AQA Physics PHYA5 - Thursday 18th June 2015 [Exam Discussion Thread]

OP

Original post by TheSmartOne

Question for those doing astro;

At what point does Tan(theta) become equal to theta in angular magnification? My teacher was very vague about it, and I've read somewhere that it is for angles below 6 degrees? Can someone confirm/correct this? Thanks.

I know what you mean. It's the small angle approximation:
"a useful simplification of the basic trigonometric functions which is approximately true in the limit where the angle approaches zero."

This concept also applies in Simple Harmonic Motion. We assume the equation:

[br]T=2\pi \sqrt{\dfrac{l}{g}}[br]

Is relevant for a simple pendulum, provided the angle of the thread to the vertical does not exceed about 10 degrees.

I have heard 10 degrees being quoted as the upper limit before, which is approximately

\frac{\pi}{18}

radians.

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Reply 882

JJBinn

Can somebody explain this from the Jun 14 paper? I don't get it:

Explain how measurements of a type 1a supernova can be used to determine how faraway it is from the Earth. (3)

The mark scheme says the peak abs magnitude for 1a supernovae is the same. And that you can then measure the apparent magnitude at this peak and therefore work out distance. I don't understand how you would know when the absolute magnitude was at its peak?

Reply 883

OP

Original post by JJBinn

Can somebody explain this from the Jun 14 paper? I don't get it:

Explain how measurements of a type 1a supernova can be used to determine how faraway it is from the Earth. (3)

The mark scheme says the peak abs magnitude for 1a supernovae is the same. And that you can then measure the apparent magnitude at this peak and therefore work out distance. I don't understand how you would know when the absolute magnitude was at its peak?

Type 1a supernovae are used because they have a known peak absolute magnitude (of about -19.3).

If you plot a graph of intensity against time for the supernovae, the peak brightness may be worked out.

To find distances in space, astronomers use objects called "standard candles." Standard candles are objects that give a certain, known amount of light. Because astronomers know how bright these objects truly are, they can measure their distance from us by analyzing how dim they appear.

Type Ia supernovae occur in a binary system — two stars orbiting one another. One of the stars in the system must be a white dwarf star, which attracts matter from the other star (a giant or even another white dwarf) before exploding.

When the white dwarf reaches 1.4 solar masses, a nuclear chain reaction occurs, causing the white dwarf to explode.

Because the chain reaction always happens in the same way, and at the same mass, the brightness of these Type Ia supernovae are also always the same.

Using the inverse square law, or the equation

m-M=5log\left(\frac{d}{10}\right)

, along with measurements of the apparent magnitude from Earth, astronomers can compute the distance to the supernova and thus to the supernova's home galaxy.

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Reply 884

JJBinn

Original post by CD223

Type 1a supernovae are used because they have a known peak absolute magnitude (of about -19.3).

If you plot a graph of intensity against time for the supernovae, the peak brightness may be worked out.

To find distances in space, astronomers use objects called "standard candles." Standard candles are objects that give a certain, known amount of light. Because astronomers know how bright these objects truly are, they can measure their distance from us by analyzing how dim they appear.

Type Ia supernovae occur in a binary system — two stars orbiting one another. One of the stars in the system must be a white dwarf star, which attracts matter from the other star (a giant or even another white dwarf) before exploding.

When the white dwarf reaches 1.4 solar masses, a nuclear chain reaction occurs, causing the white dwarf to explode.

Because the chain reaction always happens in the same way, and at the same mass, the brightness of these Type Ia supernovae are also always the same.

Using the inverse square law, or the equation

m-M=5log\left(\frac{d}{10}\right)

, along with measurements of the apparent magnitude from Earth, astronomers can compute the distance to the supernova and thus to the supernova's home galaxy.

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Thanks, this is incredibly helpful. I never felt the textbook was very good with surpernovae. This is really clear on type 1a, thanks

Reply 885

OP

Original post by JJBinn

Thanks, this is incredibly helpful. I never felt the textbook was very good with surpernovae. This is really clear on type 1a, thanks

No problem! I agree. The textbook isn't great on types of supernovae.

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Reply 886

JJBinn

Original post by CD223

No problem! I agree. The textbook isn't great on types of supernovae.

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Do you know why the relative abundance of hydrogen and helium compared to other elements is evidence for TBBT? the text book just explains the ratios but not actually why the relative abundance is supportive of the BBT.

Reply 887

OP

Original post by JJBinn

Do you know why the relative abundance of hydrogen and helium compared to other elements is evidence for TBBT? the text book just explains the ratios but not actually why the relative abundance is supportive of the BBT.

The ratios predicted in the conditions of the early universe (14:2 proton to neutron, 3:1 H :h:

e) match the relative abundance seen today, suggesting the theory is valid.

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Reply 888

OP

Anyone know the key characteristics of a beta source in an experiment? I get that they are stopped by 3cm of Aluminium, but I've read that there is an exponential attenuation with distance in air. Is this something we are likely to need to know?

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Reply 889

_Caz_

Original post by CD223

Anyone know the key characteristics of a beta source in an experiment? I get that they are stopped by 3cm of Aluminium, but I've read that there is an exponential attenuation with distance in air. Is this something we are likely to need to know?

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No the only relationship with distance we need is the gamma one I think. :smile:

Reply 890

_Caz_

I don't suppose any astro person has a good flow diagram for stellar evolution?

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Reply 891

OP

Original post by _Caz_

I don't suppose any astro person has a good flow diagram for stellar evolution?

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What about this?

ImageUploadedByStudent Room1432230118.553463.jpg

Reply 892

frankiejayx

Original post by CD223

What about this?

ImageUploadedByStudent Room1432230118.553463.jpg

Can the white dwarf produce a Supernova also?

Reply 893

TheSmartOne

How do you rearrange m-M=5log d/10 for d?

I've got d=10e^[(m-M)/5] but my answer is still wrong. Help?

Reply 894

Lau14

11

Original post by TheSmartOne

How do you rearrange m-M=5log d/10 for d?

I've got d=10e^[(m-M)/5] but my answer is still wrong. Help?

the log is base 10, not ln (log of base e), so you should have 10^etc instead of e^etc

Reply 895

Lau14

11

Original post by frankiejayx

Can the white dwarf produce a Supernova also?

Yes, if a white dwarf is in a binary system with another star that goes red giant and so increases in size rapidly then the white dwarf can pull in enough material from the other star (due to its gravity) to restart fusion. This releases enough energy for a supernova! It's called a type 1a supernova and this is the one you have to know as a standard candle (as discussed further up this page). This isn't part of the standard life cycle of a star though.

Reply 896

OP

Original post by TheSmartOne

How do you rearrange m-M=5log d/10 for d?

I've got d=10e^[(m-M)/5] but my answer is still wrong. Help?

[br]m-M=5\log{\left(\dfrac{d}{10}\right)}[br]

[br]\therefore \dfrac{m-M}{5}=\log{\left(\dfrac{d}{10}\right)}[br]

[br]\therefore 10^{\frac{m-M}{5}}=\dfrac{d}{10}[br]

[br]\therefore 10 \times 10^{\frac{m-M}{5}}= d[br]

Reply 897

OP