Physics Predictions/Thoughts (1 Viewer)

[carrotsss]

Is this raw mark to hsc mark? or hsc mark to atar equivalent?

Raw to aligned (aka hsc)

 

[Hsc1111]

75 english would be around 83
65 ext 2 would be around 89

A 68 last year in English advanced was an 81 so surely its a bit higher?

 

[carrotsss]

A 68 last year in English advanced was an 81 so surely its a bit higher?

not really, it would’ve been around 84-85 last year but this year was a tad easier

 

[carrotsss]

With growing impatience one does Question 34 on Chat GPT and gets his:

a.

Let's analyze the energy changes as the satellite moves from point P to point Q. To do this, we'll consider the initial and final states, as well as the conservation of energy.

1. Initial State (Point P):
   • Potential Energy (U_P) = -2.389 x 10^10 J
   • Total Energy (E_P) = -1.195 x 10^10 J
   • Kinetic Energy (K_P) is not given, but it changes by 5.232 x 10^8 J when the engines are fired.
2. Final State (Point Q):
   • The satellite is at a distance of 6.850 x 10^6 m from Earth's center.

Now, let's consider the energy changes:

• Total Energy (E): The total energy of a satellite in a circular orbit remains constant. This is a result of the conservation of mechanical energy. Therefore, at point P and point Q, the total energy (E) should remain the same.
• Potential Energy (U): The potential energy of an object in orbit depends on its distance from the center of the Earth. As the satellite moves from point P to point Q, it gets closer to Earth's center, reducing its potential energy. This means that the potential energy at point Q (U_Q) will be more negative (i.e., lower) than at point P (U_P).
• Kinetic Energy (K): The kinetic energy of the satellite increases when the engines are fired at point P. This is evident from the given increase in kinetic energy (5.232 x 10^8 J). However, as the satellite moves from point P to point Q, its speed will decrease due to gravitational attraction, causing its kinetic energy to decrease. Therefore, the kinetic energy at point Q (K_Q) will be lower than at point P (K_P).

In summary, as the satellite moves from point P to point Q:

• The potential energy decreases (gets more negative).
• The kinetic energy decreases.
• The total mechanical energy remains constant.

These changes in energy are a result of the satellite's changing distance from Earth's center and the interplay between potential and kinetic energy in a gravitational field. The satellite loses some of its initial kinetic energy to gravitational potential energy as it moves from a higher orbit to a lower orbit closer to the Earth's surface.

b.

To find the kinetic energy of the satellite at point Q, we can use the law of conservation of energy, which states that the total mechanical energy of the satellite remains constant throughout its motion.

Given information:

• Mass of the satellite (m) = 400 kg
• Radius of the initial circular orbit (r_P) = 6.700 x 10^6 m
• Potential energy at point P (U_P) = -2.389 x 10^10 J
• Total energy at point P (E_P) = -1.195 x 10^10 J
• Increase in kinetic energy at point P (ΔK_P) = 5.232 x 10^8 J
• Radius at point Q (r_Q) = 6.850 x 10^6 m

First, let's calculate the kinetic energy at point P (K_P). We know that the total energy at point P is the sum of the potential and kinetic energies:

E_P = U_P + K_P

-1.195 x 10^10 J = -2.389 x 10^10 J + K_P

Now, solve for K_P:

K_P = -1.195 x 10^10 J + 2.389 x 10^10 J K_P = 1.194 x 10^10 J

So, the kinetic energy of the satellite at point P is 1.194 x 10^10 J.

Now, let's calculate the kinetic energy at point Q (K_Q). Since the total mechanical energy is conserved, we can write:

E_P = E_Q

At point Q, the potential energy is given by:

U_Q = -G * (M * m) / r_Q

Where:

• G is the gravitational constant (approximately 6.674 x 10^-11 N·m^2/kg^2).
• M is the mass of the Earth (approximately 5.972 x 10^24 kg).

