September 19, 2020, 07:01:56 pm

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#### twelftholmes

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##### Re: HSC Physics Question Thread
« Reply #3915 on: June 29, 2020, 10:04:09 pm »
0
Hey!!
I have a question from Jamon Windeyer's awesome physics topic tests if that's okay

Here is the Q: https://imgur.com/a/EfhvX8o
Here is my attempted answer: https://imgur.com/a/kXLoz0e

So basically, in the back of the answers Jamon has put that we can solve this either using conservation of energy and work, or calculate the acceleration and then approach it as a projectile motion question. I don't feel confident with latter and my class hasn't done module 5 yet (we're doing the modules in the order 6, 8, 7, 5 oop) so I chose the former which I feel like I understand much at this point. However I got it wrong and I'm wondering if it's because of an error in my working out or understanding using the method I did, or if it's because I'm better off approaching it like a projectile motion question (like Jamon did in the back of the book, however I didn't understand that which is why I'm here haha). If someone wouldn't mind helping me figure out how to get the correct answer that'd be awesome.  (using the method I did if that's possible).

#### blasonduo

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##### Re: HSC Physics Question Thread
« Reply #3916 on: June 29, 2020, 11:21:27 pm »
+7
Hey!!
I have a question from Jamon Windeyer's awesome physics topic tests if that's okay

Here is the Q: https://imgur.com/a/EfhvX8o
Here is my attempted answer: https://imgur.com/a/kXLoz0e

So basically, in the back of the answers Jamon has put that we can solve this either using conservation of energy and work, or calculate the acceleration and then approach it as a projectile motion question. I don't feel confident with latter and my class hasn't done module 5 yet (we're doing the modules in the order 6, 8, 7, 5 oop) so I chose the former which I feel like I understand much at this point. However I got it wrong and I'm wondering if it's because of an error in my working out or understanding using the method I did, or if it's because I'm better off approaching it like a projectile motion question (like Jamon did in the back of the book, however I didn't understand that which is why I'm here haha). If someone wouldn't mind helping me figure out how to get the correct answer that'd be awesome.  (using the method I did if that's possible).

Hey!
Just looking at the question given, it originally says half a meter apart, but the diagram shows it as a meter apart. (So there's a typo in the question!)

What distance did Jamon use? This may be the error, because at a quick glance I don't see anything wrong with your working!

EDIT: So I've checked out how Jamon has done his answer. In the first part he uses d = 0.5, but in the second part of the working he uses d = 1.

$\ v^2 = 0^2 + 2 \times 3.52\times 10^{15}$
There should be an additional factor of 0.5.

Doing this does indeed give the same answer that you have worked out
« Last Edit: June 29, 2020, 11:45:01 pm by blasonduo »
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#### twelftholmes

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##### Re: HSC Physics Question Thread
« Reply #3917 on: June 30, 2020, 06:17:03 pm »
+1
awesome, that's a relief to hear! thanks, I was so confused trying to figure out where I went wrong haha. It's okay I understand that even in textbooks there is human error

#### shekhar.patel

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##### Re: HSC Physics Question Thread
« Reply #3918 on: August 04, 2020, 09:58:55 pm »
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Hi,

How did we determine the direction of torque on an object like in this question. Is there a rule?

Thank you.

#### Bri MT

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##### Re: HSC Physics Question Thread
« Reply #3919 on: August 05, 2020, 09:20:04 am »
+2
Hi,

How did we determine the direction of torque on an object like in this question. Is there a rule?

Thank you.

Hey, if you look at the diagram you should be able to see that the component of the force perpendicular to the radius is pointed down. Downwards force to the right of the pivot point = clockwise rotation

#### mrsc

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##### Re: HSC Physics Question Thread
« Reply #3920 on: August 08, 2020, 05:43:24 pm »
0
Hey just wondering the logic behind this question. The answer was B, but I somehow don't seem to get why. Is it because decreasing the number of secondary coils reduces the emf in them which causes an increased deflection due to the change in magnetic flux?

#### fun_jirachi

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##### Re: HSC Physics Question Thread
« Reply #3921 on: August 08, 2020, 09:12:19 pm »
+5
Hey there!

The galvanometer detects changes in current - so when we look at the transformer equation $\frac{V_P}{V_S} = \frac{I_S}{I_P} = \frac{N_P}{N_S}$, we can deduce that to increase the deflection in the galvanometer ie. increasing $I_S$, we can do a few things: increase the number of primary coils, decrease the number of secondary coils, increase the primary voltage, etc. We also can rule out C and D because the primary function of a resistor is to reduce current (not what we want here!) and copper cores perform worse as compared to iron cores. Hence, the answer is B.

Hope this helps
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#### Coolmate

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##### Re: HSC Physics Question Thread
« Reply #3922 on: August 11, 2020, 07:44:31 pm »
0
Hey Physics Gang!

