HSC Physics Marathon 2016 (1 Viewer)

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Drsoccerball

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Ohh so do you mean I should talk about the re-entry ? Because stoping acceleration upon launching would seem counterintuitive :/ Was my answer complete rubbish or did I get some marks



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No Multi-staged rockets are used when it is launched. Did you understand my explanation? Also the middle part was incorrect so you would've lost some marks.
 

Glyde

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No Multi-staged rockets are used when it is launched. Did you understand my explanation? Also the middle part was incorrect so you would've lost some marks.
I don't think so :/ wouldn't dumping empty fuel tanks allow for greater acceleration in the upward direction ?


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Drsoccerball

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I don't think so :/ wouldn't dumping empty fuel tanks allow for greater acceleration in the upward direction ?


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You're thinking about it wrong. Fair enough you would get a greater acceleration but since we attempt to reduce g force as much as possible we waste all the fuel in one tank rather than using one tank for the whole journey as this would provide a constant acceleration according to So what we do is we waste all the mass in one and the acceleration rapidly decreases. When it is at reasonable levels we use the next tank so that the people don't die.

Since
 

Glyde

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You're thinking about it wrong. Fair enough you would get a greater acceleration but since we attempt to reduce g force as much as possible we waste all the fuel in one tank rather than using one tank for the whole journey as this would provide a constant acceleration according to So what we do is we waste all the mass in one and the acceleration rapidly decreases. When it is at reasonable levels we use the next tank so that the people don't die.

Since
Ohhhh so if we are constantly burning fuel from the same rank the whole way , the ship would be experiencing too many g's causing a black out. Hence dumping them will keep the acceleration at a constant level instead and that constant acceleration would be under say 6 g's to stop people's dying



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leehuan

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You're thinking about it wrong. Fair enough you would get a greater acceleration but since we attempt to reduce g force as much as possible we waste all the fuel in one tank rather than using one tank for the whole journey as this would provide a constant acceleration according to So what we do is we waste all the mass in one and the acceleration rapidly decreases. When it is at reasonable levels we use the next tank so that the people don't die.

Since
1+a/g confused the crap out of me. I used sigma(a)/g where the sum of all forces could only be from gravity and thrust (i.e. no 'normal reaction'). So on Earth's surface I treated sigma(a) as 9.8 instead.
 

Drsoccerball

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Ohhhh so if we are constantly burning fuel from the same rank the whole way , the ship would be experiencing too many g's causing a black out. Hence dumping them will keep the acceleration at a constant level instead and that constant acceleration would be under say 6 g's to stop people's dying



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No ahaha. We don't even know if we could lift off with such low acceleration. We use it so that its as if you have tiny rocket launches throughout the journey. From A to B we have one rocket launch and when the fuel stops being used it stops accelerating and a reduces drastically reducing the g forces experienced. From B to C we have another rocket launch which starts from basically 0 acceleration... Lets say that the acceleration of A to B is X and B to C is also X. The max g force experienced would be = However if this was done in one journey it woud be ...
 

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Omg I get it now!!!!!!!! Thanks sooooo much !!! I had no idea that the rockets stopped in mid air and then turned on there next rocket


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Glyde

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Sorry it took me so long


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Drsoccerball

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1+a/g confused the crap out of me. I used sigma(a)/g where the sum of all forces could only be from gravity and thrust (i.e. no 'normal reaction'). So on Earth's surface I treated sigma(a) as 9.8 instead.






 

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To add to this, the reason why it is sometimes plus or minus is because the acceleration (a) is a vector. If can be either pointing downwards which is negative and why it is negative in your first example, or it can be pointing upwards which is why it is positive in your second example.
 

leehuan

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Although that's the most accurate interpretation, when I think about that I get confused. I'm just saying instead I actually imagine we are accelerating at 9.8ms^-2 right now instead.
 

leehuan

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Bump
I saw this question from Syd Grammar's old trial paper and thought was a great question:
"The Earth exerts the exact same gravitational force on the Sun as the Sun does on the Earth due to Newtons 3rd Law". Assess the above statement (4 marks)
 

Glyde

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According to Newtons 3rd Law, 'Every reaction will have an equal and opposite reaction' we are aware that when two objects interact with each other that there will be a force between them that is the same in magnitude however opposite in direction. Hence it is logical to assume that the Earth and Sun will follow this rule. That is the gravitational pull of the sun on the earth will be the same magnitude as the gravitational pull of the earth on the sun. We can also show that the Sun and Earth are attracted towards each other using the formulae 'F=(Gm1m2)/r^2' where the mass (m) are representative of the earth and sun.


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Glyde

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I wouldn't be surprised if I am completely wrong



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leehuan

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According to Newtons 3rd Law, 'Every reaction will have an equal and opposite reaction' we are aware that when two objects interact with each other that there will be a force between them that is the same in magnitude however opposite in direction. Hence it is logical to assume that the Earth and Sun will follow this rule. That is the gravitational pull of the sun on the earth will be the same magnitude as the gravitational pull of the earth on the sun. We can also show that the Sun and Earth are attracted towards each other using the formulae 'F=(Gm1m2)/r^2' where the mass (m) are representative of the earth and sun.


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We call the formula Newton's Law of Universal Gravitation. Also, it's every action will have an equal and opposite reaction, so be careful of this. I'll let someone else put on a mark.
 

Glyde

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Okay, yes I new that I just wasn't thinking about what I was writing haha thank you


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Drsoccerball

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New question : Explain why the object used for slingshot motion (planet, star, etc...) needs to be moving.
 

leehuan

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New question : Explain why the object used for slingshot motion (planet, star, etc...) needs to be moving.
LOL.

Sorry, the question is genuine but that seemed like logic to me.
 

RachelGreen

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According to Newtons 3rd Law, 'Every reaction will have an equal and opposite reaction' we are aware that when two objects interact with each other that there will be a force between them that is the same in magnitude however opposite in direction. Hence it is logical to assume that the Earth and Sun will follow this rule. That is the gravitational pull of the sun on the earth will be the same magnitude as the gravitational pull of the earth on the sun. We can also show that the Sun and Earth are attracted towards each other using the formulae 'F=(Gm1m2)/r^2' where the mass (m) are representative of the earth and sun.


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3/4! Marking criteria says to don't forget to make a judgement of the statement at the end
 

Glyde

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Could you please explain what you mean by this or provide an example


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