Login

Welcome, Guest. Please login or register.

March 29, 2024, 10:38:26 pm

Author Topic: HSC Physics Question Thread  (Read 1030730 times)  Share 

0 Members and 1 Guest are viewing this topic.

arunasva

  • Forum Regular
  • **
  • Posts: 54
  • 93.90
  • Respect: +1
Re: Physics Question Thread
« Reply #1950 on: March 25, 2017, 04:26:23 am »
0
What do you have to do if you do not get a concept at all ? I have 2
Projectile Motion and Special Relativity (Math stuff the theory is alright) I tried practising everything but couldnt do anything without looking at the answers.
:3

jamonwindeyer

  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 10150
  • The lurker from the north.
  • Respect: +3108
Re: Physics Question Thread
« Reply #1951 on: March 25, 2017, 10:22:55 am »
+1
What do you have to do if you do not get a concept at all ? I have 2
Projectile Motion and Special Relativity (Math stuff the theory is alright) I tried practising everything but couldnt do anything without looking at the answers.

Getting used to the Math questions is just practice - It can take a bit, but eventually you will start noticing patterns and getting better!! ;D

A few things that might help:

- Guide to Math Questions
- Guide to Rocket Launches
- Guide to Relativity

bluecookie

  • Trendsetter
  • **
  • Posts: 125
  • Respect: +1
Re: Physics Question Thread
« Reply #1952 on: March 27, 2017, 11:13:05 pm »
0
Ok im confused. The formula for transformers are:
vp/vs=is/ip (if u let the power of the primary equal the power of the secondary que law of conservation and sub p=iv and crossmultiply)
They say if you increase vp by factor, is decreases by same factor and vice vercea, but it's just not making sense to me atm. If you multiply vp by a factor of k, you must multiply RHS by factor of k so they can cancel out and the equation becomes the same. So you multiply k to the numerator and isn't is increased by a factor of k as well? Not decreased?

bluecookie

  • Trendsetter
  • **
  • Posts: 125
  • Respect: +1
Re: Physics Question Thread
« Reply #1953 on: March 27, 2017, 11:19:36 pm »
0
On similar note. Does np>ns apply for step up or step down? (and vice vercea for ns>np)

arunasva

  • Forum Regular
  • **
  • Posts: 54
  • 93.90
  • Respect: +1
Re: Physics Question Thread
« Reply #1954 on: March 27, 2017, 11:51:20 pm »
+1
On similar note. Does np>ns apply for step up or step down? (and vice vercea for ns>np)

When you increase the Voltage of the primary Current of the Primary decreases. Vp/Vs = Is/Ip. Yeah if you multiply both numerators by K you will increase the value of the Voltage of the Primary and Current Current Of The Secondary. Therefore current of the primary(the denominator Ip) decreases because the numerator Is gets bigger than Ip. So you increased the voltage for the Primary coil and the Current in that coil (Ip) decreased while the current in the other coil Increased (Is)
:3

kiwiberry

  • HSC LECTURER
  • Forum Obsessive
  • ***
  • Posts: 315
  • Respect: +97
Re: Physics Question Thread
« Reply #1955 on: March 28, 2017, 08:07:49 am »
+1
On similar note. Does np>ns apply for step up or step down? (and vice vercea for ns>np)

np>ns for step down because emf is proportional to the number of coils, so a smaller number will induce a smaller emf in the secondary coil :) and vice versa for step up
HSC 2017: English Adv (93) | Maths Ext 1 (99) | Maths Ext 2 (97) | Chemistry (95) | Physics (95)
ATAR: 99.85

jamonwindeyer

  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 10150
  • The lurker from the north.
  • Respect: +3108
Re: Physics Question Thread
« Reply #1956 on: March 28, 2017, 08:32:40 am »
+1
Just on above as well, it depends if you keep power constant! If you increase voltage without decreasing current/power, that is different to an increase in voltage but a decrease in current. If you increase voltage and decrease current (to make power a constant),  then the output must have the same power too - This means that the voltage at the output will increase, and therefore the current at the output will decrease - Always think conservation of energy!

Awesome answers above, cheers guys

lsong

  • Adventurer
  • *
  • Posts: 16
  • Respect: 0
Re: Physics Question Thread
« Reply #1957 on: March 28, 2017, 02:55:15 pm »
0
Hi, one of the dot points in the Motors and Gen syllabus asks to identify how transmission lines are insulated from supporting structures. I can only find one out of the three (which is the number that my teacher asked for). Is there three possible examples for this?
Any ideas would be great thanks!

jamonwindeyer

  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 10150
  • The lurker from the north.
  • Respect: +3108
Re: Physics Question Thread
« Reply #1958 on: March 28, 2017, 03:17:54 pm »
0
Hi, one of the dot points in the Motors and Gen syllabus asks to identify how transmission lines are insulated from supporting structures. I can only find one out of the three (which is the number that my teacher asked for). Is there three possible examples for this?
Any ideas would be great thanks!

