Physics Predictions/Thoughts (1 Viewer)

[tywebb]

That was ChatGPT by the way. And that just leaves Question 34. Come on surely it won't take 3 hours to do just 1 question!

 

[carrotsss]

That was ChatGPT by the way. And that just leaves Question 34. Come on surely it won't take 3 hours to do just 1 question!

Tbf that response definitely wouldn’t get full marks, it doesn’t go nearly specific enough to the HSC syllabus and specific experiments/scientists but it’s nonetheless impressive

 

[Hsc1111]

chem was slightly easier than usual asw, maybe 86 or so

Is this a band 6? What will be the cut off for a band 5?

 

[carrotsss]

Is this a band 6? What will be the cut off for a band 5?

Around 70

 

[Hsc1111]

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

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

 

[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

 

[tywebb]

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

a.

To analyze the qualitative energy changes as the satellite moves from point P to point Q, we can consider the satellite's potential energy, kinetic energy, and total mechanical energy. The total mechanical energy (E) of the satellite remains constant unless acted upon by an external force. The formula for total mechanical energy is:

E = Potential Energy (U) + Kinetic Energy (K)

1. At point P:
   • The initial potential energy is -2.389 x 10^6 J.
   • The initial kinetic energy is unknown (but it's a part of the total energy, which is -1.195 x 10^10 J).
   • The initial total mechanical energy is the sum of these two values: E = -2.389 x 10^6 J + (unknown).
2. The satellite's engines are fired, increasing its kinetic energy by 5.232 x 10^8 J. After the engine is shut down:
   • The potential energy remains the same.
   • The final kinetic energy is K = (initial kinetic energy) + 5.232 x 10^8 J.
   • The final total mechanical energy is E = (unchanged potential energy) + (new kinetic energy).
3. At point Q (diametrically opposite to P, 6.85 x 10^6 m from Earth's center):
   • The potential energy will be different from the initial value, as the satellite is at a different distance from Earth's center.
   • The kinetic energy will be different from the value at point P, as the satellite's speed will change.
   • The total mechanical energy at Q is E = (new potential energy at Q) + (new kinetic energy at Q).

Now, let's analyze the qualitative changes:

• As the satellite moves from P to Q, the distance from Earth's center increases (6.85 x 10^6 m at point Q compared to 6.7 x 10^6 m at point P). Therefore, the potential energy will increase since the gravitational potential energy depends on the distance from the center of the Earth.
• The satellite's speed will change as it moves from P to Q, affecting its kinetic energy. The kinetic energy will depend on the speed, and since the engine was shut down at point P, it won't receive any more energy input. Therefore, the kinetic energy may decrease or remain the same but won't increase as it did when the engines were fired.

In summary, as the satellite moves from P to Q, its potential energy will increase due to the change in distance from Earth's center, and its kinetic energy may decrease or remain constant. The total mechanical energy should still remain constant throughout the trajectory, as long as there are no external forces acting on the satellite.

b.

To show that the kinetic energy of the satellite at point Q is 1.194 x 10^10 J, we can use the conservation of mechanical energy. At point P, the satellite has a certain potential and kinetic energy. When its engines are fired, its kinetic energy increases, but the total mechanical energy remains constant. As the satellite moves to point Q, the total mechanical energy will still be conserved.

Let's denote the initial kinetic energy at point P as K_P and the potential energy at P as U_P. We are given that the total energy at P is -1.195 x 10^10 J, so:

Total Energy at P (E_P) = K_P + U_P = -1.195 x 10^10 J

When the engines are fired, the kinetic energy increases by 5.232 x 10^8 J, so the new kinetic energy at P (K'_P) is:

K'_P = K_P + 5.232 x 10^8 J

The potential energy at P remains the same, so U_P is unchanged.

