February 24, 2020, 06:07:23 am

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#### TheEagle

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##### Re: VCE Physics Question Thread!
« Reply #2340 on: January 19, 2020, 12:47:42 am »
0
I’m not quite sure what you mean. The area under a force-distance graph is work done, and by scaling a field strength-distance  graph by mass the are is thus the same as under a force-distance,

I don't get why we must scale a field strength - distance graph's area by mass to attain the change in energy. Whereas, the area under a force - distance graph is equal to energy without multiplying by mass?

#### Tau

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##### Re: VCE Physics Question Thread!
« Reply #2341 on: January 19, 2020, 12:55:40 am »
+3

I don't get why we must scale a field strength - distance graph's area by mass to attain the change in energy. Whereas, the area under a force - distance graph is equal to energy without multiplying by mass?

By definition we have that $W=\int F dx$, and for a Force-distance graph we can work directly with that. However, gravitational field strength is defined by $g=F/m$ - force per unit mass - if we multiply $g$ by $m$ we get our $F$ and can then use the first equation (well, counting squares really ) to find the work done.
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#### redleafbun

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##### Re: VCE Physics Question Thread!
« Reply #2342 on: January 20, 2020, 12:33:39 pm »
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YEP thank you for your suggestion! I have managed to solve two questions after reviewing chapter 2. I am just a bit confuse, I thought the magnitude of magnetic fields are dependent on electric fields. So as the first question mentions where there is no current, I assume the answer is 0. And why would the direction be into the page, shouldn't it be going downwards using the right hand rule? Or is this question irrelevant to the right hand rule?

Hey redleafbun
You seem to have posted multiple questions - are you having trouble with all three, or is there one particular question that has you confused?
I am more than happy to answer your question, but before I do its important for you to have a go at it first and try to explain your current aproach to the question. Don't worry about being wrong as thats all part of the learning process.
Having a go at the question first and showing us your thinking process helps people answering your question figure out what specifically you don't quite get so that we can better help you towards understanding the question.

Interms of recomendations to help develop your understanding of fields, If you are currently a bit confused as to what exactly you need to know it might be a good first step to have a look at the study design here which outlines precisely that (if there are any points which you don't quite understand, feel free to ask about it here .
If you see some content listed in the study design that you aren't confident on, it might be good to revisit your textbook and re-read over/make notes on those topics. In addition to reading over the textbook/making notes, I found that making your own diagrams when possible really helped. I think that a fair amount of feild questions (e.g. determining the direction of induced currents) requires the ability to visualise the problem - which I found drawing diagrams helped with.
While going over content in this way, If there are any really confusing bits that you encounter which you can't understand on your own, myself (and others) are happy to help explain stuff to you, just try to be specific with what you do and don't understand so that we can help you more effectivley.
After going over the bits of content that you are stuggling with, I would recoment jumping back into answering questions (the most effective ones would be exam style questions if you have access to any - These are things like Atar notes topic tests or checkopints).

#### thisissara

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##### Re: VCE Physics Question Thread!
« Reply #2343 on: January 20, 2020, 03:35:59 pm »
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Hey guys,

Could someone help me establish the difference between Banked Tracks and Inclined Planes? I.e in regards to forces etc! Tysm!
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#### DrDusk

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##### Re: VCE Physics Question Thread!
« Reply #2344 on: January 20, 2020, 04:38:36 pm »
+5
Hey guys,

Could someone help me establish the difference between Banked Tracks and Inclined Planes? I.e in regards to forces etc! Tysm!
Banked tracks are like those tilted race tracks that you see for professional bicycle races whereas an inclined plane is basically just a hill. Difference is that on a Banked track you are going in a circle and an inclined plane your just going up or down. Obviously this means the biggest difference in terms of forces is that a Banked track has a Centripetal force unlike an inclined plane.
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#### puzzledstudent

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##### Re: VCE Physics Question Thread!
« Reply #2345 on: January 26, 2020, 10:59:42 pm »
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So I am starting Year 11 and I haven't finished the holiday homework sent out :[
I'm struggling so hard with this one question and I would be very thankful if anyone has the time to help me solve it and explain it.

A net force of 20.0 N acts on a 5.00 kg mass for 6 seconds. The original velocity of the mass is 100 m s-1 east. Calculate the velocity after the 6seconds if
a) The net force is to the east
b) The net force is to the west

Thank you soSO MUCH  in advance!

#### Bri MT

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##### Re: VCE Physics Question Thread!
« Reply #2346 on: January 26, 2020, 11:14:24 pm »
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So I am starting Year 11 and I haven't finished the holiday homework sent out :[
I'm struggling so hard with this one question and I would be very thankful if anyone has the time to help me solve it and explain it.

A net force of 20.0 N acts on a 5.00 kg mass for 6 seconds. The original velocity of the mass is 100 m s-1 east. Calculate the velocity after the 6seconds if
a) The net force is to the east
b) The net force is to the west

Thank you soSO MUCH  in advance!

Hi! Welcome to the forums

Have you used the constant acceleration formulas before?

