studiot

I sugest you reconsider your answer to part (i), which I have underlined. Setting aside the inefficiency of light conversion (typically less than 10%), and given the configuration mentioned what % of the light from the source reaches the reflector? This should then help with ideas for part (ii)

When certain individuals are firing on the range the safest place to stand is by the target!

I agree, Doe anti mean in some way inverted? A mirror image perhaps? Take a simple plane mirror. It appears to 'invert', left to right but not top to bottom Of opposite 'sign'? mass has no sign Some version of chirality perhaps? Just as we cannot assign an absolute coordinate system to space we cannot then assign a chiral structure either.

The horizontal distance between A and the impact point is x, not 50+x. It is 50+x from the launch point to the impact point. I get 50+x = 98.29825t horizontal flight y = xtan42 = 0.9x Triangle y = 68.829t - 4.9t2 vertical flight Three equations, three unknowns. Easy solution over to you

You are correct that the origin of magnetism in material matter arises from the motion of its electrons. This is because electrons carry an electric charge. Any moving electric charge generates a magnetic field, and the electrons are always in motion. The electric current in a copper wire is due to electrons in motion and this also generates a magnetic field due to this, however not all material objects carry electric current and yet respond to magnetic fields and some actually produce their own field. Except for ferromagnetic materials (Iron, nickel and a few other substances) no substance exhibits magnetic effects unless it is situated in an externally generated field. Ferromagnetic materials can form what are known as permanent magnets. That is they remain 'magnetised' when there is no external magnetic field. So they can generate a magnetic field in their own right. Copper is not a ferromagnetic material and cannot do this. The magnetic effect in copper wires is due to electrons that are 'free'. That is they are not part of any particular atom and move along the wire, whilst (obviously) the copper atoms stay in place. Electrons belonging to particular atoms are very important since not only do they determine the chemical properties of that atom, they also determine its magnetic properties. The electrons in an atom are circling round the nucleus of the atom in quite specific manner, not "loosely round an object". Because they are travelling in orbit around the nucleus they generate a (small) magnetic field. In the atoms of many substances there are as many electrons going one way as there are going the other so the effects cancel out and there is no net magnetic field. However in iron atoms there are four electrons more going one way than there are going the other around the iron nucleus so these four electrons generate quite a powerful net magnetic field. Finally some atoms have one or two unbalanced electrons orbiting, but the atoms are arranged randomly so that the fields of one atom cancel the fileds of nearby atoms with a net zero result. How are we doing so far?

Hello, you have posted this in homework help, but you don't seem to have a homework question. Are you really asking for a better explanation of magnetism or do you have an associated question. Please clarify.

So the projectile travels a horizontal distance = (50 + x) where x is to be determined ( and a vertical distance of y, also to be determined). Have a look back at your first projectile question and draw a sketch similar to my one there and then look at the equations I started with. Can you find a pair of simultaneous equations from this that you can solve for x and y?

There is another approach, that leads to the same answer, but you said you were considering energy. Since the 200N force is constant it imparts constant acceleration to the box. So if you draw a velocity-time graph; The box starts at time t0 = 0 at zero velocity v0=0 and is then accelerated by the constant force with acceleration a1 for a distance of 10 metres until it enters the friction zone at a velocity of v1. The box is then subject to a constant acceleration a2 (which works out negative) due to the difference between the 200N force and the opposing frictional force for a further distance of 10 metres until it reaches v2. You have enough information to solve the system this way. Have you covered this method in your studies?

part c? Well if this was an exam I wouldn't get may marks since I didn't read the question properly. Red face. You have correctly identified that velocity is a vector and may be resolved into horizontal and vertical components. Further this is also true of acceleration and distance. There is only one acceleration, that due to gravity which is vertical and downwards. There is no horizontal acceleration (ie it is zero horizontally) So we used the formula s = ut + 1/2 f t2 to solve parts a and b. Note I have used s for general distance, and f for acceleration. This allow the use of a, b, c for angles instead of Greek letters and also allows us to put in the distance in the appropriate direction. To solve part c we introduce a second formula v = u + ft where v is the velocity at time t and u is the initial velocity. Horizontally Vx = 15cos(50) + 0*t = 9.6418 m/s Vertically Vy = 15sin(50) + (-9.81)*(2.153) = -9.630 m/s ie downwards. Total velocity V = sqr(Vx2 + Vy2) = 13.627 m/s You got all this but I see you were unsure about the angle denoting the direction of travel. This angle, call it b, is in the fourth quadrant since the travel is down and to the right. We obtain a fourth quadrant angle by calculating a first quadrant angle © from arctan(Vy/Vx), ignoring any signs and subtracting from 360. b = 360 - c This is consistent with the velocity diagram I have shown superimposed on top of the trajectory diagram below. Have I completed the job this time?

I'm glad there's someone else on the planet still prepared to read books.

Talking of books you might like the following (Cambridge 2008) On Space and Time Edited by Professor Majid Essays by Connes, Heller, Majid, Penrose, Polkinghorne, Taylor on the title subject.

