Physics Predictions/Thoughts (1 Viewer)

[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

 

[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

 

[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.

I changed the Chat GPT in post #444 from last night's version for Q34 and it looks better now. But still wait for eddie's version. That might be a little more succinct. ChatGPT can be unreliable and isn't restricted by any syllabus.

 

[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)

 

[chilli 412]

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)

what if there are only 2 people in a physics class (this is just a thought experiment)
person A is ranked first
person B is ranked second

in the hsc exams, person A absolutely bombs the exam and gets like 20%
person B gets 80%

won't person B's hsc mark be made up of his exam mark (80%) and the exam mark of 2nd place (20%)? that doesnt seem fair

 

[carrotsss]

what if there are only 2 people in a physics class (this is just a thought experiment)
person A is ranked first
person B is ranked second

in the hsc exams, person A absolutely bombs the exam and gets like 20%
person B gets 80%

won't person B's hsc mark be made up of his exam mark (80%) and the exam mark of 2nd place (20%)? that doesnt seem fair

unless person A was considered such an outlier that it was ignored (which is fairly possible), then yes both people would get 50% overall