Now, the total energy at point Q is the sum of the potential and kinetic energies:

E_Q = U_Q + K_Q

E_Q = (-G * (M * m) / r_Q) + K_Q

We know that E_P = E_Q (total energy is conserved), so:

1.194 x 10^10 J = (-G * (M * m) / r_Q) + K_Q

Now, solve for K_Q:

K_Q = 1.194 x 10^10 J + (G * (M * m) / r_Q)

Substitute the known values and calculate K_Q:

K_Q = 1.194 x 10^10 J + [(6.674 x 10^-11 N·m^2/kg^2) * (5.972 x 10^24 kg * 400 kg) / (6.850 x 10^6 m)]

K_Q ≈ 1.194 x 10^10 J

So, the kinetic energy of the satellite at point Q is approximately 1.194 x 10^10 J, which matches the total energy at point P, as expected due to the conservation of energy.

c.

To understand the motion of the satellite after it passes through point Q, it's essential to consider the conservation of mechanical energy and how the satellite's energy changes during its journey.

Given information:

• Mass of the satellite (m) = 400 kg
• Radius of the initial circular orbit (r_P) = 6.700 x 10^6 m
• Potential energy at point P (U_P) = -2.389 x 10^10 J
• Total energy at point P (E_P) = -1.195 x 10^10 J
• Increase in kinetic energy at point P (ΔK_P) = 5.232 x 10^8 J
• Radius at point Q (r_Q) = 6.850 x 10^6 m

The conservation of mechanical energy states that the total mechanical energy of the satellite remains constant as it moves in the gravitational field of Earth.

At point P, the satellite's total energy (E_P) is a combination of potential and kinetic energy:

E_P = U_P + K_P

E_P = -2.389 x 10^10 J + K_P

The satellite's kinetic energy at point P (K_P) is 1.194 x 10^10 J, as previously calculated.

Now, as the satellite moves from point P to point Q, its potential energy decreases (gets more negative) because it gets closer to Earth's center. The change in potential energy (ΔU) is given by:

ΔU = U_Q - U_P

Where U_Q is the potential energy at point Q.

Since energy is conserved, the decrease in potential energy (ΔU) should be equal to the increase in kinetic energy (ΔK) from point P to point Q:

ΔU = ΔK

-ΔU = -ΔK

Substitute the values:

• (U_Q - U_P) = - (K_Q - K_P)

Now, we know that K_P is 1.194 x 10^10 J, and we want to find K_Q at point Q, so:

(U_Q - U_P) = (1.194 x 10^10 J - K_Q)

Now, plug in the values for U_P, U_Q, and solve for K_Q:

(-2.389 x 10^10 J - U_P) = 1.194 x 10^10 J - K_Q

K_Q = 1.194 x 10^10 J + (2.389 x 10^10 J - U_P)

K_Q = 1.194 x 10^10 J - (2.389 x 10^10 J + U_P)

K_Q ≈ 1.194 x 10^10 J - (-1.195 x 10^10 J)

K_Q ≈ 1.194 x 10^10 J + 1.195 x 10^10 J

K_Q ≈ 2.389 x 10^10 J

So, the kinetic energy of the satellite at point Q is approximately 2.389 x 10^10 J.

The satellite's kinetic energy at point Q is significantly higher than at point P due to its closer proximity to Earth's center. With this increase in kinetic energy, the satellite will continue to move away from Earth's center, following a path that takes it to a higher orbit or possibly escaping Earth's gravitational influence, depending on the specific conditions and velocity it achieves at point Q. The satellite is essentially in an elliptical or hyperbolic trajectory depending on its final velocity, and it will continue along this path, influenced by Earth's gravity, unless acted upon by external forces.

Haven’t read through the rest but (a) is just wrong, it’s saying both energies will decrease and yet somehow that total energy will be constant

 

[Interdice]

Umm. COuld someone explain to me why EMF is considered a force? I thought it was a voltage?

