# Physics Predictions/Thoughts (1 Viewer)

#### 大きい男

##### Active Member
- when a paddle wheel was placed inside an evacuated CRT, the paddle wheel spun - implying the cathode rays had momentum and thus mass.
Technically, this did not prove that cathode rays had mass, although we now know this. The paddlewheel turns due to the radiometric effect, also demonstrated by the Crooke's Radiometer.

#### psmao

##### Member
In summary: cathode rays were argued to be either waves or particles.

Evidence for the particle nature:
- when a paddle wheel was placed inside an evacuated CRT, the paddle wheel spun - implying the cathode rays had momentum and thus mass.
- when cathode rays were passed through electric and magnetic fields, they deflected
- they travelled significantly slower than light.
- they left at right angles to the surface of the cathode

Evidence for wave nature:
- rectilinear propagation (deduced from the shadows produced from the rays, refer to the Maltese across experiment)
- they passed through thin metal foils without damaging them
maybe its worthy to mention that initially scientists used electric field to deflect but the technology at that time was not accurate enough to detect the deflection

#### Arrowshaft

##### Well-Known Member
Technically, this did not prove that cathode rays had mass, although we now know this. The paddlewheel turns due to the radiometric effect, also demonstrated by the Crooke's Radiometer.
While I agree with you on the radiometric effect, the transfer of momentum also had a part to play. It may not be concrete evidence, but evidence nonetheless. Also, I would think delving into the radiometric would be too much for HSC physics.

#### Arrowshaft

##### Well-Known Member
Technically, this did not prove that cathode rays had mass, although we now know this. The paddlewheel turns due to the radiometric effect, also demonstrated by the Crooke's Radiometer.
Also I’m impressed you know this considering your HSC is in 2020.

#### TheOnePheeph

##### Active Member
Hey guys, a question - for the applications of step up and step down transformers, how much detail have you gone to? Because I dont know whether it is worth memorising the structure of powerstations and use of pylons, etc. or just knowing the basic concept of stepping up voltage to reduce current and hence power loss by
$\bg_white P_{loss}=I^2R$
Yeah I'm not mentions pylons or anything like that, just the basic idea like you said with power loss in wires.

#### Arrowshaft

##### Well-Known Member
Yeah I'm not mentions pylons or anything like that, just the basic idea like you said with power loss in wires.
Ok thats good, I swear if the HSC screws us over though hahaha

#### TheOnePheeph

##### Active Member
Ok thats good, I swear if the HSC screws us over though hahaha
Yeah, rip any chance of me getting b6 if they expect stuff like this lol.

#### Faye19

##### New Member
Can someone solve this qtn plz

#### nasr7601

##### New Member
N(1) = 1000, N(2) = 500, N(3) = 100; V(1) = 240V, V(2) = 120V, V(3) = 24V; I(1) = 0.5A, I(2) = 1A, I(3) = 5A
So I guess, If the number of windings are halved in the first transformer and then fifthed, do some reverse problem solving to realise that current equates to 1A (so C).

The assumption is that the secondary coil of the first transformer is the same number of windings as the primary coil of the second transformer.

Gl for tomorrow

#### lplsz2000

##### New Member
Can someone solve this qtn plz
View attachment 27537
From the rightmost circuit, the total power should be 120W, from the ratio of the coils of the first transformer, the voltage in the second circuit should be 120V, P=IU, so the current should be 1A?

#### Arrowshaft

##### Well-Known Member
Can someone solve this qtn plz
View attachment 27537
We are required to use the identities

$\bg_white \dfrac{V_p}{V_s}=\dfrac{N_p}{N_s}=\dfrac{I_s}{I_p}$

To find the voltage in the secondary coil to Transformer 1,

$\bg_white V_s=\dfrac{N_s}{N_p}\times V_p=\dfrac{1000}{500}\times240=120~V$

Now, in Transformer 2, this section behaves as the primary coil, hence;

$\bg_white \dfrac{V_p}{V_s}=\dfrac{I_s}{I_p}\implies I_p=\dfrac{I_s\times V_s}{V_p}=5.0\times\dfrac{24}{120}=1.0~A$

So, the ammeter would read 1.0 A, thus (C).

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#### Faye19

##### New Member
Thank you all so much, it's clear now

#### Arrowshaft

##### Well-Known Member
View attachment 27539
Note that dark spots occur during destructive interference, that is,

$\bg_white d\sin\theta=\left(m+\dfrac{1}{2}\right)\lambda$

Using this, we can see that it is a second order dark spot, so $\bg_white m=1$ (this is because the first order dark spot occurs at exactly half of the path difference, i.e. $\bg_white m=0$).
Thus, we get

$\bg_white d\sin\theta=\dfrac{3}{2}\lambda$

You should also know from theory, that

$\bg_white d\sin\theta$
is the definition of the path difference, hence the path difference is $\bg_white \dfrac{3}{2}\lambda$, i.e. (C).

#### Faye19

##### New Member
Thank you so much

#### nasr7601

##### New Member
I hope to hear back from @Arrowshaft in regards to his 98 in physics lmao

#### Faye19

##### New Member
And the next qtn plz

#### TheOnePheeph

##### Active Member
We know that increasing the intensity has no effect on the maximum kinetic energy of the emitted electrons, so A and C are instantly out. If you use the wavelength to find frequency of the light with c=f lambda, you will find that the frequency is 6.66×10^14 Hz. Since this is below the threshold frequency for metal Y, no photoelectrons are emitted, meaning doubling the intensity will have no effect. Since it is above the threshold frequency for metal X, however, the number of electrons emitted will increase, due to increased intensity, so B is the answer.