HSC Physics Question Thread

[david.wang28]

Hey guys,

Just after a basic run-down on the pros and cons of an AC and DC motor, and what would be ideal for an appliance like a vacuum cleaner. (i.e. Brushless or induction motors)

Cheers in advance

I think AC motors would be ideal for a vacuum cleaner, because it can handle varying voltages; you can vary the suction strength (correlating to the strength of the motor).

[david.wang28]

Hello,
I am stuck on Q 1.3, Q 2.3 a) and Q 2.3 b) in the attachments below. Can anyone please help me out? Thanks

[DrDusk]

Hello,
I am stuck on Q 1.3, Q 2.3 a) and Q 2.3 b) in the attachments below. Can anyone please help me out? Thanks

[david.wang28]

Ahhh, thanks man 

[Joseph.Ryan]

I have a question which I think is fairly simple but I am not able to get the correct answer:

A double slit is illuminated by light of two wavelengths, 600nm and the other unknown. The two interference patterns overlap with the third dark band of the 600nm wavelength coinciding with the fourth bright band from the central band of the pattern for the light of unknown wavelength. What is the value for the light of unknown wavelength?

Answer: 450nm

[DrDusk]

I have a question which I think is fairly simple but I am not able to get the correct answer:

A double slit is illuminated by light of two wavelengths, 600nm and the other unknown. The two interference patterns overlap with the third dark band of the 600nm wavelength coinciding with the fourth bright band from the central band of the pattern for the light of unknown wavelength. What is the value for the light of unknown wavelength?

Answer: 450nm

Its just making use of the formula dsin(theta) = m*lambda
For the first wave if we sub in the wavelength and take m as 3 as it is the 3rd dark band we get dsin(theta) = 1800

Then just do the same for the second wave. SInce we know what dsin(theta) is now(this will be the same for both waves) we can do m*lambda = 1800.

Now we take m and 4 as it is the 4th bright band giving us lambda = 1800/4 which equals 450nm

[Joseph.Ryan]

Yes, but for the third dark spot it is destructive interference, which means m should be 2.5, not 3.

[jamonwindeyer]

Yes, but for the third dark spot it is destructive interference, which means m should be 2.5, not 3.

Cool! So for the overlap to occur, \(\theta\) is the same in both equations. Since \(d\) is a constant, that means \(d\sin{\theta}\) is the same for both uses of the equation. Those two uses are:



(Notice I have used \(m+0.5=2.5\) in the second one because the first dark band is arguably at \(m=0\), meaning the third dark band is at \(m=2\).

Solving:



And that should be it ps - welcome to the forums!

[Joseph.Ryan]

Thanks for the help!

The only problem is that my textbook said the answer was 450nm  - should I just assume it is wrong?

[jamonwindeyer]

Thanks for the help!

The only problem is that my textbook said the answer was 450nm  - should I just assume it is wrong?

You're welcome! And yep, I think so - Seems they've mistakenly done both calculations with the bright band as was done above. Great question regardless, but it looks like their working didn't match the question in this case

[Jefferson]

Hi everyone, Just a quick question.

The moon is a satellite of Earth, which is a satellite of the Sun.

Then, is the moon a satellite of the Sun?

[DrDusk]

Hi everyone, Just a quick question.

The moon is a satellite of Earth, which is a satellite of the Sun.

Then, is the moon a satellite of the Sun?

I would say this is more so an English question than a Physics question, but I don't think the Moon is a satellite of the Sun.
A satellite in a perfect world is defined as an object that has a defined circular trajectory around something else, and the moon doesn't have a defined circular trajectory around the sun.

And actually in reality the orbits are elliptical

[david.wang28]

Hello,
I have trouble with a question in the link below (I have no idea how to answer the question). Can anyone please help me out? Thanks

[jamonwindeyer]

Hello,
I have trouble with a question in the link below (I have no idea how to answer the question). Can anyone please help me out? Thanks

Hey! This is a loaded question, particularly Part A - Have you learned Proton-Proton Chain reactions and are you comfortable with them? That's where you need to start, want to make sure we're on the same page

[david.wang28]

I understand proton-proton chain reactions and I know about them, but I don't understand how to do this question.