HSC Physics Question Thread

[blasonduo]

OHHH I GET 1 - IT SAID TRANSFORMER MY BAD
16 I READ WRONG OML.

15 - WHY IS C WRONG?

Remember, between this junction is a Depletion zone, and this depletion zone creates a potential difference between it. This is what stops the electrons from continuously moving from the n-type to the p-type. When an electron gains energy, it will move to a place that is "easiest" to move to. Jumping that depletion zone is a lot more work than moving through the external circuit, so the electrons will always move into the external circuit. PLUS if they were to jump the depletion zone, the solar cell wouldn't do its purpose!

[CyberScopes]

Hey could someone explain q20 from 2015
the answer is B

Δx = Uxt
so Ux = Δx/t = 70/3.5 = 20

To find Uy, assume Vy = 0 at max height, and take t as half (3.5/2 = 1.75):
Vy = Uy + ayt
0 = Uy + (-9.8 )(1.75)
Uy = 17.15

Now θ = tan^-1(Uy/Ux) = tan^-1(17.15/20)
= 40.61

The closest answer is 40 degrees, which is B.

[beau77bro]

It goes around the whole circuit, you need to think of the depletion zone like a voltage source. It is setting up an electric field which 'pushes' electrons around the circuit, just like a voltage source would. The electric field is directed from N to P, so electrons will flow in the opposite direction around the whole circuit

im still struggling with this? so the Photon strikes the n-type, which produces a positive hole and a ejects an electron. now the positive hole is attracted to the negative portion of the depletion zone (the p-type) and the electron is attracted to the positive portion of the n-type. so how do they get there? does the electron just go straight to the positive or does it go all the way around the circuit. or it the positive hole that goes all the way around the circuit to the negative part, or does it just travel striaght to the negative through the junction?

i understand that it sets up a positve to negative electric field but which way. how do the electrons and holes move? WAIT LEGENDBLASONDUO JUST ANSWERED OMG THANK YOU!!!! but wait. isnt the electron just emitted in the N-type so it can go straight to the positive part of the depletion zone?

[blasonduo]

im still struggling with this? so the Photon strikes the n-type, which produces a positive hole and a ejects an electron. now the positive hole is attracted to the negative portion of the depletion zone (the p-type) and the electron is attracted to the positive portion of the n-type. so how do they get there? does the electron just go straight to the positive or does it go all the way around the circuit. or it the positive hole that goes all the way around the circuit to the negative part, or does it just travel striaght to the negative through the junction?

i understand that it sets up a positve to negative electric field but which way. how do the electrons and holes move? WAIT LEGENDBLASONDUO JUST ANSWERED OMG THANK YOU!!!! but wait. isnt the electron just emitted in the N-type so it can go straight to the positive part of the depletion zone?

It could, but why would it? There would be no purpose!

[khadeeja_]

 2016 q14.. 

[beau77bro]

But isn't it the simple mode of travel - the n-type is the one that is exposed to sun, so its the one the produces the extra electron - why wouldn't it go straight to the positive with the positive hole going through the external to the negative part of the depletion zone?

[blasonduo]

(Image removed from quote.)

But isn't it the simple mode of travel - the n-type is the one that is exposed to sun, so its the one the produces the extra electron - why wouldn't it go straight to the positive with the positive hole going through the external to the negative part of the depletion zone?

Are you saying why the electrons can't go directly to the P-type? IF so, it is because it physically cannot. The potential difference is negative in the p-type, and electrons are repelled by that!

[jamonwindeyer]

2016 q14.. 

Kepler's Law of Periods states:



If we quadruple the mass of earth on the RHS, then that factor of four needs to come to the LHS. If the radius is set, it must come from the orbital period. For the ratio \(\frac{r^3}{T^2}\) to get four times larger, the period needs to be cut in half:



So the answer is B

[beau77bro]

Are you saying why the electrons can't go directly to the P-type? IF so, it is because it physically cannot. The potential difference is negative in the p-type, and electrons are repelled by that!

no i'm saying they will just go straight to the N-type, and not pass through the external circuit or the depletion zone. is that what happens?

[CyberScopes]

2016 q14.. 

So you have two satellites orbiting two planets, with the only difference being one has mass 4 times the other. Using Keplers Law of Periods:
r³/t² = GM/4π²

We know r is the same so lets just set it to 1, and we dont actually care about G, π, etc, since its all just comparison, so lets simplify this equation:
1/t² = M

Ok lets chuck in M = 1 for earth:
1/t² = 1
t = 1
So lets say t_Earth = 1

Now lets chuck M = 4:
1/t² = 4
t² = 1/4
t_X = 1/2

So since were tryna get the value of T in terms of Earth values we can rewrite as such:
t_X = 1/2
-> Sub t_Earth = 1
t_X = T/2

Your answer should be B 

[CyberScopes]

Kepler's Law of Periods states:



If we quadruple the mass of earth on the RHS, then that factor of four needs to come to the LHS. If the radius is set, it must come from the orbital period. For the ratio \(\frac{r^3}{T^2}\) to get four times larger, the period needs to be cut in half:



So the answer is B

Dammit why are you so quick give us a chance 

[khadeeja_]

So you have two satellites orbiting two planets, with the only difference being one has mass 4 times the other. Using Keplers Law of Periods:
r³/t² = GM/4π²

We know r is the same so lets just set it to 1, and we dont actually care about G, π, etc, since its all just comparison, so lets simplify this equation:
1/t² = M

Ok lets chuck in M = 1 for earth:
1/t² = 1
t = 1
So lets say t_Earth = 1

Now lets chuck M = 4:
1/t² = 4
t² = 1/4
t_X = 1/2

So since were tryna get the value of T in terms of Earth values we can rewrite as such:
t_X = 1/2
-> Sub t_Earth = 1
t_X = T/2

Your answer should be B 

Thanks so much and to Jamon too :')

[khadeeja_]

how about 2016 q18
and 2014 q15 ( i never know how to solve these.. )

[CyberScopes]

how about 2016 q18

The way I did this may be different to some people but here:

You need to find velocity so that the effects of gravity do not pull the bike down as it does a spin. The bike is moving in a circle so thats centripetal force. So equate weight force and centripetal force to find the min velocity required to maintain its circular motion:

F=mg
F=mv²/r
mg = mv²/r
g = v²/r
v² = gr
v = square root of g*r = root 9.8*3.6
= 5.93...
= C

[khadeeja_]

The way I did this may be different to some people but here:

You need to find velocity so that the effects of gravity do not pull the bike down as it does a spin. The bike is moving in a circle so thats centripetal force. So equate weight force and centripetal force to find the min velocity required to maintain its circular motion:

F=mg
F=mv²/r
mg = mv²/r
g = v²/r
v² = gr
v = square root of g*r = root 9.8*3.6
= 5.93...
= C

Thanks omg, you made it so much simpler