repeat some or all of these for each channel. Which one of these (if any) can be considered as earth? Which one do you connect the screen earth of a connecting cable to? What if that cable is an blanaced or XLR connector? How do you avoid earth loops with this amplifier? I think you will find that you will find that you need all of my post#9 to properly answer these questions and wire this amp up correctly. Another feature of some circuits is an earth or ground plane. I think the OP is reasonably asking for guidance that will carry him into the future and that there are many variations that come to light when you delve deeply into the subject.

My PP3 battery does not have a terminal marked zero. It only has terminals marked + and - I think it is important to be clear and precise about statements as many learners take forward incorrect or partially correct ideas as a result of loose talk.

Be careful of associating an earth with low resistance (or impedance). This is not always the case. For example what is the resistance of the granite underlying New York? Secondly, earths are (or should be) designed to carry no (zero) current in normal circuit operation.

It's more than that. A reference point is no use unless it can be relied upon not to change when other things do.

An earth (british english) or ground (american english) is a circuit element. The idealisation only exists in theory, but the planet Earth can come pretty close for some purposes. Most circuit elements have at least two terminals also called nodes. For example a resistor has two, a transistor has three external terminals. An earth has only one terminal. Its properties are such that however much current (or charge) passes into or out of the earth, the potential of the earth node does not change. This potential may not be actually zero, but we usually find it convenient to call it so. This is like using sea level as a zero for elevation. sea level is not actually at zero distance above the centre of the earth but we call it zero and reckon elevations above sea level as positive and depressions or depths below sea level as negative.. This propertiy of maintaining itself at a fixed voltage has three main uses. Firstly as a protective earth, as in house wiring. Secondly since the voltage never varies there can be no electromagnetic radiation (radio waves) propagating in an earth. So it is used for screening and shielding. Thirdly for your purpose we can use it as a voltage reference point for the rest of the circuit since, no matter what happens in the circuit, its potential does not change. Any other point in the circuit is liable to undergo change of voltage due to circuit action whether wanted or unwanted. In this respect, 'earths' are better than sea level because there is no equivalent phenomenon to tides in electricity.

Hello, h4tt3n, a most interesting post. Your smelting process sounds similar to that employed in the Blackdown hills in Somerset from the (pre Roman) iron age to medieval times. I'm not sure how thick you ingot was but it seems a bit thicker than they could manage. You say this iron is soft and ductile, Lump mass of cast iron is relative insensitive to rust and survives for centuries as bollards etc on sea walls. But cast iron is brittle, not ductile. A structural factor that promotes corrosion is lack on homogeneity so perhaps your product is particularly homogenous? Have this happened to many samples. After the break I will look out our blackdown local experts to compare notes.

I should have added that if the load that provides the sideways push or pull is applied at an angle, the load resolves into two components. One component is parallel the direction of the friction and is the only component that opposes it. The other component is parallel to the direction of the normal reaction and does not affect the friction. This component adds (subtracts from) to the normal forces acting, but does not appear in the relation friction = coefficient times the normal reaction ( or the force required to maintain equilibrium if not at limiting friction). Also, when the object is moving, there is a different situation in action and equilibrium no longer applies.

Sorry quoting or copy pasting no longer works, no doubt a Christmas present from Microsoft. However you should have been taught that Limiting Friction = coefficient times the Normal Reaction. The Normal Reaction is always at right angles to the frictional force. However the frictional force is variable. That is it is only as large as it needs to be to oppose motion. So it is always exactly equal but opposite to the pulling or pushing force up to the 'Limit' when the object starts to move. At the limit the frictional force equals the coefficient times the normal force. Before this the frictional force is less than this.

Does it and do you? How far does the fixed end of the vine move? I make it zero distance. So what is the work done at the fixed end by the wind? If F is the total force applied to a total length L of vine then we can say that each metre is subject to a force of F/L Since F is constant and L is constant F/L is constant or fixed. Please note I have used L for the fixed total length and l as a variable so that an integratiion can be performed. I am sorry if this has confused you. But remember that distance is also a function of l and work is the product of all the little forces acting over a variable distance. Work = a fixed force times a variable distance. = F times a function of l So the integration is [math]W = \int_0^L {\frac{F}{L}} (afunctionofl)dl = \frac{F}{L}\int_0^L {(afunctionofl)dl} [/math] Have another look at my post to see if you can sort out what that function is. Hint horizontal distance is due to the sine, vertical distance is due to the cos, but is not so simple to integrate. I'm away for a week now so I will have to let others sort out this question with you. good luck

This is really difficult to discuss without doing the problem completely for lightburst. However I feel that discussing path integrals along curved paths is overkill for this question and someone should have said so. I have already indicated that it cannot be done without considering potential energy. The detail is just geometry, in terms of the variables given. To examine the work done by the wind in terms of the geometry and the variables given first realise that the weight (Tarzan + Jane) play no part in this, as stated by the problem. Presumably becuase their dimensions are small compared to the length of the rope L. So let us consider and elemental length of rope, dl, at a distance l from the pivot. The wind force per unit length = F/L and is horizontal. Each elemental length dl travels a horizontal distance lsin(theta) + lsin(phi) against (or with in part b) the force F/L dl so the work done is F/L * l(sin(theta) + sin(phi)) dl If we integrate this from l = 0 to l = L we have the total work done by the wind. OK lightburst, it is up to you to turn this into a polished solution. Note you cannot just claim a wind force F at some unknown point in the diagram as you have done, although you could replace my integral by a suitable averaging process and a single force, but you would then have to calculate that.