For the exam you'd only need to know one, the only example I ever learned was ceramic insulating discs :P but hopefully someone can lend a hand with more ideas! :)

lsong

  • Adventurer
  • *
  • Posts: 16
  • Respect: 0
Re: Physics Question Thread
« Reply #1959 on: March 28, 2017, 05:19:21 pm »
+1
For the exam you'd only need to know one, the only example I ever learned was ceramic insulating discs :P but hopefully someone can lend a hand with more ideas! :)

Ah ok. Yeah in tutoring I only learnt about the one, might have to ask my teacher.
Thank you though!  ;D

Sukakadonkadonk

  • Forum Regular
  • **
  • Posts: 99
  • Respect: 0
Re: Physics Question Thread
« Reply #1960 on: March 28, 2017, 08:41:15 pm »
0
Hey guys,

Could someone please explain this question for me please? Especially the potential energy part (how they did it in the solutions).

"An object is stationary in space and located at a distance 10 000 km from the centre of a certain planet. It is found that 1.0 MJ of work needs to be done to move the object to a stationary point 20 000 km from the centre of the planet.
Calculate how much more work needs to be done to move the object to a stationary point 80 000 km from the centre of the planet."


Thanks!
« Last Edit: March 28, 2017, 08:43:49 pm by Sukakadonkadonk »

jamonwindeyer

  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 10150
  • The lurker from the north.
  • Respect: +3108
Re: Physics Question Thread
« Reply #1961 on: March 28, 2017, 08:54:06 pm »
0
Hey guys,

Could someone please explain this question for me please? Especially the potential energy part (how they did it in the solutions).

"An object is stationary in space and located at a distance 10 000 km from the centre of a certain planet. It is found that 1.0 MJ of work needs to be done to move the object to a stationary point 20 000 km from the centre of the planet.
Calculate how much more work needs to be done to move the object to a stationary point 80 000 km from the centre of the planet."


Thanks!

Hey hey! I can't see the solution, could you upload it if you haven't already? That way I can explain the approach they have taken for you ;D

Sukakadonkadonk

  • Forum Regular
  • **
  • Posts: 99
  • Respect: 0
Re: Physics Question Thread
« Reply #1962 on: March 28, 2017, 09:10:18 pm »
0
Hey hey! I can't see the solution, could you upload it if you haven't already? That way I can explain the approach they have taken for you ;D

Ok, here is the solution they gave.
Is there an easier way to do it?

jamonwindeyer

  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 10150
  • The lurker from the north.
  • Respect: +3108
Re: Physics Question Thread
« Reply #1963 on: March 28, 2017, 10:00:10 pm »
+1
Ok, here is the solution they gave.
Is there an easier way to do it?

Cool! They've done it the best/easiest way, let me explain! ;D

We know that a certain amount of work, 1MJ, is needed to move the object from 10,000km to 20,000km. This energy goes directly to a change in GPE. So, the 4 lines at the top are saying, "We know that the difference in GPE at the two heights is equal to work done. Let's use that." In this case, we use it to find the grouped value of the constants \(GMm\).

So now we have \(GMm\) - And we need to find the energy required to move from 20,000km to 80,000km. This is now just formula work - The last 3 lines involve calculating the difference in GPE at those two radii ;D

(Note - They do factor out the \(GMm\), don't let that confuse you, it is still just GPE! :)

Does that help give it some context?

Sukakadonkadonk

  • Forum Regular
  • **
  • Posts: 99
  • Respect: 0
Re: Physics Question Thread
« Reply #1964 on: March 29, 2017, 07:30:04 am »
+1
Cool! They've done it the best/easiest way, let me explain! ;D

We know that a certain amount of work, 1MJ, is needed to move the object from 10,000km to 20,000km. This energy goes directly to a change in GPE. So, the 4 lines at the top are saying, "We know that the difference in GPE at the two heights is equal to work done. Let's use that." In this case, we use it to find the grouped value of the constants \(GMm\).

So now we have \(GMm\) - And we need to find the energy required to move from 20,000km to 80,000km. This is now just formula work - The last 3 lines involve calculating the difference in GPE at those two radii ;D

(Note - They do factor out the \(GMm\), don't let that confuse you, it is still just GPE! :)

Does that help give it some context?


Yes, thank you very much  :)