  # Gareth56

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279

## Everything posted by Gareth56

1. So not a case of simply adding the masses of the individual nucleons up!! Many thanks
2. Thanks for the clarification but I'm still a tad confused because the book states the following:- "To see how much mass is consumed, we compare the combined mass of four hydrogen atoms (the ingredients) to the mass of one helium atom (the product) 4 hydrogen atoms = 6.692 x 10^-27kg - 1 Helium atom = 6.645 x 10^-27kg Mass lost = 0.048 x 10^-27 kg What I don't follow is that when determining the mass of the 4 hydrogen atoms I assume you just multiply the mass of a proton i.e. 1.673 x 10^-27kg by 4 to get 6.692 x 10^-27kg, so why can't you just add up the masses of the two protons and the two neutrons which make up the nucleus of a Helium atom to get the mass of the Helium atom (ignoring the masses of the electrons). Is it a case that the book doesn't go into the detail of why the mass of the Helium atom is less than what you would expect by just adding up the masses of the nucleons? Thanks
3. Whist reading a book on how nuclear fusion takes place in stars I noted that 4 hydrogen atoms are converted 1 Helium atom with the resulting mass difference being converted into energy via E = mc^2. In their calculation (ignoring the mass of the electron) they gave the mass of 4 hydrogen atoms as 6.692 x 10^-27kg which I concur with as 1 proton = 1.673 x 10^-27kg therefore 4 protons = 4 x 1.673 x 10^-27kg = 6.692 x 10^-27kg But it's the mass of the He atom that I'm unsure about. It states in the book that the mass of the He atom = 6.645 x 10^-27kg. However I understand a (normal) He atom (again ignoring the mass of the electrons) to comprise of 2 protons & 2 neutrons making a mass calculation of:- (2 x 1.673 x 10^-27kg) + (2 x 1.675 x 10^-27kg) = 6.696 x 10^-27kg This is again shown by the RAM of He to be 4.0026 which again gives the mass of the He atom as 1.673 x 10^-27kg x 4.0026 = 6.696 x 10^-27kg. Is the mass of He given in the book wrong? Thanks.
4. That's what the explanation went on to say in the book but what was confusing me was the Galileo experiment and having seen (in the flesh as it were) the Leaning Tower of Pisa it does seem (to me anyway) that it would have been high enough to show that heavier objects do drop faster in the presence of air.
5. I read in my physics book the following which is confusing me slightly. “Consider a pair of tennis balls, one a regular hollow ball and the other filled with iron pellets. Although they are the same size, the iron filled ball is considerably heavier than the regular ball. If you hold them above your head and drop them simultaneously, you’ll see that they strike the ground at the same time. But if you drop them from a greater height- say from the top of a building you’ll note the heavier ball strikes the ground first. Why?” It then goes onto explain why and I can follow and understand the argument with no problem BUT what is confusing me is that I’ve heard that if you drop two objects (of differing mass) from the same height then they’ll hit the ground at the same time due to the heavier one having more inertia as performed by Galileo Galilei's dropping objects from the leaning tower of Pisa in order to prove that all objects fall at the same rate, whatever their mass. So I’m not too sure why the text in the book tells me that two objects of differing mass will hit the ground at different times. Am I misunderstanding the text in the book or what?
6. Also to be fair there is a "nice" way to give confirmation and a somewhat condescending way to do it. IMHO the latter was the case here. However I'm obliged for the confirmation. I don't think we need dwell on the matter any further. Merged post follows: Consecutive posts merged If you'll excuse my ignorance could you expand on that?
7. Thank you. It may have been obvious to you buy sadly not that obvious to me hence the reason I asked. However thank you for taking the time and trouble to indulge such a dimwit!
8. I came across this in my physics book when discussing friction:- "If the wheels of a car are rolling normally without slipping or sliding, the bottom of the tyre is at rest against the road at each instant; so the friction force the road exerts on the tyres is static friction." Does that mean that when I'm travelling at say 70mph in my car the bottom of each tyre just for a fraction of a second is at rest with respect to the road surface? If so it must be for a very very short period of time!!
