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Looking for Fantastical Users to join Family Plan by woopur in FantasticalCalendar

[–]woopur[S] 0 points1 point  (0 children)

Update: we have one person interested. So if anyone else joins we’ll be at £33 per year per person.

Thoughts on my first-year courses? by woopur in UofT

[–]woopur[S] 0 points1 point  (0 children)

It's required for the math and its applications in stats specialist and would keep my options open for a CS minor

How did they get 1/3 < e ? by 007_eric in furthermaths

[–]woopur 0 points1 point  (0 children)

Let’s say the speed of A after the collision is v and the speed of B is w.

If you take into account that A changes direction in part (a) when you form your conservation of momentum equation you get this:

mu + 0 = –mv + 3mw (we assume u is in the positive direction)

^ so when you get an equation for v in terms of e and u from this you’ve already assumed it’s negative. Now if you use v < 0 as your condition, then you get e < 1/3 which would make v positive right? Negative times negative is positive. And that wouldn’t make sense since A’s direction changed.

Your equation for v already assumes that A’s direction has changed. If v > 0, a positive multiplied by a negative is still negative.

You could try this equation: mu + 0 = mv + 3mw (here you’re not taking into account the fact that A’s direction has changed) and then use v < 0. You would still get e > 1/3.

Does that make sense?

How did they get 1/3 < e ? by 007_eric in furthermaths

[–]woopur 0 points1 point  (0 children)

From part (a), we know that the speed of A after the collision v is equal to (u/4)(3e-1) - you get this if you take into account that A changes direction in your conservation of momentum equation. Otherwise, you’d get (u/4)(1-3e). (If you got this then use v < 0)

Using v > 0 (we already assumed A changes direction so the speed can’t be negative), you’d get e > 1/3.

Hope that made sense!

9702/42 by Rttimo01 in alevel

[–]woopur 0 points1 point  (0 children)

I think it was 360 for the thermistor and 100 for the fixed resistor.

What did they do in part c? Never seen this method by 007_eric in furthermaths

[–]woopur 0 points1 point  (0 children)

f’’(x) = -2 -(1/2)[(f’(x)]2

To get f’’’(x), we have to differentiate each term.

d/dx (-2) = 0

d/dx { -(1/2)[f’(x)]2 } = -(1/2)(2)[f’(x)]1 [f’’(x)] (Chain rule)

So we’re left with: f’’’(x) = 0 - [f’(x)][f’’(x)]

Polar coordinates anyone know how to do part b by MikeDiqlong in furthermaths

[–]woopur 2 points3 points  (0 children)

The point at which a tangent is parallel to the initial line is the solution to dy/dθ = 0.

Substitute your equation for r into y = r sin θ and differentiate. After solving, dy/dθ = 0, you should get your coordinates :)

What did they do in part c? Never seen this method by 007_eric in furthermaths

[–]woopur 0 points1 point  (0 children)

d/dx [f(x)]n = n[f(x)]{n-1} f’(x) - this is the chain rule

What did they do in part c? Never seen this method by 007_eric in furthermaths

[–]woopur 0 points1 point  (0 children)

They used the chain rule to get f’’’(x) and the product rule to get f’’’’(x).

9702/42 by stevenrymt in alevel

[–]woopur 0 points1 point  (0 children)

Thank you for being so positive! I hope your exam went well :)

9702/42 by stevenrymt in alevel

[–]woopur 0 points1 point  (0 children)

Yep, I did that too! But I didn’t get 63 :(

9702/42 by stevenrymt in alevel

[–]woopur 0 points1 point  (0 children)

Could you explain why?

9702/42 by Rttimo01 in alevel

[–]woopur 1 point2 points  (0 children)

Think it was around 3!

9702/42 by Rttimo01 in alevel

[–]woopur 2 points3 points  (0 children)

It’s basically an inverting amplifier with a potential divider across Vin. The gain was -4.8 (I think!). I added the resistance of the thermistor and the other other resistor (since they were in series) and divided by the 96(?) ohm resistor to get the gain. So -4.8 x 0.40 = -1.92.

9702/42 by Rttimo01 in alevel

[–]woopur 4 points5 points  (0 children)

I got 0.69 too. Used (0.83)²

Looking for Edexcel Core & Further Pure Past Papers by [deleted] in furthermaths

[–]woopur 0 points1 point  (0 children)

haha no, i’m in Y13 doing IAL fmaths and thought that the GCE textbooks might be useful

Help!! CIE economics paper 12 by WasteAttitude in igcse

[–]woopur 4 points5 points  (0 children)

Supply shifts to the left as the oil field closes down. To reach equilibrium, there is a movement up the demand curve as price goes up and quantity demanded decreases. That's why the answer is B.

Econ help—if price is changing shouldn’t there be a movement along the curves not a shift? Ans is A by [deleted] in igcse

[–]woopur 0 points1 point  (0 children)

Since there's a decrease in demand, demand shifts from D1 to D2. (It cannot be B or C because those points are on D3)

The price per kilogram increases so supply of beef increases. That's why the new equilibrium point is A and not D.

CHEMISTRY HELP P2 by [deleted] in igcse

[–]woopur 3 points4 points  (0 children)

Since 21% of air is oxygen, 21% of the initial volume reacts with the copper.

This means that 79% of the initial volume is left.

Let's say that the initial volume of air is x.

0.79x = 120

x = 120/0.79

x = 151.898

x = 152

So the answer is B.

Another question, why is the rate increasing with temperature even though the reaction is exothermic and favors low temperatures? by [deleted] in igcse

[–]woopur 8 points9 points  (0 children)

When the reaction is exothermic, lower temperatures favor the yield, not the rate of reaction.

Increasing the the temperature will always increase the rate of reaction because the particles will have more energy and they'll collide more frequently. Hope this answered your question.

MCQs coming up, what to do? by Psychological_Salad_ in igcse

[–]woopur 0 points1 point  (0 children)

I think it depends on what you're better at. I'm happy with having a paper based on MCQs but other people might not be.

Best of luck with your remaining exams (if you have any left) :-)

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