Unit 4 help

Zenarthra

3

These are currently the questions im still confused about, any help would be appreciated!

ThankS!

Do you have the markscheme or answers to some of them?

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(edited 9 years ago)

Reply 2

krisshP

14

Original post by Zenarthra

!

I'll try my best, the other harder ones I suppose someone else needs to do them.

Q4-v=flambda. Work out f to get number of cycles per second, then divide by 100 to get number of cycles in 10 miliseconds as milli=10^-3 so 10 milliseconds=10 X 10^-3=10^-2.

Q5iii- equate centripetal force F=mv^2/r to gravitational force F=GMm/r^2. You can cancel the ms and then you can rearrange and solve for M, the mass of Jupiter.

Q4-See this
http://www.schoolphysics.co.uk/age14-16/Mechanics/Circular%20motion/text/Circular_motion/images/5.png

Q11-

Q2-V₀=12V

1/4 energy -->1/4 charge on capacitor plates-->C=Q/V so V=Q/C, with C constant, (1/4)Q means (1/4)V=(1/4)(12)=3V answer. I hopes I'm right here, not 100% sure though, check the answer/MS.

Q1-M's acceleration is in opposite direction to that of O when M is at top and O is at bottom and vice versa (M at bottom and O at top), so T/2. That's the only hint I can give for this q as it is to do with visualising in a way.

Q4-

Lambda/4=0.40m -->lambda=1.6m

v=flambda
=(400)(1.6)
=640ms^-1

Q9-

but that answer isn't there, so not sure :frown:

.

Q4-v=flambda
8=5lambda
Lambda=1.6km

1.6km -->2pi
2km -->2.5pi

So 0.5pi or pi/2 phase difference.

Q2-yes, I agree with you there, after being confused and then looking at what you wrote there. 5RC too short means hard to measure time due to human reaction time as stopwatch used. 5RC too long means you in experiment forever... :frown:

So

5RC=600, just choosing a sensible realistic value here, I chose 10mins=600s

R=600/[5(30 X 10^-6)]

R=4MOhms

Q8-

So l coefficient divided by M coefficient should give 2, but it makes no options right, so not sure :frown:

.

Q21-

Q2-strange, I got D, not A. Look

Q8-

Q12-

DONE AS MUCH AS I CAN!!! The rest is either not in my spec or too difficult for me.

(edited 9 years ago)

Reply 3

Jaydude

11

Original post by krisshP

Q4-v=flambda. Work out f to get number of cycles per second, then divide by 100 to get number of cycles in 10 miliseconds as milli=10^-3 so 10 milliseconds=10 X 10^-3=10^-2.

Q5iii- equate centripetal force F=mv^2/r to gravitational force F=GMm/r^2. You can cancel the ms and then you can rearrange and solve for M, the mass of Jupiter.

Q4-See this
http://www.schoolphysics.co.uk/age14-16/Mechanics/Circular%20motion/text/Circular_motion/images/5.png

So then what would you say for Q4??

I think the question is tricking a person into thinking "oh it must go off a tangent to the circle" like the picture you posted, but the keyword is parabola. In the horizontal direction the track of the object won't curve, ie no parabola but in the vertical direction due to g there will vertical parabola, a curvature towards the ground so I thought it's probably D. Thoughts?

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Reply 4

krisshP

14

Original post by Jaydude

So then what would you say for Q4??

I think the question is tricking a person into thinking "oh it must go off a tangent to the circle" like the picture you posted, but the keyword is parabola. In the horizontal direction the track of the object won't curve, ie no parabola but in the vertical direction due to g there will vertical parabola, a curvature towards the ground so I thought it's probably D. Thoughts?

Posted from TSR Mobile

I see what you are saying, but the parabola will be "one-sided" and in reality I doubt much of a parabola will be seen.

Reply 5

Jaydude

11

Original post by krisshP

I see what you are saying, but the parabola will be "one-sided" and in reality I doubt much of a parabola will be seen.

I agree to an extent. But I still think it would be noticeable in the vertical plane, hence option D. Where's the mark scheme when you need it! :redface:

Op post the answers so we can determine the other answers with certainty too.

Q 24: For the transformer question I think there is greater flux linkage in the second coil due to more turns, so flux of second coil divided by flux of first is greater than one.
I think there will be greater induced emf (current) in second coil too so again greater than one. So I'd go with option C.

Posted from TSR Mobile

(edited 9 years ago)

Reply 6

Zenarthra

OP

3

Original post by Jaydude

Do you have the markscheme or answers to some of them?

Posted from TSR Mobile

Was a mix from many different past papers, i will go through and find the answers though. :tongue:

Reply 7

Zenarthra

OP

3

Original post by krisshP

I'll try my best, the other harder ones I suppose someone else needs to do them.

