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PAT Solutions.

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Seems like no one is doing the 07 paper. Great, calculators can be used for the physics part that year. I'll start off with Q10:

(a)

Time taken to fly from his base to the airport =DistanceTime taken=300170=11317h=1 h 46 min= \dfrac{\mathrm{Distance}}{ \mathrm{Time \ taken}} = \dfrac{300}{170} = 1 \dfrac{13}{17} \mathrm{h} = 1 \ \mathrm{h}\ 46 \ \mathrm{min}

Estimated arrival time = 1046 h

(b)

Drawing a diagram out is helpful for this part.

Distance the plane should have travelled due north in 1 hr = 170 km

As it travelled only 153 km, less than 170 km, and 10 deg from north, it means the wind is made up of 2 perpendicular components, 1 pointing south and the other pointing east.

Actual distance travelled due north by the plane =153sin80=150.6756 km=153\sin80^{\circ} =150.6756 \ \mathrm{km}

Actual distance travelled due east by the plane =153cos80=26.5682 km=153\cos80^{\circ} =26.5682 \ \mathrm{km}

Distance travelled due south by the wind = Distance the plane should have travelled due north in 1 hr - Actual distance travelled due north by the plane = 170 - 150.6756 = 19.3 km

Distance travelled due east by the wind = Actual distance travelled due east by the plane = 26.6 km

Using Pythagoras Theorem, the wind speed is 19.32+26.62=32.9 km h1\sqrt{19.3^2+26.6^2} = 32.9 \ \mathrm{km \ h^{-1}}
For 07 Q13:

Since this is a series circuit and the current through the circuit is 0.4 A, the current through the non-linear resistor is 0.4 A.

0.4=0.05V3V=2 V0.4 =0.05V^3 \Rightarrow V= 2 \ \mathrm{V}

P.D. across the fixed resistor = 9 - 2 = 7 V

Resistance of the fixed resistor =VI=70.4=17.5Ω= \dfrac{V}{I} = \dfrac{7}{0.4} = 17.5 \Omega

Thus, the value of the resistance of the fixed resistor should be 17.5Ω17.5\Omega for the current to be 0.4 A.
Hey - mind if I join in? :smile:

2008 - Q. 10.

Spoiler

2008 - Q. 11

a.)

Spoiler


b.)

Spoiler

Original post by mf2004
Sounds more likely to be Material sciences


I haven't seen the e-mail yet, could well be, but if that was the case I wouldn't bother. I was just wondering if anyone else had received something to that effect.
I have no idea how to spoiler stuff, but with regards to the question about the wind speed, I got fairly confused (and ended up not bothering with it) because surely the affect of the windspeed is dependent on the drag that it creates on the plane... some planes will be more aerodynamic than others so will be affected differently. I think I'll do the 2010 paper in a while, it looks like most of the solutions are on here as well! SWEET :smile:
(edited 12 years ago)
Reply 186
Original post by _Say_Goodbye_

Original post by _Say_Goodbye_
I have no idea how to spoiler stuff, but with regards to the question about the wind speed, I got fairly confused (and ended up not bothering with it) because surely the affect of the windspeed is dependent on the drag that it creates on the plane... some planes will be more aerodynamic than others so will be affected differently. I think I'll do the 2010 paper in a while, it looks like most of the solutions are on here as well! SWEET :smile:


The air is moving relative to the ground, and the plane is moving relative to the air. If you could have a plane floating and not moving, the air wouldn't actually be pushing on it because there's nothing to push it back. When it's moving there's probably a lot more to consider but seeing as they don't expect you to know advanced aerodynamics just having done AS Physics, there's no need.
(edited 12 years ago)
I might go back and look at the question... if I can find it. It just seemed strange to me, how the hell should I know how wind is going to affect a plane, which we are just imagining? I'm pretty sure if it was made out of paper and there was a windspeed of 32km/h it would struggle! I see your point though, I guess it would make sense to make those assumptions, kind of :tongue:
Reply 188
Original post by alexthomas93
2008 Q4:

Spoiler



Can you not just solve the inequality, so x^2 + 2 < 3- 3x^2, then 4x^2 < 1, so X^2<1/4 so x<+ or- 1/2, so -1/2<x<1/2
i got the email and its for materials science and you have to write an extra statement about why you would like to study it. Seem like it would be good for people who got an A in chemistry last year and are carrying it on to A2 :smile:
Original post by Benniboi1
i got the email and its for materials science and you have to write an extra statement about why you would like to study it. Seem like it would be good for people who got an A in chemistry last year and are carrying it on to A2 :smile:


Are you thinking about doing it? I'm doing Chemistry and got an A... but it just seems the most painstakingly, mindnumbingly dull thing in the world to me. I don't think I'll say yes to that!