9. Many thanks for your time and trouble. It's 99.99% clear now so if you could indulge a dimwit for one more time and explain the last two steps i.e. from -( mC - mE)g = (mE +mC)a to (mE - mC)g = (mE +mC)a I would be eternally grateful. I understand that in Step 4 you multiplied both sides by -1 but not too sure how you got rid of the minus sign outside the left hand side bracket in Step 5 Just popped to my local library to get Engineering Mathematics by Stroud which I understand is a good book for this sort of thing.
10. many thanks Swansont. Can I just ask how do you arrive at $(m_C - m_e)g$ from adding $- m_eg$ & $+ m_Cg$ Sorry about the LaTex this is something in addition to basic maths manipulation that I need to master.
11. My issue is that I haven't a clue how the answer in the book was arrived at as they've assumed a certain level of knowledge on the subject and left a couple of steps out.
12. As I said I changed the signs of the bottom line then added so I got F(subT) - m(subE)g = -m(subE)a -F(subT) + m(subC)g = -m(subC)a and got:- (m(subE) - m(subC))g = (m(subE) - m(subC))a which is wrong but when it comes to dealing with letters instead of numbers my head explodes in confusion. I mean how do you subtract -m(subE)g from +m(subC)???? I need numbers!!!!! I'm sorry D H but your post was too advanced for my simple brain
13. In my physics book it has the following 2 equations:- F(subT) - m(subE)g = -m(subE)a F(subT) - m(subC)g = +m(subC)a To get rid of F(subT) it says to subtract the first equation from the second. here's where I get confused, I was to told (years ago) do this change the signs of the bottom equation and then add both equations but when I try that I don't get their answer of:- (m(subE) - m(subC))g = (m(subE) + m(subC))a --------------(3) Could anyone describe the process, as I don't get what's on the right hand side of equation (3) Thanks
14. I thought that when you got (1.33T1)sin53 which I presume is the same as (1.33 x T1)0.8, then everything inside the bracket had to be multiplied by everything outside the bracket.
15. Thanks for the pointer but sadly what's throwing me is the (1.33T1)sin53 term I just don't know what to do with that. The closest I've got is T1 = 51.7N
16. I've hit a brick wall in trying to see how in a text book they went from:- T1xsin37deg + (1.33T1)(sin53deg) -100N = 0 straight to T1 = 60.1N Any help on the intervening steps would be appriciated. Thanks
17. For interest I didn't ignore the "new word" that had been introduced to me by swanshot to my vocabulary. I also did check out the You Tube website wherupon I observed the videos that you allude to above, however I was hoping that there would be a video of a demo using the tracks as in the diagram above...that's all
18. Thanks to all. I've been trying to find a video clip of this as a demo but to no avail
19. I'm not sure I follow the acceleration argument. I thought the last "uphill" part of the track in B would have cancelled out the "downhill" part behind it. Also I'm not too sure what track C is supposed to signify.
20. AmI correct in thinking that both balls reach the ends of the tracks at the same time due to them having the same initial PE? The intervening route doesn't matter.
21. I thought I was just asking whether the ball was rotating or revolving
22. When a [spinning] cricket ball is flying through the air is it rotating or is it revolving? I ask because here in the UK a cricket sports presenter is claiming that if you watch the cricket ball in super slow motion and in high definition on this particular sports programme you can "see the revolutions of the ball". I think he should be saying "you can see the rotations of the ball" because the ball is rotating about its axis of rotation similar to the Earth rotating about its axis of rotation once every 24 hours BUT revolves around the Sun every 365 days. Who is correct him or moir?
23. The only bit of it I can think of that could be vulnerable to low pressure is the battery. You couldn't even use this on a stroll up Mt. Everest!!
24. According to the Apple website the maximum operating altitude of the iShuffle is 3000 m (10,000 feet). Why should the altitude affect the operation of a solid state bit of electronics.
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