(Hint: the east vs west difference is about whether acceleration is in the same direction as initial velocity)

1. Define a direction to be positive
2. Write down your values including units
3. Find the magnitude of acceleration using f=ma
4. Identify the constant acceleration formula matching the info you have
5. Sub and solve

For part b,
1. Multiply the acceleration by -1
2. Sub this into the formula from before & solve

If any part of it doesn't make sense or you'd like to confirm your answer feel free to reply & I'm happy to help further

#### puzzledstudent

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##### Re: VCE Physics Question Thread!
« Reply #2347 on: January 26, 2020, 11:35:48 pm »
0
Hi thanks!
Ok let me assume East is positive.
F-20N M-5kg T-6s V-100ms-1
f=ma
20=5*a
a=4m/s

What I don't get is the net force to the east, west bit. Because it says the original velocity is 100ms-1 east. So doesn't this mean the velocity for east is 100ms-1?

Uhh I think I've done this last year but I've completely forgotten about every single information LOL
x[

#### DrDusk

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##### Re: VCE Physics Question Thread!
« Reply #2348 on: January 27, 2020, 12:29:04 am »
+2
What I don't get is the net force to the east, west bit. Because it says the original velocity is 100ms-1 east. So doesn't this mean the velocity for east is 100ms-1?
If a force acts towards the East on an object that is travelling East how will the object have the same velocity after 6 seconds? It's going to accelerate.
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#### Bri MT

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##### Re: VCE Physics Question Thread!
« Reply #2349 on: January 27, 2020, 08:04:06 am »
+3
Hi thanks!
Ok let me assume East is positive.
F-20N M-5kg T-6s V-100ms-1
f=ma
20=5*a
a=4m/s

What I don't get is the net force to the east, west bit. Because it says the original velocity is 100ms-1 east. So doesn't this mean the velocity for east is 100ms-1?

Uhh I think I've done this last year but I've completely forgotten about every single information LOL
x[

No worries!

First up, your acceleration isn't 4 metres per second, it's 4 metres per second per second. Additionally, usually initial velocity would be indicated with u rather than v, and then we use v for final velocity.

As DrDusk said, the object is undergoing acceleration. By definition,  this means that its velocity is changing.  Because we've calculated the acceleration, we know that the velocity will change by 4 metres per second each second.

In part A, with acceleration and initial velocity in the same direction, we know that the speed will increase by 4 m s^-2 . We also know how many seconds this acceleration is being applied for & the initial velocity. This allows us to find the final velocity.

#### puzzledstudent

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##### Re: VCE Physics Question Thread!
« Reply #2350 on: January 28, 2020, 10:53:59 pm »
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I honestly still have no idea..
If anyone could take the time to make a worked solution for me, I would really appreciate it. I'm not lazy, I just do not understand anything you are saying etc "second per second". I'm new to Physics and I feel like I'm going to fail it at this pace. I hope if anyone has used Jacaranda 1/2 Physics, can they please tell me whether it could be so complicated? If it is, I'm definitely dropping it =[

#### Bri MT

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##### Re: VCE Physics Question Thread!
« Reply #2351 on: January 28, 2020, 11:21:31 pm »
+3
I honestly still have no idea..
If anyone could take the time to make a worked solution for me, I would really appreciate it. I'm not lazy, I just do not understand anything you are saying etc "second per second". I'm new to Physics and I feel like I'm going to fail it at this pace. I hope if anyone has used Jacaranda 1/2 Physics, can they please tell me whether it could be so complicated? If it is, I'm definitely dropping it =[

You calculates the acceleration to be 4 m/s^2. Aside from getting the units wrong your working here was fine.

This means that each second, the velocity changes by 4 metres per second (i.e. change of 4 metres per second per second)

In part a we start off at 100 m/s east. This is the same direction as our acceleration. After 1 second, the new velocity will be 104 m/s east. After another second, it will be 108 m/s east. After the 3rd second, it will be 112 m/s east... until eventually after 6 seconds it will be 100 m/s +(4m/s^2)(6s) = 124 m/s

In part be we start off at 100 m/s west. This is the opposite direction to our acceleration. After 1 second the velocity will be 96 m/s. After another second it will be 92 m/s. After 3 seconds it will be 88 m/s.... after 6 seconds it will be 100 m/s - (4 m/s^2)(6) = 76 m/s west

In both cases, we can use the forumla v = u + at to arrive at our answers. In this approach, we can jump straight to

a)
v = u + at
= 100 + (4)(6)
= 124
v= 124 m/s east

b)
v = u + at
= -100 + (4)(6)
= -76
v = 76 m/s west

I hope this clarifies things a bit

#### BitcoinEagle

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##### Re: VCE Physics Question Thread!
« Reply #2352 on: January 29, 2020, 12:03:52 pm »
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Hi, I was wondering if I could have some help with this physics motion question. Thanks!

As a result of being hit from behind by a toy truck, a 250 g toy car, initially at rest rolls 10 m across a floor that applies a constant retarding force of 0.75N to it. The car stops 6 seconds after being hit.

a)If the truck was in contact with the car for 0.12 seconds, calculate the impulse given to the car.
b) Calculate the momentum of the car just after being hit.
c) Calculate the instantaneous speed of the car just after being hit.
d) Calculate the average force applied to the car during the collision.
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#### SmartWorker

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##### Re: VCE Physics Question Thread!
« Reply #2353 on: January 29, 2020, 12:43:47 pm »
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Hey everyone,

How does significant figures work, like how many sig figs are there in 13.07 and 13.0 and 13 and 2.00, 2.0, 3.03 x10^8

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#### DrDusk

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##### Re: VCE Physics Question Thread!
« Reply #2354 on: January 29, 2020, 07:23:27 pm »
+1
Hey everyone,

How does significant figures work, like how many sig figs are there in 13.07 and 13.0 and 13 and 2.00, 2.0, 3.03 x10^8