cooldudeman
Badges: 17
Rep:
?
#1081
Report 6 years ago
#1081
(Original post by MisterE1)
I wouldn't mind a blank copy! I just want to know the questions! :P


Posted from TSR Mobile
I knowwww

Posted from TSR Mobile
0
reply
angel2
Badges: 0
Rep:
?
#1082
Report 6 years ago
#1082
To the person with the paper (Robmarkey) ,
Please upload it soon this wait is killing me:headfire::please:....Thank you
2
reply
saba146
Badges: 0
Rep:
?
#1083
Report 6 years ago
#1083
(Original post by Robmarkey)
The graph was negative....yea I did say change In flux I've got the paper here.
can you upload it please??
0
reply
kingm
Badges: 2
Rep:
?
#1084
Report 6 years ago
#1084
(Original post by Robmarkey)
The graph was negative....yea I did say change In flux I've got the paper here.
:damnmate:
0
reply
cooldudeman
Badges: 17
Rep:
?
#1085
Report 6 years ago
#1085
(Original post by MisterE1)
as with all questions where we can't decide what the answer is, we'll have to wait till someone gets hold of the paper! I personally got 2/3!
Yo did you get the paper from your school yet?

Posted from TSR Mobile
0
reply
mmss
Badges: 0
Rep:
?
#1086
Report 6 years ago
#1086
i thought the paper was coming up this morning? any update?
0
reply
MisterE1
Badges: 0
Rep:
?
#1087
Report 6 years ago
#1087
(Original post by cooldudeman)
Yo did you get the paper from your school yet?

Posted from TSR Mobile
No the teacher I had didn't have a copy


Posted from TSR Mobile
0
reply
The H
Badges: 1
Rep:
?
#1088
Report 6 years ago
#1088
I want to steal my paper and correct my mistakes
0
reply
kingm
Badges: 2
Rep:
?
#1089
Report 6 years ago
#1089
(Original post by The H)
I want to steal my paper and correct my mistakes
Is there any way we can track them down? How quickly do schools send them off?
1
reply
The H
Badges: 1
Rep:
?
#1090
Report 6 years ago
#1090
(Original post by kingm)
Is there any way we can track them down? How quickly do schools send them off?
Haha I don't know
0
reply
cooldudeman
Badges: 17
Rep:
?
#1091
Report 6 years ago
#1091
Ah damn so no one is gonna upload the paper and then..

Posted from TSR Mobile
0
reply
MisterE1
Badges: 0
Rep:
?
#1092
Report 6 years ago
#1092
(Original post by Robmarkey)
I'm filling in the answers but I got to check them so there not uploaded with silly mistakes, it will be up first thing tomorrow section B anyway.but I can assure you that the graph is negative!
PPPLLLEEEAASSEE upload it!
1
reply
angel2
Badges: 0
Rep:
?
#1093
Report 6 years ago
#1093
(Original post by Robmarkey)
I'm filling in the answers but I got to check them so there not uploaded with silly mistakes, it will be up first thing tomorrow section B anyway.but I can assure you that the graph is negative!
look I'm sure you are very busy and fed up of all this nagging but can you please just take a few minutes and upload it....

you don't need to write in answers (we have textbooks ) but please I want to know how unit 4 has gone before unit 5.

PLEASE UPLOAD THE WHOLE PAPER ASAP.

Thank you...sorry for nagging but these are stressful times :ashamed2::bawling::cry2::cry::hmmm::hmmmm2::nooo::zomg:
0
reply
AdamRuby
Badges: 3
Rep:
?
#1094
Report 6 years ago
#1094
The graph showed on y axis magnetic flux and x axis time. The graph was a -cos graph. Im pretty sure it was defiantly a MINUS cos graph. The gradient of a -cos graph is +sin. So the graph we had to draw was a positive sine graph.

I hope that is right? I got confused with the work out max emf question cause I didn't notice until our exam guy was taking the sheets off us that the time was measured in milliseconds (oh crap) so I didn't even put my max value on the graph haha! Only reason why I remember it
0
reply
fizzbizz
Badges: 4
Rep:
?
#1095
Report 6 years ago
#1095
(Original post by AdamRuby)
The graph showed on y axis magnetic flux and x axis time. The graph was a -cos graph. Im pretty sure it was defiantly a MINUS cos graph. The gradient of a -cos graph is +sin. So the graph we had to draw was a positive sine graph.

I hope that is right? I got confused with the work out max emf question cause I didn't notice until our exam guy was taking the sheets off us that the time was measured in milliseconds (oh crap) so I didn't even put my max value on the graph haha! Only reason why I remember it
Regarding the graphs, I thought the same, however induced emf is equal to the negative of the gradient of the flux line!

 e.m.f.= -\frac{\delta\phi}{\delta t} = -[\frac{\delta}{\delta t}[-cos(t)]]=-[sin(t)]=-sin(t)
0
reply
cooldudeman
Badges: 17
Rep:
?
#1096
Report 6 years ago
#1096
Can anyone else get the paper? My teacher didn't have it

Posted from TSR Mobile
0
reply
AdamRuby
Badges: 3
Rep:
?
#1097
Report 6 years ago
#1097
(Original post by fizzbizz)
Regarding the graphs, I thought the same, however induced emf is equal to the negative of the gradient of the flux line!

 e.m.f.= -\frac{\delta\phi}{\delta t} = -[\frac{\delta}{\delta t}[-cos(t)]]=-[sin(t)]=-sin(t)
Ahhhh so it is! Damn!
0
reply
fizzbizz
Badges: 4
Rep:
?
#1098
Report 6 years ago
#1098
(Original post by AdamRuby)
Ahhhh so it is! Damn!
It's killing me, along with the radians question and the very final question

I'm scared I've missed the grade A
0
reply
cooldudeman
Badges: 17
Rep:
?
#1099
Report 6 years ago
#1099
(Original post by AdamRuby)
Ahhhh so it is! Damn!
What did you lot get for the value of peak emf. I thought I done it accurately by finding max gradient and got 102. Wtf that's the one q that I thought I done carefully.

Posted from TSR Mobile
0
reply
fizzbizz
Badges: 4
Rep:
?
#1100
Report 6 years ago
#1100
(Original post by cooldudeman)
What did you lot get for the value of peak emf. I thought I done it accurately by finding max gradient and got 102. Wtf that's the one q that I thought I done carefully.

Posted from TSR Mobile
I believe the anticipated method was to use the fact that peak induced emf is equal to  BAN\omega (as given in the data and formulae booklet)

BAN could be found by reading the peak value of magnetic flux linkage from the graph, and angular speed could be found by seeing how long one oscillation took..

The peak BAN value was 0.55 Wb turns, and the angular speed was  \frac{2\pi}{40\times 10^{-3}}

Plugging these values into the equation gives a value of 86.4V for the induced emf!

It sucks becuase the first thing I thought of was using the BANw formula, but we weren't explicitly given B, A and N so I thought it wasn't applicable
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • University of Bristol
    Undergraduate Open Afternoon Undergraduate
    Wed, 23 Oct '19
  • University of Exeter
    Undergraduate Open Day - Penryn Campus Undergraduate
    Wed, 23 Oct '19
  • University of Nottingham
    Mini Open Day Undergraduate
    Wed, 23 Oct '19

Have you made up your mind on your five uni choices?

Yes I know where I'm applying (135)
62.79%
No I haven't decided yet (46)
21.4%
Yes but I might change my mind (34)
15.81%

Watched Threads

View All