Edexcel Physics Unit 4 - 21st June 2011

yh its jus 1 mark:/ bt i dont get it why they've added all the masses.in annihilation qs we always subtract mass of those on the right frm mass of those on the left.tht gives us d mass defect which we substitute into E=mc^2 for energy released:/

There's absolutely NO mass defect in annihilation of electron/positron. They aren't hadrons, they are leptons. They feel no strong force. They are also seperated. To find out the energy released you simply use 2mc^2.

(edited 12 years ago)

Reply 81

sciencefreak00

PLEASE SOME ONE HELP ME WITH THIS QUESTION FROM REVISION GUIDE. :\

Page 13 Quick questions Q3. :\

The question is here:

A ceiling fan is turning at a rate of 100 revolutions per minute. A spider is clinging to a blade of the fan. If the spider experiences a centripetal acceleration greater than 3g, it will lose its grip on the blade and be flung off. How far from the centre of the fan can the spider go safely?

Reply 82

sciencefreak00

THE ANSWER IS 1.05m , and thats weird.

HELP

OP

Original post by sciencefreak00

PLEASE SOME ONE HELP ME WITH THIS QUESTION FROM REVISION GUIDE. :\

Page 13 Quick questions Q3. :\

The question is here:

A ceiling fan is turning at a rate of 100 revolutions per minute. A spider is clinging to a blade of the fan. If the spider experiences a centripetal acceleration greater than 3g, it will lose its grip on the blade and be flung off. How far from the centre of the fan can the spider go safely?

OMG ! I had the SAME DOUBT! =|

Reply 85

greenwoodfan93

Hey was doing the Jan 2010 past paper, can't work out where I've gone wrong on question 5:

"Figure 1 shows a vertical plane square coil of 50 turns, carrying a current of 3.0 A. The
length of each side of the coil is 4.0 cm. Figure 2 shows a view of this coil from above
within a horizontal magnetic field of flux density 0.20 T."

(In the figure they show a wire at 30 degrees to the direction of the magnetic field.)

The question is find out the force on a side of the coil so I did:

F=BIlsin30
F=0.2x3x0.04x0.5
F=0.012N - however this isn't even an option - they give the answer of 1.2N

I have no idea where I went wrong, and I'm starting to wonder if they made an error and forgot to put cm's into m's. Any help?

Reply 86

sciencefreak00

Original post by vampire-love4ever

omg ! I had the same doubt! =|

pleaseee some one solve our doubt. :frown:

Reply 87

Vampire-Love4ever

OP

13

Original post by sciencefreak00

pleaseee some one solve our doubt. :frown:

I tried it like billion times but no chance. I will try it again and see if I can solve it this time. :redface:

Reply 88

Summerdays

10

Original post by greenwoodfan93

Hey was doing the Jan 2010 past paper, can't work out where I've gone wrong on question 5:

"Figure 1 shows a vertical plane square coil of 50 turns, carrying a current of 3.0 A. The
length of each side of the coil is 4.0 cm. Figure 2 shows a view of this coil from above
within a horizontal magnetic field of flux density 0.20 T."

(In the figure they show a wire at 30 degrees to the direction of the magnetic field.)

The question is find out the force on a side of the coil so I did:

F=BIlsin30
F=0.2x3x0.04x0.5
F=0.012N - however this isn't even an option - they give the answer of 1.2N

I have no idea where I went wrong, and I'm starting to wonder if they made an error and forgot to put cm's into m's. Any help?

You need to multiply your answer by 50 :smile:

The more turns the wire has, the more "open surfaces" the wire has, and thus the greater the force.

Reply 89

greenwoodfan93

Original post by Summerdays

You need to multiply your answer by 50 :smile:

The more turns the wire has, the more "open surfaces" the wire has, and thus the greater the force.

Cheers, forgot about that and was using the wrong angle for the coil. Got it now. :smile:

Reply 90

aym3n

Original post by Vampire-Love4ever

I tried it like billion times but no chance. I will try it again and see if I can solve it this time. :redface:

haha the ans is 0.03m ...the revision guide has the ans wrong :tongue:

Reply 91

englishman129

7

Original post by greenwoodfan93

Cheers, forgot about that and was using the wrong angle for the coil. Got it now. :smile:

i dont get that question either - can you gimme a working of how u got it in end plls?

Reply 92

Vampire-Love4ever

OP

13

Original post by aym3n

haha the ans is 0.03m ...the revision guide has the ans wrong :tongue:

0MGGGGGG. Really? How did you know? :|
Btw, another question with a wrong answer? :p:

Reply 93

aym3n

Original post by Vampire-Love4ever

0MGGGGGG. Really? How did you know? :|
Btw, another question with a wrong answer? :p:

lol a friend posted this question on yahoo answers and all the people got this ans

lol there is a question in which u have to find the force... the question mentions the diameter but tht diameter is actually the radius... i think its question 1 on page 15 :tongue:

not sure though :tongue:

Reply 94

fatima00

Original post by Summerdays

There's absolutely NO mass defect in annihilation of electron/positron. They aren't hadrons, they are leptons. They feel no strong force. They are also seperated. To find out the energy released you simply use 2mc^2.

no mass defect?:O no mass defect is wen the 2 collide bt do not produce any other particles right?cz thn all their mass changes to energy..nd in the q i askd..they produced 5 particles.so shudnt v consider the masses of all the particles formed?

Reply 95

Summerdays

10

that's not mass defect. Mass defect is when you are specifically talking about a nucleus having less energy than its constituent particles because the separated particles have more potential energy than if they were together as a nucleus, which means hat they have more mass based on the energy mass equivalance. But electrons and positions aren't hadrons! What you're describing is the energy mass equivalance that comes from annihilation, but not mass defect. Don't get the two confused.

Reply 96

Vampire-Love4ever

OP

13

Original post by aym3n

lol a friend posted this question on yahoo answers and all the people got this ans

lol there is a question in which u have to find the force... the question mentions the diameter but tht diameter is actually the radius... i think its question 1 on page 15 :tongue:

not sure though :tongue:

Thanks ALOT! God bless! :h:

Reply 97

aym3n

Original post by Vampire-Love4ever

Thanks ALOT! God bless! :h:

no problem at all :smile:

Reply 98

gozatron

16

Original post by englishman129

i dont get that question either - can you gimme a working of how u got it in end plls?

Original post by greenwoodfan93

Hey was doing the Jan 2010 past paper, can't work out where I've gone wrong on question 5:

"Figure 1 shows a vertical plane square coil of 50 turns, carrying a current of 3.0 A. The
length of each side of the coil is 4.0 cm. Figure 2 shows a view of this coil from above
within a horizontal magnetic field of flux density 0.20 T."

(In the figure they show a wire at 30 degrees to the direction of the magnetic field.)

The question is find out the force on a side of the coil so I did:

F=BIlsin30
F=0.2x3x0.04x0.5
F=0.012N - however this isn't even an option - they give the answer of 1.2N

I have no idea where I went wrong, and I'm starting to wonder if they made an error and forgot to put cm's into m's. Any help?

This is a very sneaky question, exactly why I hate this exam board. Essentially, the diagram is tricking you. It shows the side PQ is at 30 degrees perpendicular to the field BUT they ask for QS. Now QS is vertically down so is perpendicular to the field regardless how PQ is orientated.

So it's just:

F = Bilsin90
F = 0.2 x 3 x 0.04 x 50
F = 1.2

Reply 99