Gosh, Gwiyomi, you sure must like doing arithmetic. I make t = 2.153 seconds and x = 20.76 metres So pretty close to your results. For your information about projectile problems, The path of the projectile is fixed by the launch speed and angle alone. If you launch up and to the right as positive (usual x and y axes) then the position at any time t is given by a pair of equations x = Vcos(a)t y = -1/2gt2 + Vsin(a)t where a is the launch angle, and V is the launch speed. That is the x and y coordinates at any time in the flight are given by those two equations. It is only a couple of simple lines to derive these so you should not try to remember them, just how to get them. If you enter your value of y = 2 (and V and a) into the second equation it yields a quadratic in t. The two solutions correspond to going up and coming down. Knowing t you can substitute into the first equation to find the x distance at this time.

edit I noticed an typo in my asymptotic series, sorry [math]A(z) = {a_0} + \frac{{{a_1}}}{z} + \frac{{{a_2}}}{{{z^2}}}.... = \sum {_0^\infty } {a_n}{z^{ - n}}[/math]

I'm sorry to tell you that you are not right. It would be very helpful to link to the article in question. Are you thinking of real or complex variables? Power series have the form [math]P(z) = {a_0} + {a_1}{z_1} + ... = \sum {_0^\infty {a_n}{z^n}} [/math] They may be convergent or divergent or convergent within some 'radius of convergence'. Asymptotic series have the form [math]A(z) = {a_0} + \frac{{{a_1}}}{z} + \frac{{{a_2}}}{z} + ... = \sum {_0^\infty {a_n}{z^{ - n}}} [/math] They do not converge after a finite number of terms, they are also infinite series. However as each successive term involves a higher power of the the variable the series get closer and closer to the desired function as n tends to infinity. That is the difference between the asymptotic expansion series and the desired function tends to zero as n tends to infinity. If you compare the two forms you can see that a power series converges for |z| < 1, all other things being equal, whilst an asymptotic series will converge when |z| > 1

To save you some trouble, the inverse (in your sense) of the sine function is not the cosine function it is the "angle whose sine is the given value" and often called the arcsin. So the pair is:- if y = sin(x) then x = arcsin(y) edit the arcsin(x) is also commonly written sin-1(x) - they refer to the same thing. Just to be clear the reciprocal of the sin is the cosecant. 1/sin(x) does not equal cosin(x) it equals cosec(x) I would warn you further that what you are proposing is a tail chasing exercise IMHO.

Then the force is not constant and you need calculus to solve the problem.

Well the only way this is possible is for you to have misread the problem (or they wrote it down incorrectly) Should you original statement have said x = t2/3 that is time squared not cubed?

You are told that the force causes the body to move along the x axis so that distance, x = t3/3. Do you think this force is constant during this movement? What do you know about the relationship between motion, distance covered, velocity and acceleration? Do you know anything about the relationship between force, mass and acceleration? If you do not know these things I will help, but why are you attempting this problem?

You haven't done any calculus in your working. Did you see my edit in my last post, asking if you understand calculus?

This question, like all your others, belongs in the homework section. However the rules here are that you must show you working, right or wrong, to receive help. So post your calculation. Edit I think you need calculus for this one, have you done any?

Sorry I should have included one of these.

And everything went blank?

Considering what you have revealed about your status elsewhere I guess that you have not yet studied vectors in detail. So you may be asking what is the dot product all about, as distinct from what is the particular dot product. If that is the case then the following may help. Ordinary numbers (scalars) have one single type of product. 5 x 3 is always 15 and that is all there is to it. Vectors, on the other hand have three distinct types of product. The simplest is the product of a scalar and a vector eg (aZ ) which results in another vector a times the magnitude of Z but along the same line of action. This product is called the multiplication of a vector by a scalar (it is not the so called scalar product) The second also results in another vector and comes from multiplying two vectors together. This product is more complicated as it results in a single new vector that is at right angles to the plane containing the original vectors and of magnitude Z = X x Ysin(a) where a is the angle between them. This product is called the vector or cross product. The product you are asking about is called the scalar or dot product. The result of the dot product is a scalar (with no direction) of magnitude m = X . Y cos(a) where a is again the angle between them In your original question F and ds are both vectors. Note I have used bold to show vectors, a common convention. Does this help?

What I gave before was a definition not a formula. The formulae connect measurable quantities (observables) like the focal length of the lens, the size of the object and its distance from the lens. If you know these formulae you can use them to deduce the size and therefore the magnification of the arrangement. In many applications the image is virtual or at infinity or both so the image distance is not an observable. I am not sure what you mean by 'differentiate between angular and linear magnification'. You measure the observables and use an appropriate formula to calculate the magnification required. In many cases the angular magnification reduces to the same as the linear. So you really do need to describe the setup for further information.

Note that imatfaal said the tangent of one and the angle. We are assuming the small angle approximation here and further both should be tangents. There are three formulae, depending upon where the image is and its type. Are you studying these formulae The angular magnification is also known as the magnifying power. M and is defined as: [math]{\rm{M = }}\frac{{{\rm{visual}}\;{\rm{angle}}\;{\rm{subtended}}\;{\rm{by}}\;{\rm{image}}}}{{{\rm{visual}}\;{\rm{angle}}\;{\rm{subtended}}\;{\rm{by}}\;{\rm{object}}\;{\rm{when}}\;{\rm{viewed}}\;{\rm{directly}}}}[/math]