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# NSAA Help with physics and maths questions watch

Can someone explain questions 10, 14 and 17 to me please!

Question 79, 88 and 77 too plz

Need help on 85 too!
2. Also 27, 28 and 36 please!!!
3. (Original post by AspiringUnderdog)

Can someone explain questions 10, 14 and 17 to me please!
For 10, try visualise how they've made the cylinder. (The circumferences of the circles on the cylinder=5). Using C=2πr, you can find the value for r.

Then use the formula πr^2l... You can find the answer. Sorry if I've been of no help
4. (Original post by ChallengeLover)
For 10, try visualise how they've made the cylinder. (The circumference of the circle on the cylinder =5). Using C=2πr, you can find the value for r.

Then use the formula πr^2l... You can find the answer. Sorry if I've been of no help
No that makes sense so the radius is 5/2. I assumed that the diameter would be 5 and so took the radius as 5/2 when it should be 5/2pi. Thanks a lot!

If you know how to do any of the others then that would be appreciated!
5. (Original post by AspiringUnderdog)
No that makes sense so the radius is 5/2. I assumed that the diameter would be 5 and so took the radius as 5/2 when it should be 5/2pi. Thanks a lot!

If you know how to do any of the others then that would be appreciated!
Glad I could help- I'm having a little look at 14. Will reply if someone else doesn't beat me to it.
6. Alright- this one is a little harder to spot at first.

The angle RQT=(180-x)/2.

So the exterior angle is then (180-x)/2 because the interior+exterior=180° and the side PQ is a straight line so has angle 180°. 360=180°+exterior+interior. I've started waffling I believe.

Use the formula n=360/exterior angle to find the answer. Should get a fraction within a fraction
7. I'll have a crack at the others... I'll be a little slow because I'm in yr 12
8. Not too sure here but the area traveled should be the area under the graph which would be 5*3*4=60cm per second. 60 seconds in a minute so 60*60cm traveled kn a minute. Answer should be 3600cm

For 28, I said that the total force in one half of the rope is (5*0.8)+(5*10)=54. Then I said 54*2=108 since the rope must have equal tensions on both sides but I'm not too sure here. Maybe someone else can clarify here.

36- can't make head nor tail how to do this.

That's what I have done- can't say whether it's right but I hope I've helped. If Ive not explained something properly, do ask...
9. Sorry that my working is awful but I'll try to explain 36 as best as I can (if I've even got it right idek).

So, you know v=fλ

You've been given v= 320

You have two distances of which you need to find the highest common factor

And then use this to think about at which distances the sound could be heard (my diagram has nodes and antinodes the wrong way round sorry :/)

When you've found the highest common factor you then need to double this to get λ (there are two points in one wavelength at which the amplitude of the sound is at its max).

From this wavelength you can easily find the frequency of the sound
10. (Original post by ChallengeLover)
Not too sure here but the area traveled should be the area under the graph which would be 5*3*4=60cm per second. 60 seconds in a minute so 60*60cm traveled kn a minute. Answer should be 3600cm

For 28, I said that the total force in one half of the rope is (5*0.8)+(5*10)=54. Then I said 54*2=108 since the rope must have equal tensions on both sides but I'm not too sure here. Maybe someone else can clarify here.

36- can't make head nor tail how to do this.

That's what I have done- can't say whether it's right but I hope I've helped. If Ive not explained something properly, do ask...
The first one makes sense to me. I kind of just read it wrong I suppose and ignored the amplitude assuming that it was the whole wave and not a particle.

The second one I realise how you got 54. I misread that too so didn't even get that :/. Why does it need to be doubled I'm not quite clear with that?
11. (Original post by ChallengeLover)
Alright- this one is a little harder to spot at first.

The angle RQT=(180-x)/2.

So the exterior angle is then (180-x)/2 because the interior+exterior=180° and the side PQ is a straight line so has angle 180°. 360=180°+exterior+interior. I've started waffling I believe.

Use the formula n=360/exterior angle to find the answer. Should get a fraction within a fraction
That appears to be correct. Is the formula n=360/exterior angle always the way to find the number of sides? Maybe it's from GCSE but I don't recall it.though
12. (Original post by AspiringUnderdog)
The first one makes sense to me. I kind of just read it wrong I suppose and ignored the amplitude assuming that it was the whole wave and not a particle.

The second one I realise how you got 54. I misread that too so didn't even get that :/. Why does it need to be doubled I'm not quite clear with that?
54 is the tension in one.

If the tension on either side of the wheel were different, the rope would snap (or so I think) so I doubled it to find the tension in the whole rope when I had it for half of the system
13. (Original post by AspiringUnderdog)
That appears to be correct. Is the formula n=360/exterior angle always the way to find the number of sides? Maybe it's from GCSE but I don't recall it.though
Yes it's from GCSE. I learnt it as something to do with a spider crawling to make a circle when it turns at each exterior angle but not sure where that came from.
14. (Original post by ImprobableCacti)
Sorry that my working is awful but I'll try to explain 36 as best as I can (if I've even got it right idek).

So, you know v=fλ

You've been given v= 320

You have two distances of which you need to find the highest common factor

And then use this to think about at which distances the sound could be heard (my diagram has nodes and antinodes the wrong way round sorry :/)

When you've found the highest common factor you then need to double this to get λ (there are two points in one wavelength at which the amplitude of the sound is at its max).

From this wavelength you can easily find the frequency of the sound
That's crazy but that makes sense. Thanks a lot, dude!
15. (Original post by ChallengeLover)
54 is the tension in one.

If the tension on either side of the wheel were different, the rope would snap (or so I think) so I doubled it to find the tension in the whole rope when I had it for half of the system
(Original post by ChallengeLover)
Yes it's from GCSE. I learnt it as something to do with a spider crawling to make a circle when it turns at each exterior angle but not sure where that came from.
The wheel one I kind of get but have doubts about. I suppose I need to also look through some GCSE content. I know that similar shapes comes up but I mostly know that.
16. (Original post by AspiringUnderdog)
That's crazy but that makes sense. Thanks a lot, dude!
It's fine dw
17. (Original post by ImprobableCacti)
Sorry that my working is awful but I'll try to explain 36 as best as I can (if I've even got it right idek).

So, you know v=fλ

You've been given v= 320

You have two distances of which you need to find the highest common factor

And then use this to think about at which distances the sound could be heard (my diagram has nodes and antinodes the wrong way round sorry :/)

When you've found the highest common factor you then need to double this to get λ (there are two points in one wavelength at which the amplitude of the sound is at its max).

From this wavelength you can easily find the frequency of the sound
Now that's how you explain something. I'll take notes for next time...
18. (Original post by ChallengeLover)
Now that's how you explain something. I'll take notes for next time...
Haha thank you I do try
19. (Original post by ImprobableCacti)
Haha thank you I do try
If you don't mind me asking, what year are you in?
20. (Original post by ChallengeLover)
If you don't mind me asking, what year are you in?
Year 13 (I'm studying Maths, FM, Physics and Chem) , you?

I am not studying Chem xD I meant Computer Science hahaha

Updated: October 21, 2017
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