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AQA Physics PHYA4 - Thursday 11th June 2015 [Exam Discussion Thread]

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Can someone explain what these equations find?

Are the emf equations average emf induced and is the 2nd average flux linkage?
Also when should I use each of them (e.g. for a fixed rotation or continuous rotation?)

Reply 2181

bwr19

Original post by gcsestuff

Sorry this is the other question ImageUploadedByStudent Room1432987214.375637.jpg

Posted from TSR Mobile

Potential Gradient, Field Strength and Force Per Unit Mass all have units Nkg^-1.

Potential has the units Jkg^-1

Reply 2182

bwr19

Original post by gcsestuff

ImageUploadedByStudent Room1432987088.510587.jpg

Can I have some help with these 2, I don't know what the formula/units for potential gradient are.

And also for question 1 I've always just accepted this, how do you work it out :smile:

Posted from TSR Mobile

Basically, I don't know how good you are with maths, but, lets assume that the displacement, x, is a cosine wave.

Velocity is dx/dt - the rate of change of displacement. In other words, this is the gradient of the displacement-time graph. If you differentiate a cosine wave, you get a negative sine wave (you can verify this by looking at the gradient on the displacement-time graph when it is a 0, max, and min).

Acceleration is dv/dt or d²x/dt², so the rate of change of velocity, or the rate of rate of change of displacement. If you differentiate the negative sine graph, you get a negative a cosine graph. This is the gradient of the velocity-time graph which you can verify for yourself.

So a negative sine graph is pi/2 radians out of phase with a cosine graph.

Hope that makes sense, but it might be a bit too mathsy for physics. That's how I remember it, anyway.

Reply 2183

bwr19

Original post by gcsestuff

Finally.

ImageUploadedByStudent Room1432987716.262116.jpg

ImageUploadedByStudent Room1432987726.909437.jpg

ImageUploadedByStudent Room1432987737.959531.jpg

I thought that the centripetal force acted towards the centre of oscillation? But the mark schemes answer says t=mv^2/l+mg

Thanks for any help!

Posted from TSR Mobile

Your problem arises when you wrote

T + \frac{mv^2}{r} = mg

Remember that centripetal force is a resultant force. So the centripetal force is provided by the resultant force of the weight and the tension. In other words:

T - mg = \frac{mv^2}{r}

(Remember that forces are vectors. The centripetal force acts in the same direction as the tension, and the weight in the opposite direction hence the negative sign).

Hope this helps.

Reply 2184

JJBinn

Original post by CD223

And question 12:
ImageUploadedByStudent Room1432984825.479479.jpg

Sorry I don't have the others! Let me know if you still need help when I'm back at home :smile:

Posted from TSR Mobile

Thanks, don't worry about it! I don't quite understand how you got to your last two boxes of working on this question though?

Reply 2185

bwr19

Original post by gcsestuff

ImageUploadedByStudent Room1432986814.002696.jpg

Can anyone help me with 13, I know it acts downwards and I know field strength is constant but how do I work out the potential; thanks :smile:

)

Posted from TSR Mobile

For electric field strength:

E = \frac{V}{d} = \frac{50}{5.0 \times 10^{-3}} = 1.0 \times 10^4

Acting downwards since that is the direction a positive test charge would move when placed in the field.

For electric potential energy:

Because this is a uniform field, then the equipotentials are spread out uniformly - in other words, the potential changes linearly. There is a potential difference of 50V between the plates, and it's uniform, so directly in the centre the potential will be 25V since it's half way between the plates. Make sense?

So the answer is B.

Reply 2186

noseypo

How is the resultant electrostatic Force found between two oppositely charged ions?

Reply 2187

bwr19

Original post by JJBinn

Thanks, don't worry about it! I don't quite understand how you got to your last two boxes of working on this question though?

He just substituted in the values he had previously worked out.

\displaystyle g_e = \frac{GM_e}{R^2_e}

and

\displaystyle g_m = \frac{GM_m}{R^2_m}

\displaystyle \frac{g_e}{g_m} = \frac{G\left (\frac{M_e}{M_m} \right )}{\left ( \frac{R_e}{R_m} \right )^2}

And we know:

\displaystyle M_e = \rho \times \frac{4}{3}\pi R_e^3

and

\displaystyle M_m = \rho \times \frac{4}{3}\pi R_m^3

Then we can just sub in some of the ratios that the question told us, and solve.

(edited 8 years ago)

Reply 2188

bwr19

Original post by noseypo

How is the resultant electrostatic Force found between two oppositely charged ions?

\displaystyle F = \frac{Qq}{4\pi \varepsilon_0 r^2}

(edited 8 years ago)

Reply 2189

SuperMushroom

How would you go about answering Question 10 on this paper from the multiple choice section, i just cant quite understand it

http://filestore.aqa.org.uk/subjects/AQA-PHYA4-1-W-QP-JAN10.PDF

Thanks

Reply 2190

noseypo

Original post by bwr19

\displaystyle F = \frac{Qq}{4\pi \varepsilon_0 r^2}

Thanks, is it positive if they are of opposite charge?

Reply 2191

bwr19

Original post by SuperMushroom

How would you go about answering Question 10 on this paper from the multiple choice section, i just cant quite understand it

http://filestore.aqa.org.uk/subjects/AQA-PHYA4-1-W-QP-JAN10.PDF

Thanks

Here you go. The way of doing is it to describe the g and r of the mystery planet in terms of the g and r of Earth. And then things neatly cancel away.

Reply 2192

bwr19

Original post by noseypo

Thanks, is it positive if they are of opposite charge?

The value would be negative if they are of opposite charge. However, the exam usually only asks for the magnitude of the force, so you can ignore the negative sign in that case.

Reply 2193

SuperMushroom

Original post by bwr19

Here you go. The way of doing is it to describe the g and r of the mystery planet in terms of the g and r of Earth. And then things neatly cancel away.

Hi thanks for that however you answered question 11, the question i don't quite understand is question 10

Reply 2194

bwr19

Original post by SuperMushroom

Hi thanks for that however you answered question 11, the question i don't quite understand is question 10

Whoops! That'll teach me not to rush. Here's question 10 for you:

Image (38).jpg

Reply 2195

SuperMushroom

Original post by bwr19

Whoops! That'll teach me not to rush. Here's question 10 for you:

Image (38).jpg

Thanks for taking your time to do that :biggrin:

, Are you a student who is also taking the exam?

Reply 2196

bwr19

Original post by SuperMushroom

Thanks for taking your time to do that :biggrin:

, Are you a student who is also taking the exam?

I am indeed.

I'm feeling pretty good about most of the course apart from magnetic fields, induction, and transformers. My teacher left all that stuff until the last week or two, so we rushed through it and none of us really understand it. :s-smilie:

How about you? How are you feeling for it?

Reply 2197

SuperMushroom

Original post by bwr19

How about you? How are you feeling for it?

Ahh ok i see, im pretty confident with answering all the section B questions, the thing that i am struggling with the most is the Multiple choice questions, just going to keep practicing them and hopefully i can get the hang of them :biggrin:

I'm relying to get most of my marks from Physics 5

Reply 2198

FN510

Original post by Disney0702

Yes they are!

You have to make sure you put any numerical answer in the correct number of significant figures.

Unfortunately they do not usually tell you what is the correct number of significant figures is, it is something that you'll have to figure out when you tackle the questions.

It is usually 2 or 3 significant figures.

People usually do not score the final answer mark just because they didn't round it to the correct significant figure even if their answer is correct.