 

[esybeast]

Umm. COuld someone explain to me why EMF is considered a force? I thought it was a voltage?

emf causes current to flow in closed loop, the current per Lenz's law will induce a magnetic field that opposes change of flux that caused it. so it exerts an opposing force.

 

[iheartcookies1012]

what do u guys think b6 would be this yr? 87?

 

[matiaschong123]

anyway back to 1-32. eddie latexed it which looks much nicer than his scribbles and also makes a much smaller filesize!

also this will be updated later today with 33 and 34 added.

i’m pretty sure nesa wont accept ur answer for 30bi and ii bc nesa had an identical question in their sample hsc questions and they stated that if there is a slit there would be no current induced and no force exerted

 

[Jamesssss10]

If I talked about Thomson and Millikan for the 9 marker is that valid? Started writing about Geiger Marsden then realised after 3/4 of the page I was off track.

 

[Alamjot]

If I talked about Thomson and Millikan for the 9 marker is that valid? Started writing about Geiger Marsden then realised after 3/4 of the page I was off track.

You could of talked about anyone if u related it to the question

 

[Jamesssss10]

You could of talked about anyone if u related it to the question

Yeah hopefully my English adv skills came in helpful

 

[Rattlehead15]

what do u guys think b6 would be this yr? 87?

Probably but I hope its a little lower because Im on track for raw 85

 

[Aminza21]

Aye
Q5 multi choice asks which has the greatest travel time yeah. And all the intervals are equal distance. And if velocity is greater at larger radii then the “greatest travel time” would be the lowest velocity coz t = d/v … so ain’t the answer B (not D)

 

[carrotsss]

Aye
Q5 multi choice asks which has the greatest travel time yeah. And all the intervals are equal distance. And if velocity is greater at larger radii then the “greatest travel time” would be the lowest velocity coz t = d/v … so ain’t the answer B (not D)

total energy is constant throughout (by LOCE), and gravitational potential energy is higher when it’s further away, and hence kinetic energy and by extension velocity is lower to maintain total energy

 

[Goldattacker40000]

For Q27a why do they discuss other features that are not related to temperature and chemical composition. Talking about density and translational and rotational velocities seems to not answer the question.

 

[Engineer_192931]

For Q27a why do they discuss other features that are not related to temperature and chemical composition. Talking about density and translational and rotational velocities seems to not answer the question.

Yep I only talked about chemical composition and surface temp

 

[wizzkids]

Umm. COuld someone explain to me why EMF is considered a force? I thought it was a voltage?

EMF is not a force. EMF is a field potential or field potential difference. The units of EMF are joules per coulomb. The EMF of a point in an electric field is defined as the work done to bring 1 unit of positive test charge from infinity to that point within the electric field. The electric field potential is the electric potential energy per unit of positive charge. The electric field potential at infinite distance from any charge is defined as zero.

 

[Interdice]

EMF is not a force. EMF is a field potential or field potential difference. The units of EMF are joules per coulomb. The EMF of a point in an electric field is defined as the work done to bring 1 unit of positive test charge from infinity to that point within the electric field. The electric field potential is the electric potential energy per unit of positive charge. The electric field potential at infinite distance from any charge is defined as zero.

Look at teh solutions. When it says "find a force" they are finding the emf. idk what's going on. I hope I get at least 1 mark

 

[salihaaaameeer]

Raw to aligned (aka hsc)

hey so I had a question, im ranked 2nd in physics however the persons that's ranked first is only 0.5% away from me. what would happen if they for example fail the exam and I end up with a, for example, aligned mark of 80? would they affect my mark

 

[carrotsss]

hey so I had a question, im ranked 2nd in physics however the persons that's ranked first is only 0.5% away from me. what would happen if they for example fail the exam and I end up with a, for example, aligned mark of 80? would they affect my mark

not really too much, your internal mark will just be a bit behind the highest, so a bit below 80 (assuming there is a much greater spread for other rankings)