Now let's get onto revision. Can anyone explain fairly succinctly about 'the phases of the moon and eclipses'? Something I haven't really studied nor known a lot about. I've tried to find something on the internet without much luck and my teacher didn't really have a clue!!
Original post by _Say_Goodbye_
Are you thinking about doing it? I'm doing Chemistry and got an A... but it just seems the most painstakingly, mindnumbingly dull thing in the world to me. I don't think I'll say yes to that!

Now let's get onto revision. Can anyone explain fairly succinctly about 'the phases of the moon and eclipses'? Something I haven't really studied nor known a lot about. I've tried to find something on the internet without much luck and my teacher didn't really have a clue!!


I haven't read much about the course :P but I got a C in Chemistry last year thanks to many reasons and dropped it, so theres no point in me applying!
Reply 192
Original post by wcp100
2010-Q25

v)

Spoiler




For the horizontal kinetic energy I'm not sure how you've got 500j

Surely it's

Kex=12×10×1002=50000j Ke_x = \frac{1}{2} \times 10 \times 100^{2} = 50000j ?
Reply 193
^^ thats what i got too.

also, would anyone mind posting solutions to the multiple choice qs for the 2010 PAT paper? if its no trouble.

thanks :biggrin:
For 2008 - Q. 10, did anyone else get 8π(6+33):278\pi(6+3\sqrt{3}):27 to some lesser or greater degree of simplification?
Reply 195
Original post by Underwood94

Original post by Underwood94
For 2008 - Q. 10, did anyone else get 8π(6+33):278\pi(6+3\sqrt{3}):27 to some lesser or greater degree of simplification?


I got 8π(3+23):938 \pi (3 + 2 \sqrt{3}) : 9 \sqrt{3} :dontknow:
(edited 12 years ago)
Reply 196
Original post by AYEP
^^ thats what i got too.

also, would anyone mind posting solutions to the multiple choice qs for the 2010 PAT paper? if its no trouble.

thanks :biggrin:


Here you go!

2010 PAT: Multiple Choice

12)

Spoiler


13) Seems easy, but I can't do it.

14)

Spoiler


15)

Spoiler


16)

Spoiler


17) Really don't understand the Q!

18)

Spoiler


19)

Spoiler


20) Not sure whether to use Kepler's 3rd Law or inverse squared rule?

21)

Spoiler

(edited 12 years ago)
Original post by und
I got 8π(3+23):938 \pi (3 + 2 \sqrt{3}) : 9 \sqrt{3} :dontknow:


Yep, same thing I think :biggrin: - I just multiplied by 3\sqrt{3} to get rid of the roots one on side.
On the multiple choice answers, I got all the same answers as the above poster. 13, my answer V^2 x V2/ V1 x R2 wasn't there... so I gave up.

Q.17 I think the answer will be B, Exactly 180 degrees, there is no reason why it shouldn't be!!

Q 20 I thought would be to do with ratios, it may well be wrong though, so I gave the answer as C, 20 km/s

Q.21 I thought that the fish would be lower than it seems to be, due to the refraction of the light. Due to 1.33x 0.75 equalling 1 metre, I went with C, 1m

Thanks for loads of these answers guys, much appreciated!
Reply 199
Original post by _Say_Goodbye_
On the multiple choice answers, I got all the same answers as the above poster. 13, my answer V^2 x V2/ V1 x R2 wasn't there... so I gave up.

Q.17 I think the answer will be B, Exactly 180 degrees, there is no reason why it shouldn't be!!

Q 20 I thought would be to do with ratios, it may well be wrong though, so I gave the answer as C, 20 km/s

Q.21 I thought that the fish would be lower than it seems to be, due to the refraction of the light. Due to 1.33x 0.75 equalling 1 metre, I went with C, 1m

Thanks for loads of these answers guys, much appreciated!


Yep, sorry, I meant C for 21. For Q20 I was thinking 20km/s but like you, I'm not sure, I used Kepler's third law.

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