Now, as the satellite moves to point Q, the total mechanical energy at point Q (E_Q) will still be the same as at point P since there are no external forces acting on the satellite:

E_P = E_Q

We know the potential energy depends on the distance from Earth's center, and at point Q, the satellite is 6.850 x 10^6 m from the center, so the potential energy at Q (U_Q) is different from the potential energy at P (U_P):

U_Q = -G * (m * M) / r_Q, where r_Q = 6.850 x 10^6 m

Now, we can write the equation for the total mechanical energy at point Q:

E_Q = K_Q + U_Q

Since E_P = E_Q, we can equate the expressions for the total mechanical energy at points P and Q:

K'_P + U_P = K_Q + U_Q

Now, substitute the values:

(K_P + 5.232 x 10^8 J) + U_P = K_Q + U_Q

We already know the values of U_P, K_P, and U_Q from the given information. We can solve for K_Q:

(K_P + 5.232 x 10^8 J) + (-2.389 x 10^6 J) = K_Q + U_Q

Now, calculate U_Q:

U_Q = -G * (m * M) / r_Q U_Q = -6.67430 x 10^(-11) N·m²/kg² * (5.972 x 10^24 kg * 400 kg) / (6.850 x 10^6 m) U_Q = -2.091 x 10^9 J

Now, substitute the values into the equation:

(K_P + 5.232 x 10^8 J) + (-2.389 x 10^6 J) = K_Q + (-2.091 x 10^9 J)

K_Q = (K_P + 5.232 x 10^8 J) - (-2.389 x 10^6 J) + 2.091 x 10^9 J

K_Q = (K_P + 5.232 x 10^8 J + 2.389 x 10^6 J + 2.091 x 10^9 J)

K_Q = K_P + 1.194 x 10^10 J

Now, you can see that the kinetic energy at point Q (K_Q) is indeed 1.194 x 10^10 J. This shows that the satellite's kinetic energy at point Q is 1.194 x 10^10 J, as required.

c.

After the satellite passes through point Q, which is diametrically opposite to point P, its motion will be influenced by the conservation of angular momentum and energy. Let's explain the motion of the satellite after it passes through point Q:

1. Initial Circular Orbit (Before Point P):
   • The satellite was in a circular orbit around Earth with a radius of 6.700 x 10^6 m.
   • Its motion in this orbit was characterized by a constant speed and angular momentum, and its kinetic energy was balanced by the gravitational potential energy, resulting in a stable circular orbit.
2. Firing the Engines at Point P:
   • At point P, the satellite's engines were fired, increasing its velocity and kinetic energy while maintaining its angular momentum.
   • This increased kinetic energy allowed the satellite to temporarily move to a higher, elliptical orbit with a larger semi-major axis.
3. Trajectory to Point Q:
   • The satellite followed a trajectory to reach point Q, which is 6.850 x 10^6 m from Earth's center.
   • As the satellite moves from P to Q, its speed decreases due to the increase in the distance from Earth's center.
   • The kinetic energy decreases while angular momentum remains conserved. However, it's important to note that the total mechanical energy remains constant during this motion.
4. Point Q:
   • At point Q, the satellite has reached its highest point in the elliptical orbit.
   • Its kinetic energy is lower than it was at point P, and its potential energy has increased significantly.
   • The satellite is momentarily at rest at this point before starting to move back towards Earth.
5. Post-Q Motion:
   • After reaching the highest point at Q, the satellite will start descending back towards Earth.
   • As it descends, its speed will increase due to the decreasing distance from Earth's center, and it will regain kinetic energy.
   • Its angular momentum remains constant, so the motion will be in an elliptical path, and the satellite will continue to orbit Earth.

In summary, after passing through point Q, the satellite will descend from its highest point, and its kinetic energy will increase as it approaches Earth. It will continue to follow an elliptical orbit, similar to its motion between points P and Q, with the angular momentum conserved throughout the entire trajectory.

EDIT: I changed this Chat GPT from last night's version. This one looks better.

 

[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

 

[tywebb]

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

Well you read more of it than I did. I just Chat GPT'd and copy-pasted.

 

[Interdice]

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

 

[tywebb]

just wrong

maybe we'd better wait more than 3 hours for eddie.

 

[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?

 

[tywebb]

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 probably be updated later with 33 and 34 added.

 

[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