VCE Physics Question Thread!

[KiNSKi01]

Woah thanks Syndicate for the help!

We were never really taught why momentum can be conserved yet KE can be lost...sort of just had to accept it 

Never really thought I should just do some calculations to prove it to myself ahah

[KiNSKi01]

Yo can someone help me with 11.11 and 11.13 e

11.11 I thought pressure was inversely proportional to volume and therefore I don't understand why the pressure increases. I get 0.514 metres cubed by doing PxV is a constant 

11.13 How are you meant to answer this question when all three of pressure volume and temperature change

EDIT: Whoops I just realised all you do is PV=nRT for 11.13

[Shadowxo]

Yo can someone help me with 11.11 and 11.13 e

11.11 I thought pressure was inversely proportional to volume and therefore I don't understand why the pressure increases. I get 0.514 metres cubed by doing PxV is a constant 

11.13 How are you meant to answer this question when all three of pressure volume and temperature change

EDIT: Whoops I just realised all you do is PV=nRT for 11.13

You're right, pressure is inversely proportional to volume but you have to keep the other variables (temp and number of moles) constant. So as the pressure inside the tyre increases, the air that started off in the tyre decreases in volume (gets squeezed into a smaller space by the new air).

For the air inside the tyre, we want to find the new volume it occupies (to then find how much volume the new air takes up). Since it's the same air, the number of moles is constant so we can use PV=constant
P₁V₁=P₂V₂
V₂=P₁V₁/P₂
=105*0.150/360
=0.04375 m³
Note that since it's P1/P2 you can use whichever units you choose for pressure but they must be the same.

Volume of new air inside the tyre (at pressure of 360kPa) is the remaining volume = 0.150 - 0.04375 = 0.10625 m³
Volume at atm pressure: use the same equation as number of moles & temp stays the same
V₁=P₂V₂/P₁
=360*0.10625/105
=0.364 m³

Also note that you need to take into account sig figs if applicable

[KiNSKi01]

Cheers Shadowxo!!

Didn't consider having to subtract the difference between the  volumes

[Unsplash]

I'm having trouble with this question. So far I have got, Fnet = 360 + R * sin(20) where R is the reaction force. I have also got mg = R * cos (20) but cannot figure out what to do next. No matter what way I substitute it in, I always get two variables. I have a feeling it has something to do with "only just" taking the corner. Could someone point me in the right direction?

[Syndicate]

I'm having trouble with this question. So far I have got, Fnet = 360 + R * sin(20) where R is the reaction force. I have also got mg = R * cos (20) but cannot figure out what to do next. No matter what way I substitute it in, I always get two variables. I have a feeling it has something to do with "only just" taking the corner. Could someone point me in the right direction?

I will reply soon with the worked solution, however, your starting step would be to resolve for the horizontal forces of the system.

Note this is not a banked track, as the surface is horizontal, and "only just" implies that the system is in equilibrium.

Therefore, mv^2/r = mgsin(20) (the horizontal component of weight force) + 360

[Vaike]

Therefore, mv^2/r = mgsin(20) (the horizontal component of weight force) + 360

Unfortunately, the question doesn't seem to supply the velocity of the cyclist nor the radius of the bend, and I can't see any way to calculate these values, so I'm not sure this approach would work.

From what I can tell, there is no way to work this out properly. Is this a textbook question FelixHarvey? I won't be surprised if so, there are quite a few that leave out necessary information.

[Unsplash]

From what I can tell, there is no way to work this out properly. Is this a textbook question FelixHarvey? I won't be surprised if so, there are quite a few that leave out necessary information.

Yes, this is a question from the Jacaranda textbook. Thanks for your help!

[KiNSKi01]

How do i approach this question?
Show that the total energy of an electron has been accelerated to a speed of 0.98c is about 4x10^-13 J

[Vaike]

How do i approach this question?
Show that the total energy of an electron has been accelerated to a speed of 0.98c is about 4x10^-13 J

Hi!
Since the particle is traveling at very high speeds, we must take relativistic effects into account; as a particle approaches light speeds it's total energy is composed of both it's rest energy, E₀ and it's kinetic energy, E_k. If you use the relevant formulae, adding them together, you get an equation for the total energy of the particle. Plug in the numbers and you should get the answer

[Syndicate]

I'm having trouble with this question. So far I have got, Fnet = 360 + R * sin(20) where R is the reaction force. I have also got mg = R * cos (20) but cannot figure out what to do next. No matter what way I substitute it in, I always get two variables. I have a feeling it has something to do with "only just" taking the corner. Could someone point me in the right direction?

The question you have posted is missing some information (there might be a diagram somewhere?). I have checked the worked solutions as well just to confirm, and they have seemed to use  v = 9 m/s and radius of 10m to get 59.4 kg for the cyclist.

[Unsplash]

The question you have posted is missing some information (there might be a diagram somewhere?). I have checked the worked solutions as well just to confirm, and they have seemed to use  v = 9 m/s and radius of 10m to get 59.4 kg for the cyclist.

No diagram that I can see. I've checked both the pdf and the online textbook through JacPlus. I've just skipped the question. Thanks for your help!

[viviianle]

please help
--
One of the fastest objects ever made on Earth was
the Galileo Probe which, as a result of Jupiter’s huge
gravity, entered its atmosphere in 1995 at a speed of
nearly 50 000 m/S. Give an estimate of the Lorentz
factor for the probe to nine decimal places. 
------
answer is 1.4*10^-8

[sweetiepi]

please help
--
One of the fastest objects ever made on Earth was
the Galileo Probe which, as a result of Jupiter’s huge
gravity, entered its atmosphere in 1995 at a speed of
nearly 50 000 m/S. Give an estimate of the Lorentz
factor for the probe to nine decimal places. 
------
answer is 1.4*10^-8

The equation we're using is: 1/sqrt(1-v^2/c^2)
c= speed of light, which is 3.0*10^8 (afaik, could be slightly off)
Therefore 1/sqrt(1-(50000)^2/(3.0x10^8)^2)
Lorentz's factor= 1.000000014 (9dp)

Disclaimer: I'm not 100% certain of my answer on this question

[KiNSKi01]

The equation we're using is: 1/sqrt(1-v^2/c^2)
c= speed of light, which is 3.0*10^8 (afaik, could be slightly off)
Therefore 1/sqrt(1-(50000)^2/(3.0x10^8)^2)
Lorentz's factor= 1.000000014 (9dp)

Disclaimer: I'm not 100% certain of my answer on this question

Yep i got same answer (1.0000000138889 exactly)