Gareth56

Here’s part of a problem from Tipler Physics for Scientists & Engineers:- In a Sci-fi short story written in the 70s, Ben Bova described a conflict between two hypothetical colonies on the Moon- one founded by the USA the other by the USSR. In the story, colonists are from each side started firing bullets at each other, only to find to their horror that their rifles had a high enough muzzle velocity that the bullets went into orbit. (a) If the magnitude of free-fall acceleration on the Moon is 1.67m/s^2 , what is the maximum range of a rifle bullet with a muzzle velocity of 900m/s? (assume the curvature at the surface of the Moon is negligible). I read this as being what is the horizontal range thus using the Range formula: R = U^2 sin(2alpha)/g where g = 1.67m/s^2 U = 900m/s I used zero degrees in my answer for the reason that the rifle was held horizontally with respect to the surface of the Moon. However in the answer the angle in the Range formula was 90degrees. Surely that’s pointing the rifle straight up with respect to the surface of the Moon. Have I misunderstood the question? If so why is the angle 90degrees?

Oh I see many thanks for the clarification. I suppose to close the matter I'm bound to ask "why are you entitled to do such an operation"? Or again is it glaringly obvious to all by the uninitiated

My problem is (other than being thick in this respect) is how does: F - Tcosx = 0 B - W-Tsinx = 0 become: [math] \frac{B-W}{F} = \frac{T}{T}\frac{\sin X}{\cos X} [/math] I've heard words like divide and ratios but I just cannot see how you divide 2 equations. I can get as far as: F = Tcosx and B - W = Tsinx But it's the steps that lead to: [math] \frac{B-W}{F} = \frac{T}{T}\frac{\sin X}{\cos X} [/math] that have me stumped.

Function ??? I though they were 2 equations.

Ok thanks but what I'm not seeing is how do you equate [math] \frac{B-W}{F} = \frac{T}{T}\frac{\sin X}{\cos X} [/math] When we initially had two separate equations. The problem is I don't know what you meant when you said "All you have to do is take the ratio" (Really sorry to appear thick here)

OK, So I have F = TcosX and B-W = TsinX ; what I'm uncertain about is the mechanics of bringing them together. I knew I should have listened in maths class!

Yes that's it. I don't know how to do that fancy equation thingy you do.

Sorry guys I made a typo, It should read:- F - Tcosx = 0 B - W-Tsinx = 0 Apologies again:embarass:

I just can't see this rearrangment, could someone please explain it? Thanks F - Tcosx = 0 B - Wsinx = 0 all that boils down to tanX = B - W/F I understand that cosX/sinX = tanX but it's the other bits I can't work out.

Thanks for your responses. So going all the way down the "pushing tree" a train moves because the loco pushes against Earth and because the Earth has a far greater mass than the loco the loco is propelled forward?

We are told that "The action and reaction forces between a pair of interacting bodies are of equal magnitude and are opposite in direction" So if a locomotive exerts a force of say 10000N on a wagon (via the coupling) we know that the wagon exerts exerts an equal force of 10000N on the locomotive in the opposite direction, why then does the train move? Doesn't each force cancel each the other out?

Many thanks.

I've often seen in equations cos^2 then some angle so for example:- cos^2 45deg. How does this work on a calculator or do you have to do something clever to work this out? Also what does cos^2 mean (other than cos x cos)? Thanks

I see. I shall always look to where I can stick my "invertible elements", hopefully I won't get arrested! Thanks again ajb and give my regards to Manchester where I was born and educated (in a place on Oxford Road!)

You guys are brilliant. It's so simple if you can "see" these things. 30+ years have dulled my eyesight somewhat Although I'm not too sure how you can just put things in like g/g, clearly it's allowed but why? Also when you say "take out the factor" do you mean take the factor outside the brackets? Thanks again

I know the following involves common factors but I just cannot see how the 1appears in the second equation could some kind soul add an explanation? = mg + m v^2/r .................(1) ??????????? = mg (1 + v^2/rg) ................(2) It's what happens between (1) & (2) that's bugging me. Maths was so long ago!!! Thanks

Many thanks for your input. What it was was I heard an advertisment on the radio about a company blowing its own trumpet for check in times at an airport and was just wondering if they had given us the whole picture Averages can be so much removed from reality I suppose. Thanks again

If you're given an average value and the sample size can you determine from those values what the maximum & minimum values were that gave the average value? Thanks

Thanks for that. I didn't look on them as being fractions

Is the answer to this equation correct? I'm unsure if it should 2g in the denominator of the answer. I think it should be just g. (U sinX)^2/g - (U sinX)^2/2g = (U sinX)^2/2g (The g and the 2g are the denominator of (U sinX)^2 respectivley)

I was just wondering also what data you would need to make a somewhat accurate estimate of which is cheaper. Presumably you would need the calorific value of the gas and the specific heat capacity of the water in joules, obviously one has to ignore the heat loss to the surroundings to a certain extent.

In these days of increasing fuel prices it got me wondering which is cheaper to use to boil 1L of water, gas or electricity? Also what data would you need? I assume the calorific value of the gas in Joules would be needed also the Specific Heat Capacity of water and would you require the volume of gas used? It's far easier to calculate the amount of electricity used you just look at how many revolutions your electricity meter makes.

Wouldn't that temperature reading include some other form of energy (friction?) from the water hitting the sensor at a certain velocity rather than just relying on gravity as what occurs in a waterfall?

Out of interest, how could you do this in the lab?

I read somewhere that water has PE at the top of a hill (waterfall) which is mostly converted into heat by the time it reaches the bottom of the hill (waterfall). So is it possible to measure the change in temperature (if there is any) for water than falls over the top of Niagra Falls and finishes at the bottom of the Falls?