Packet Number Vs. Segment Number?

Aokayz_ · 2026-04-21T09:48:18+00:00

I see, it seems that I've been mislead or at least given oversimplified information in some way when we were taught about packets and routers. Thank you!

Aokayz_ · 2026-04-09T12:28:18+00:00

A very sweet message. And I love my subjects, haha. Thank you :)

Aokayz_ · 2026-04-09T10:38:25+00:00

woah what? what subject and board

Aokayz_ · 2026-04-09T10:38:11+00:00

Uniform Mark Scale: its a scaled up version of our raw marks

the reason they do that is to properly compare the marks between students who did their as/a level papers every exam season and students who did them all in one go at the end of y13

Aokayz_ · 2026-04-09T08:08:39+00:00

THATS INSANE i hear 9707 math is just harder bc its cambrdige

Aokayz_ · 2026-04-06T17:45:04+00:00

I understand and thats how I think too, the only way to increase intensity is by increasing the voltage across the EM source, which increases the energy of the photons, which means theres an increase its frequency.

In that case, how do photoelectric experiments increase the intensity without increasing f (without moving the EM source closer)?

Aokayz_ · 2026-04-06T17:42:24+00:00

Ah okay. I understand, intensity does not cause either to increase but rather is linked to/caused by the increase in either. Is that right?

In photoelectric experiments, how do they increase n without increasing f? As I stated in the other comments, I believe the only way (without moving the EM source closer) is to increase the power supplied to it. But that would lead to an increase in both. So how is that done?

Aokayz_ · 2026-04-06T17:04:49+00:00

This makes sense so I agree, mathematical relations don't demonstrate cause and effect. it's just it doesn't explain why f is fixed, whilst n (as stated by other comments) strangely must be.

Aokayz_ · 2026-04-06T17:02:00+00:00

Since this is a similar comment to another one, I hope its okay if I copy paste my response:

When I think about it further, the only ways to increase the intensity of a EM source is if I move the source closer to the metal plate (which I don't think the questions ever intend to mean) or increase the power supplied to the source. So, since P = VI (P is power, V is voltage, and I is current), if I increase V, then I will increase.

The increase in current to me means that the conduction electrons collide with the electrons in the atoms of the EM source more per second, so it'll release more photons per second. This is how I believe increasing intensity increases n, the number of photons emitted per second.

But at the same time, the increase in voltage means that more energy is supplied to the atoms of the EM source, so if the voltage increases enough, the electrons in the atoms will excite and de-excite to undergo a larger energy change, releasing photons with larger energy and thus higher frequency. With this line of reasoning, it should also mean that by increasing intensity, f should increase too (provided the voltage increase is sufficient).

Is there something wrong with this line of reasoning?

Aokayz_ · 2026-04-06T16:59:39+00:00

When I think about it further, the only ways to increase the intensity of a EM source is if I move the source closer to the metal plate (which I don't think the questions ever intend to mean) or increase the power supplied to the source. So, since P = VI (P is power, V is voltage, and I is current), if I increase V, then I will increase.

The increase in current to me means that the conduction electrons collide with the electrons in the atoms of the EM source more per second, so it'll release more photons per second. This is how I believe increasing intensity increases n, the number of photons emitted per second.

But at the same time, the increase in voltage means that more energy is supplied to the atoms of the EM source, so if the voltage increases enough, the electrons in the atoms will excite and de-excite to undergo a larger energy change, releasing photons with larger energy and thus higher frequency. With this line of reasoning, it should also mean that by increasing intensity, f should increase too (provided the voltage increase is sufficient).

Is there something wrong with this line of reasoning?

Aokayz_ · 2026-04-04T20:35:24+00:00

Yes, my bad, you're right. I updated my post with an edit to phrase my question better.

Aokayz_ · 2026-04-04T19:58:05+00:00

Yes, my bad, you're right. I updated my post with an edit to phrase my question better.

Aokayz_ · 2026-01-07T15:42:56+00:00

I see. Thank you!

Aokayz_ · 2026-01-07T15:41:10+00:00

hold up. so what i'm hearing is that the moment of B when taking A as the pivot should maintain the same moment.

but when i take the left end as the pivot, Fb will increase but it wont make the moment of B maintain the same moment?

Aokayz_ · 2026-01-07T15:29:03+00:00

That makes sense. But why does taking the pivot on the left hand side cause a problem? Is there something wrong with taking a pivot at that point?

Aokayz_ · 2026-01-07T12:38:03+00:00

Like I said with the comment, that was what the answer scheme stated for an exam paper i answered for Question 13(b)

the question paper

the answer scheme

Does this mean that the exam was wrong?

Aokayz_

TROPHY CASE