We are clearly on a different wavelength here. Let us suppose that Jane is 0.75 metres above Tarzan and 1.00 metres above the nadir of the swing. (In the picture Jane is definitely shown as some distance above both.) Now if Jane starts with zero velocity and no wind she will execute pendulum motion with all the PE given by 50 . 1.g This will be just enough to raise her 0.75 metres above Tarzan on the other side of the river. However the wind is blowing and so Jane only reaches the level of Tarzan, ie 0.25 metres above the Nadir. So the difference in PE is lost in overcoming the wind resistance, ie 50. 0.75 . g This must equal the work done by the wind resistance If the wind is any stronger Jane will not reach Tarzan without an initial KE, to add extra energy. This extra KE will just compensate for the extra wind resistance. So the graph of launch speed v wind resistance does not pass through the origin.

You have mentioned potential energy in your equation (delta PE), but have not mentioned gravity in your list of forces? Is there not a change of elevation to take into account?

OK let us move on from a resistor and capacitor in series to a resistor and capacitor in parallel. I have shown two identical blocks of cross sectional area A and distance between the ends D. They are connected in parallel to the same (alternating) voltage source. If the resistance is R and the capacitor has capacitance C then i1 = V/R flows in the resistor [math]{i_2} = \frac{{dq}}{{dt}} = C\frac{{dV}}{{dt}}[/math] in the capacitor. We call the current through the resistor the conduction current and the current through the capacitor the displacement current. Physical current does not actually flow through the capacitor, it is just as if it did because qharge flows into the end plates and out again as the voltage alternates. Now let us suppose that the resistor is filled with a conducting medium of conductivity sigma and the capacitor is filled with a dielectric of permittivity (dielectric constant) epsilon. Now for both the capacitor and the resistor the electric field E is the voltage V divided by the separation d E = V/d. Inside the resistor the current density (symbol J1) equlas the current divided by the crossectional area and also equals the conductivity times the electric field [math]{J_1} = \frac{{{i_1}}}{A} = E\sigma [/math] The capacitance of a parallel plate capacitor is [math]C = \frac{{\varepsilon A}}{d}[/math] Also V = Ed So [math]{i_2} = C\frac{{dV}}{{dt}} = \frac{{\varepsilon A}}{d}\frac{{d(Ed)}}{{dt}} = \varepsilon A\frac{{dE}}{{dt}}[/math] We can also get an expression for the current density due to the displacement current. [math]{J_2} = \frac{{{i_2}}}{A} = \varepsilon \frac{{dE}}{{dt}} = \frac{{d\varepsilon E}}{{dt}} = \frac{{dD}}{{dt}}[/math]

Yes, you have indeed. But I think that this is the first time you have stated it (added the condition) I was observing that all other times you have eschewed the function. I don't view the [math] \mp [/math] sign as problematic. Again this is because it means either... or not both as some read it. Something cannot be both minus and plus at the same time. So saying minus or plus something implies that there are (at least) two possibilities. I consider it unwise to put in front of the radical because it implies information we may not have. So we have three situations 1 We know the answer is positive so we use + (or nothing if we like) 2 We know the answer is negative so we use - 3 We don't know which sign the answer has so we use [math] \mp [/math] Note I have use the inverted form for variety. Edit a final thought, how do you interpret the expression [math] \pm \sqrt z [/math] where z is complex since complex numbers do not associate with a sign? Obviously it the [math] \pm [/math] has to be associated with the operations of subtraction or addition but then you are assigning a different meaning to the symbol.

You surely mean extensible not flexible?

Fair comment, my finger slipped as I was getting tired of typing. How can a negative radical sign applied to the principle square root denote a positive value? You either have the positive and/or negative signs before the radical, which is then defined as always having a positive value, or you allow the value itself to be positive or negative, but not both. If the latter were to be the case the sign before the radical is worse than redundant since it can oppose the sing of the value. The radical sign by itself denotes not both roots, but either root. It is up to the employer of that radical to determine which particular one. Yes the function has an inverse. But the number 16 does not in the snese that it does not have an operation that can send it back to whence it came. Of course it has a reciprocal but that is a different thing. And no, you have not stated many times. You have argued, many times that you are not discussing functions, that they are irrelevant etc.

Let us examine the logic of this. If the roots may be positive or negative then why does the radical also need a sign? Surely if you negate the radical of a negative root you obtain a positive value? What I don't understand is why so are so wedded to the idea that [math]\sqrt {16} [/math]may not equal -4 ? You have not presented a single reason beyond 'others also do it' for this view. I can readily understand the mathematical benefits that flow from choosing only positive values for the square root function. It guarantees an inverse function and fits in with a heap of other mathematical structure about sets and functions etc. But 16 is just a number, it is not a function. and there are many ways to get to 16. 16 = 15+1 16=17-1 16 = 8*2 18= (-8)*(-2) and so on Do you also reject these?

But it is not usual, and hasn't been since at least 1880. I have no data prior to that. Maybe current vogue in some places is to call it something different. IMHO that is just counterproductive. There is so much genuinely new to uncover, it is a waste to simple rename the old.