Recent content by Tommo1

Why don't you just convert all those formulae into one. X = 2P / (kv) X = compression of the spring P = power k = spring constant v = the speed of the escaping exhaust gas Way back Dulrich gave values in his example that you can use to check it with.

This problem is a combination of two things. 1. An explosion: exhaust gases being pushed in opposite directions by both rockets. This is where the energy is going. This follows all the traditional momentum and force stuff. The gases are pushed and moving so work has been done. 2. The spring in...

Here's how to do it using only V=IR and series and parallel resistors. Combine R4 and R5 in parallel 1/Rt=1/R1 + 1/R2 This gives Rt=3.00 ohms Now add R1 + R2 in series = 18.48 ohms Then combine that with R3 in parallel to give 5.80 ohms. http://www.xphysics.co.uk/x/E1.jpg This is...

Simplify the circuit... R4||R5 = 3.00 ohms (R1 + R2)||R3 = 5.80 ohms Then work back through the circuit using the voltage divider... V1/VT=R1/RT This gives R4||R5 = 5.97 V (They both have this voltage as they are in parallel) To get the currents use the current division equation I1 =...

In reply to Xitami, Georg Ohm did okay out of the mathematical approach. It took Bavaria a while to realize it though. Stan Ulam is impressive too though! As is Stanisław Lem, an idea: explosive!

Hi Atran, this problem doesn't require algebra, geometry, calculus or trigonometry. It is only arithmetic! So get a pencil out and a scrap of paper. Here's another example... R1=10553063310 ohms R2=154064581051 ohms R (total) is still a whole number.

You engineers will be the death of mathematical exactitude! R1=25461230 ohms R2=25375670 ohms In parallel R(total)=12709189 ohms (exactly)!

Here's a one that's a bit physics and a bit maths but maybe too easy. 1/R=1/R1 +1/R2 for parallel resistors. How do you produce examples of this with whole number values only? e.g. R1=14, R2=35 giving R=10. R1=21, R2=28, R=24 gives exactly R=8. R1=1400, R2=2600 produces 910 ohms.

To solve this type of problem involves Y to delta transformations. See... http://www.allaboutcircuits.com/vol_1/chpt_10/13.html Once you convert the triangles into Y shapes it is easier.

I think that John McGeoch has to be up there with the best of them. http://en.wikipedia.org/wiki/John_McGeoch He was the guitarist for "Siouxsie and the Banshees" but performed in loads of other bands! I would include Johnny Marr(and I still do) but after trying to play McGeoch it feels...

I think that the key to this question is "art" over "maths": you must redraw the circuit in a manner that is less confusing. http://www.xphysics.co.uk/x/R1 Then redraw... [PLAIN]http://www.xphysics.co.uk/x/R2 and again... [PLAIN][PLAIN]http://www.xphysics.co.uk/x/R3 Simples. :)

I know that this is bit late for this thread but... Why don't you compensate for the wind by measuring its speed an firing the rocket at an angle into the wind. This would be similar to a sniper compensating for the wind affecting the bullet. you would need to measure the wind effect on your...