Could someone please explain to me why the answer is 'C'? Does it involve deriving another formula, or the like out of this formula: $dsin\theta = m\lambda$
Coolmate
« Last Edit: August 12, 2020, 08:36:43 am by Coolmate »
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#### 1729

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##### Re: HSC Physics Question Thread
« Reply #3923 on: August 11, 2020, 08:01:50 pm »
+6
Hey Physics Gang!

Could someone please explain to me why the answer is 'C'? Does it involve deriving another formula, or the like out of this formula: $dsin\theta = m\lambda$

Coolmate
The answer is C  because it's a dark spot, it means they're out of phase, so p2 would have to be longer than p1 by a factor of (2n(lambda)+1)/2 to create perfectly destructive interference. As you can see by the center bright spot, that's what happens when p1 and p2 are the same length. This means that the 2 dark spots closest to the center will be half a wavelength out of phase which is the closest difference that creates a dark spot. This means that it would be lambda/2, and since the dark spot we want is one dark spot further, the difference would have to be 3(lambda)/2

#### Coolmate

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##### Re: HSC Physics Question Thread
« Reply #3924 on: August 12, 2020, 09:21:10 am »
0
The answer is C  because it's a dark spot, it means they're out of phase, so p2 would have to be longer than p1 by a factor of (2n(lambda)+1)/2 to create perfectly destructive interference. As you can see by the center bright spot, that's what happens when p1 and p2 are the same length. This means that the 2 dark spots closest to the center will be half a wavelength out of phase which is the closest difference that creates a dark spot. This means that it would be lambda/2, and since the dark spot we want is one dark spot further, the difference would have to be 3(lambda)/2

Thanks 1729!
This makes much more sense now

Coolmate
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#### Coolmate

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##### Re: HSC Physics Question Thread
« Reply #3925 on: August 24, 2020, 02:44:17 pm »
0
Hi Everyone,

Could someone please step me through how to solve this question (attached) from the 2012 HSC paper (Q27, 2012)

Coolmate
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#### fun_jirachi

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##### Re: HSC Physics Question Thread
« Reply #3926 on: August 24, 2020, 05:44:54 pm »
+6
Hey

One of the first things that pops out to me is that you have that the horizontal displacement is 45m - indicating that from $s_x = u_xt$, we have that $45 = u\cos 60 \times t$. Secondly, we have that from $s_y = u_yt - \frac{1}{2}gt^2$ - thus we also have that $34 = u\sin 60 \times t - \frac{gt^2}{2}$. Rearranging and substituting the first equation into the second yields an equation in $u$, specifically $34 = 45\tan 60 - \frac{g}{2} \times \left(\frac{45}{u\cos 60}\right)^2.$ Rearrange and solve from here to get the answer of 30ms-1

The thought process I followed here was to start with the easiest equations that you can substitute info from the question into - it's important to ask yourself what do you know? It's also important to try and eliminate one variable ASAP in equations of multiple variables (you can always sub back in later to find the other once you've found a definitive value for one of them). If all else fails, literally write down everything they give you ie. $s_x = 45, s_y = 34$ and see what equations 'match up' - much like questions towards the back end of maths exams and stuff (just try something!) Definitely look to practice more of these types of questions (or change the numbers a bit) if this is the type of thing you struggle with
« Last Edit: August 24, 2020, 05:46:41 pm by fun_jirachi »
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ATAR: 99.05

UCAT: 3310 - Verbal Reasoning [740] | Decision Making [890] | Quantitative Reasoning [880] | Abstract Reasoning [800]
Subject Acceleration (2018)
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#### Coolmate

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##### Re: HSC Physics Question Thread
« Reply #3927 on: August 25, 2020, 01:12:21 pm »
+2
Hey

One of the first things that pops out to me is that you have that the horizontal displacement is 45m - indicating that from $s_x = u_xt$, we have that $45 = u\cos 60 \times t$. Secondly, we have that from $s_y = u_yt - \frac{1}{2}gt^2$ - thus we also have that $34 = u\sin 60 \times t - \frac{gt^2}{2}$. Rearranging and substituting the first equation into the second yields an equation in $u$, specifically $34 = 45\tan 60 - \frac{g}{2} \times \left(\frac{45}{u\cos 60}\right)^2.$ Rearrange and solve from here to get the answer of 30ms-1

The thought process I followed here was to start with the easiest equations that you can substitute info from the question into - it's important to ask yourself what do you know? It's also important to try and eliminate one variable ASAP in equations of multiple variables (you can always sub back in later to find the other once you've found a definitive value for one of them). If all else fails, literally write down everything they give you ie. $s_x = 45, s_y = 34$ and see what equations 'match up' - much like questions towards the back end of maths exams and stuff (just try something!) Definitely look to practice more of these types of questions (or change the numbers a bit) if this is the type of thing you struggle with

Hey fun_jirachi,

Thankyou for the assistance, it really helped me today! (I just came out of my Physics Trial exam and your help was very helpful)

Thanks again!
Coolmate
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