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Edexcel Physics Unit 1 17th May 2012

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Reply 160

Salmonidae

Original post by youlostme

when are you guys sitting for your exam(as in, in how many hours)?

about 11 hours from now! :smile:

Reply 161

nsonia95

Original post by GIR

Hi, please refer to the posts at the top of this page. I've written out a method and legend has given another valid alternative route which you can take.

Hey, yess i just read your method. It was really helpfull!! Thank you so much! :smile:

i am actually panicking for this exam and didnt realise someone had asked the same question already! sorry about that.

And thankss once again :smile:

Reply 162

LegendX

Original post by EllaBella<3

Don't take my word, but i'm pretty sure the object has to be moving in the opposite direction to the flow of the fluid to create a flow, whether it be laminar or turbulent. Say a ping pong ball was moving left and a fluid right, the flow of water would travel from left to right, being laminar up to a point on the left hand side. Say the point of turbulence is the centre of the ping pong ball, the turbulent flow would begin there and continue on the right hand side - hope this makes a little sense? Good luck!

Any chance you could draw a diagram, to show the turbulence in the middle? :colondollar:

Thanks

Reply 163

EllaBella<3

Original post by nsonia95

Pleaseee can someone explain me step by step question 13.a.ii) from the Physics past paper of Unit 1 May/June 2010
I dont understand the mark scheme that much :frown:

I dont get it why we use t equals s/v and why do we use 0.4/1.2 to substitute those values , because we are asked to find the vertical distance, and 0.4 and 1.2 are horizontal distances and components!!!
neeed hellppp pleasee ....ASAPPP!
can someone explain it clearly step by step! i will truely appreciate it!

Thanks in advance!
and good luck to all!

Hey! Ugh, I hate these questions, I've only just got my read around them .. I really hope I don't mess up tomorrow!

The way that I go about this is to find the maximum height of the projectile. You should resolve the values you are given. Finding the maximum height, you use the VERTICAL components of the velocity.

Vv=3.5xsin70; this gives you 3.3ms^-1

From this, you now have three values you can plug into a kinematics in order to find a forth value.

V=3.3 U=0 (as the vertical velocity at the maximum height is zero) A=9.81 and you are trying to calculate S.

Using V^2=U^2 + 2as you can stick in your numbers, rearrange the equation and hopefully get the right answer!

Hope this helps - good luck tomorrow! :smile:

Reply 164

Obscenedilemma

Original post by youlostme

when are you guys sitting for your exam(as in, in how many hours)?

About 11 hours from now

Reply 165

Salmonidae

Original post by LegendX

Any chance you could draw a diagram, to show the turbulence in the middle? :colondollar:

Thanks

Drum roll please its the one and only diagrammatically vague explanation of Salmonidae!!!!

Here you are, see if this helps...

turbulent etc.jpg

Reply 166

nsonia95

Ohhh

Thankkkk you ! i understand how to rearrange it into the formula....but i am still confused whyy do we use the horizontal component values when we are told to fidn the vertical distance!
anywyas, i understand the rest! and thankks for that :smile:

Reply 167

youlostme

mine is in 10 hours *sob*

Reply 168

Liver2029

Good luck for tomorrow everyone. Let's just hope it's not as bad as AQA Core 2.

Reply 169

EllaBella<3

Original post by LegendX

Any chance you could draw a diagram, to show the turbulence in the middle? :colondollar:

Thanks

Haven't got a clue how to do that .. so i've attached a picture :smile:

Basically - before the transition between flows - laminar flow must be in layers, draw atleast 3 either side to be safe with no overlaps whatsoever - beyond the point where it changes, turbulent flow causes eddy currents so I usually just draw scribbling swirls but you need to indicate the overlapping of layers etc.

http://1.bp.blogspot.com/_wL3W_4HqCNI/TIo0Ymbe4UI/AAAAAAAAADg/PthPKGWc6Sc/s320/flow_patterns.gif

Reply 170

nsonia95

Ohhhh! i gett itt ! Thankssss!!!! so we CAN use horizontal component to deduce the time for the vertical one! Thats what i was not sure aboutt!
Thankk you! It did helpp! :smile:

Reply 171

EllaBella<3

Original post by LegendX

Any chance you could draw a diagram, to show the turbulence in the middle? :colondollar:

Thanks

Found a better one!

http://www.centennialofflight.gov/essay/Theories_of_Flight/Real_Fluid_Flow/TH9G3.jpg

Reply 172

LegendX

Original post by Salmonidae

Drum roll please its the one and only diagrammatically vague explanation of Salmonidae!!!!

Here you are, see if this helps...

turbulent etc.jpg

Thanks, but sayin the points of turbulence were near the start if the ball then? Thanks again.

Reply 173

BethanyLimbo

Good luck everyone, kinda get 6PH01 it's 02 I'm going to struggle with! :confused:

Reply 174

nsonia95

Heyy thankssss a toonnn!!! This is realllyyy clear! Step by stepp as i asked!!
It really did hellppp!! :biggrin:

And i hope luck stays with me tomorrow :tongue:

Thanksss aloottttt once againnnn!!! :biggrin:

Reply 175

LegendX

Which of the following units is equivalent to the SI unit for energy?

Kgms-2
KwH
Nm-1
Ws

Is there any quick way of doing this?

Reply 176

nsonia95

Original post by Obscenedilemma

for a ii) there's two ways of going about this but I'll give you the most common way

first you need to find the horizontal component of velocity

second you need to use this value, which now becomes your V and plug it into the rearranged form of s=vt to get the time which is s/v -- why do we do this well just think of gcse maths. Speed = distance x time. You want to find time so just rearrange

third you use this value of t and plug it into s = ut + 1/2at^2

if you're familiar enough with projectiles you should be able to know what a is
it starts at rest so what can you assume
you've found t from earlier on

use the formula and a calc, you get S

Ohhh

Thankkkk you ! i understand how to rearrange it into the formula....but i am still confused whyy do we use the horizontal component values when we are told to find the vertical distance!
anywyas, i understand the rest! and thankks for that

Reply 177

tz23456

Original post by Salmonidae

Will be midnight tomorrow night :smile:

would be really helpful if u cound send me the link.btw why can't i find edexcel physics 3b may 2012 discussion thread same goes for M1?im new at this sry fr askinn a lot of question!

Reply 178

nsonia95

Original post by youlostme

first of all you should be familiar with the fact that the horizontal component of velocity doesnt change(in projectiles)....you should use this knowledge to deduce that horizontal component of velocity=horizontal range/time of flight(notice the absence of acceleration here because velocity is constant)
so thats how you find the time of flight
then, you use the formula s=ut-0.5gt^2(vertical distance=vertical component of velocity x time of flight -(0.5 x 9.8 x time of flight^2) )

hope this helped :smile:

Ohhhh! i gett itt ! Thankssss!!!! so we CAN use horizontal component to deduce the time for the vertical one! Thats what i was not sure aboutt!
Thankk you! It did helpp!