Q4-v=flambda. Work out f to get number of cycles per second, then divide by 100 to get number of cycles in 10 miliseconds as milli=10^-3 so 10 milliseconds=10 X 10^-3=10^-2.

Q5iii- equate centripetal force F=mv^2/r to gravitational force F=GMm/r^2. You can cancel the ms and then you can rearrange and solve for M, the mass of Jupiter.

Q4-See this
http://www.schoolphysics.co.uk/age14-16/Mechanics/Circular%20motion/text/Circular_motion/images/5.png

Q11-

Q2-V₀=12V

1/4 energy -->1/4 charge on capacitor plates-->C=Q/V so V=Q/C, with C constant, (1/4)Q means (1/4)V=(1/4)(12)=3V answer. I hopes I'm right here, not 100% sure though, check the answer/MS.

Q1-M's acceleration is in opposite direction to that of O when M is at top and O is at bottom and vice versa (M at bottom and O at top), so T/2. That's the only hint I can give for this q as it is to do with visualising in a way.

Q4-

Lambda/4=0.40m -->lambda=1.6m

v=flambda
=(400)(1.6)
=640ms^-1

Q9-

but that answer isn't there, so not sure :frown:

.

Q4-v=flambda
8=5lambda
Lambda=1.6km

1.6km -->2pi
2km -->2.5pi

So 0.5pi or pi/2 phase difference.

Q2-yes, I agree with you there, after being confused and then looking at what you wrote there. 5RC too short means hard to measure time due to human reaction time as stopwatch used. 5RC too long means you in experiment forever... :frown:

So

5RC=600, just choosing a sensible realistic value here, I chose 10mins=600s

R=600/[5(30 X 10^-6)]

R=4MOhms

Q8-

So l coefficient divided by M coefficient should give 2, but it makes no options right, so not sure :frown:

.

Q21-

Thank you for the help, keep them coming please. :biggrin:

Reply 8

krisshP

14

Original post by Zenarthra

Thank you for the help, keep them coming please. :biggrin:

I'll try. I thought you completely forgot about this thread somehow as you never responded through posting.

Reply 9

Zenarthra

OP

3

Original post by krisshP

I'll try. I thought you completely forgot about this thread somehow as you never responded through posting.

I nearly did because i never got a notification. :tongue:

When is your exam? :biggrin:

Reply 10

krisshP

14

Original post by Zenarthra

I nearly did because i never got a notification. :tongue:

When is your exam? :biggrin:

My edexcel unit 4 exam is on this Wednesday aaaarghhhjj :frown:

Reply 11

Light_bulb

8

I got 72ms for 9 please tell me if I'm correct.

My steps

V/2=ve^-t/rc

Figured out RC as 36x10-3/ln(.5)

Energy=.5xCxV^2

For this to drop to E/16 v=v/4 so v^2=v^2/16

Put back into formula solve for v/4

V/4=ve^-t/RC

You you RC as stated before work out t I got 72ms

Reply 12

Light_bulb

8

Its quite fiddly to do calculations on my phone and in my head. I'll have a look tomorrow at the questions when I'm on my computer just let me know which questions you need help on urgently and that you think are hard hope that I I was helpful.

Reply 13

Light_bulb

8

For question 8 use hookes law to find the value of k 10g/.25=392 then place into formula t=2pi•rootm/k the answer is 1

Reply 14

Stonebridge

13

Original post by Zenarthra

These are currently the questions im still confused about, any help would be appreciated!

ThankS!

I'm glad to see you have some responses, but please not that forum rules request that you do not post a large number of questions in one thread.
In fact, we ask just one question per thread!

There are far too many here and you have not shown any attempt at solving them yourself or stated exactly what help you need.

(edited 9 years ago)

Reply 15

Zenarthra

OP

3

Original post by Stonebridge

I'm glad to see you have some responses, but please not that forum rules request that you do not post a large number of questions in one thread.
In fact, we ask just one question per thread!

There are far too many here and you have not shown any attempt at solving them yourself or stated exactly what help you need.

What should i do, make separate threads?

Reply 16

Zenarthra

OP

3

Original post by krisshP

My edexcel unit 4 exam is on this Wednesday aaaarghhhjj :frown:

Wow, goodluck man!

Reply 17

Zenarthra

OP

3

Original post by krisshP

Q2-V₀=12V

1/4 energy -->1/4 charge on capacitor plates-->C=Q/V so V=Q/C, with C constant, (1/4)Q means (1/4)V=(1/4)(12)=3V answer. I hopes I'm right here, not 100% sure though, check the answer/MS.

They said its 6V, i did the same way as you but cant get 6V. :/