HSC Physics Question Thread

[jakesilove]

The captain of a fishing boat uses an echo sounder to determine the depth of a school of fish below the boat. The captain finds that the reflected waves return after o.15s and o.20s. The captain believes the first reflectin to be from the fish while the seconds is from the ocean floor. If the speed of sound in seawater is 1440m/s, determine the depth of the sea floor. (3 marks)

can anyone tell me the expected way of doing a question like this for year 11 physics? i looked it up and found ways which correlate to actual echo sounders, but realise i've never done this stuff in class.

any help is greatly appreciated!

If something travels for x m/s for y seconds, then it has traveled xy meters.

The sound took 0.20s to go from the boat, to the sea floor, and back. That means the sea floor is 0.10s away (half the time will be spent going down, half will be spent going up). Therefore, given the speed of sound in water, it is



away

[levendibigd]

Hi there! Could I have a hand with this syllabus dot point 'Analyse the significance of the hydrogen spectrum in the development of
Bohr’s model of the atom'.

[jakesilove]

Hi there! Could I have a hand with this syllabus dot point 'Analyse the significance of the hydrogen spectrum in the development of
Bohr’s model of the atom'.

Hey! So, before Bohr, electrons just sort of floated around in a cloud and did their own thing (in fact, they weren't even electrons; just a negative cloud!). However, some interesting observations of the hydrogen spectrum lead to a change in the theory. EM waves were directed towards Hydrogen atoms at varying frequencies, and then turned off. Light was emitted from the Hydrogen atoms, but only ever at very specific wavelengths based on the incident frequencies! If you put in 100Hz, you'd always get the same colour out! Weird, if the cloud was just random. Instead, Bohr suggested that what was happening was that electrons could only occupy very specific orbitals. When the frequency of EM radiation was being directed towards the Hydrogen, electrons had the ability to 'jump up' energy levels. When the frequency stopped, they would 'jump down', emitting light. However, since there was only specific energy levels they could occupy, there was only specific wavelengths of light that could be emitted! This explains the specificity of the Hydrogen spectrum. Let me know if you need me to clarify any of this!

Jake

[conic curve]

Hey, I've done a question two ways and I have no idea which one is right, can someone explain why one is wrong.

A spaceship in deep space fires its rocket motor that exerts a force of 8.00 x 10^4 N. The rocket continues to fire while the spaceship moves through a distance of 1.0 x 10^7 m. If the 6.0 x 104 kg spaceship had been moving at 4.5 x 10^3 m s^-1 before the rocket motor was fired, how fast is it moving now?
Method 1:
Change in kinetic energy = Fs = 8x10^11 J
Therefore, (Δv)^2=(8x10^11)/(m/2) = 2.67x10^7, Δv = 5163 ms^-1
Therefore, final v=9663ms^-1
Method 2:
Total kinetic energy = 8x10^11+3x10^4x(4.5x10^3)^2 = 1.41x10^12 J
v = (1.41x10^12/(3x10^4))^(1/2) = 6855 ms^-1

[jamonwindeyer]

Hey, I've done a question two ways and I have no idea which one is right, can someone explain why one is wrong.

Hey conic! Subtle thing here, the issue is that in Method 1, you are assuming that the mass of the rocket is constant. It is not, it is losing fuel, and so its mass is actually decreasing. Therefore, you cannot say the change in kinetic energy is purely due to the change in velocity. Your second method is more appropriate and should be giving you the correct answer!

Had to think about that one for a tad, happy to be corrected if someone spots a misinterpretation

[jakesilove]

Hey conic! Subtle thing here, the issue is that in Method 1, you are assuming that the mass of the rocket is constant. It is not, it is losing fuel, and so its mass is actually decreasing. Therefore, you cannot say the change in kinetic energy is purely due to the change in velocity. Your second method is more appropriate and should be giving you the correct answer!

Had to think about that one for a tad, happy to be corrected if someone spots a misinterpretation

I'm not entirely sure whether the second method makes up for this; I agree that the 'total kinetic energy' is correct (as it is just initial kinetic energy plus the work done, all of which would go towards increasing the speed of the shuttle), however when determining the final speed, the initial mass of the shuttle is used. Therefore, this would be the final speed if the shuttle doesn't lose mass. Which is the same assumption made in the first calculation. So, to be perfectly honest, I'm not entirely sure what the problem is here!

[jamonwindeyer]

I'm not entirely sure whether the second method makes up for this; I agree that the 'total kinetic energy' is correct (as it is just initial kinetic energy plus the work done, all of which would go towards increasing the speed of the shuttle), however when determining the final speed, the initial mass of the shuttle is used. Therefore, this would be the final speed if the shuttle doesn't lose mass. Which is the same assumption made in the first calculation. So, to be perfectly honest, I'm not entirely sure what the problem is here!

Ahhh good point, I shall have another look after my lab

[jakesilove]

Ahhh good point, I shall have another look after my lab

I can't figure out why this happens, but basically the two answers are different because, when you deconstruct them, it gives you



(ie. one calculation uses the left, one uses the right, and so obvious as the above relations is false, the answers are different.)

Not sure why at the moment!

[jamonwindeyer]

To check with a third option (assuming constant mass as we have done above):



So I'm now even more sure that the second answer is the correct one, but I still can't figure out why the first one doesn't work....

From Rui: Fixed up mistake in LaTeX leading to a wrong equation

[RuiAce]

Literally just read Jake's incorrect equation twice lol

Difference is a linear operator. Not quadratic.

[jamonwindeyer]

Literally just read Jake's incorrect equation twice lol

Difference is a linear operator. Not quadratic.

Ahhh, right you are, how subtle! I've never actually seen that pop up in HSC Physics, which is surprising, because I bet this would come up a fair bit

[jakesilove]

Literally just read Jake's incorrect equation twice lol

Difference is a linear operator. Not quadratic.

Yeah sorry that's what I was getting at (wasn't sure how to write not equal to in Latex, so just explained it in words beneath and didn't have time to apply it).

[Adriaclya]

Hi!
I'm stumped. Is trajectory and projectile motion the same thing? 0.o

[RuiAce]

Hi!
I'm stumped. Is trajectory and projectile motion the same thing? 0.o

Trajectory is just the actual path the projectile moves through.

[Klexos]

I don't know if anyone did Medical Physics but I'm having a hard time understanding the difference between T1 & T2 weighted for MRI.

Can anyone distinguish this for me?