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Multiple Choice Questions help

Hey guys, I've got some multiple choice questions I'd like some help with please!

With this one I'm confused. I thought surely because the diameter is halved, and this the area is quartered, the youngs modulus of the second sample would be 12E? The three times the length increases the YM by 3, the area is in the denominator so that would increase it by 4, so 12 times the youngs modulus, no??

This one I've just forgotten how to do, I know it's about resolving but I just can't remember, any help would be appreciated.

Honestly kinda a similar thing here too, not sure why it's A, surely others like C make sense too?

Thanks a lot.

Reply 1

phys981

For 24, EDITED - YM is constant for a given material!!!

For q20, you can use F=ma to find the resultant force.
There are two forces acting on the rocket, its weight and the thrust and these must give a resultant equal to the F=ma you calculated.

For Q18, equilibrium means forces AND moments are balanced. Choose a point in C to take moments around (for example choose the centre) and you'll find there IS a non-zero moment.

(edited 6 years ago)

Reply 2

iMacJack

Original post by phys981

For q1, remember that YM depends on cross-sectional area which goes as the square of the diameter. If the diameter changes by factor 0.5, then the area goes by 0.5² or 0.25

So YM = FL/AdeltaL

If the area is reduced to 1/4 of it's original value, the YM is 4 times greater right? Or am I being confused?

Reply 3

phys981

Yes! Sorry, and so am I (confised).

YM is a constant for a given material. If we change the lenght and area the extension might also change for a given force but YM remains the same.

(should have read properly before answering...)

Reply 4

iMacJack

Original post by phys981

So it's kinda a trick question, and the answer is based off of it being the same material and so regardless of other factors, YM is the same?

Yes.

Original post by phys981

Yes.

Any idea on this one? I got it right, it's C, but I just did it by considering the ratios, 2:1 split of the PD and the only one which goes to 4V is the third graph, kinda a BS way of doing it and I don't understand what I've done really

Reply 7

phys981

You're right, really.

Based on the ratio of resistances, the p.d. across that 10 Ohm resistormust be 4V. So, with the sliding contacts one at either end, they will cover the whole of that 4V. However, as you slide the moveable one towards the right, it covers less and less of the 10Ohm resistor so gets less and less of the 4V until it reaches P and 0V.

Reply 8

iMacJack

Original post by phys981

Yeah that does make sense, cheers bro! Electricity really isn't one of my strong points, anything you'd recommend doing to get a lot stronger at it? I've always had issues with it really.

Reply 9

phys981

Just practice really, lots and lots of questions.

If you have a decent textbook with plenty of questions in, it can be useful to practise the bits you find difficult before you attempt the real exam questions.