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.
Physics Predictions/Thoughts (2 Viewers)
- Thread starter StudyOnly
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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
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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
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Also I’m impressed you know this considering your HSC is in 2020.
TheOnePheeph
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Yeah I'm not mentions pylons or anything like that, just the basic idea like you said with power loss in wires.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
Arrowshaft
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Ok thats good, I swear if the HSC screws us over though hahaha
TheOnePheeph
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Yeah, rip any chance of me getting b6 if they expect stuff like this lol.
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
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
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
We are required to use the identities
To find the voltage in the secondary coil to Transformer 1,
Now, in Transformer 2, this section behaves as the primary coil, hence;
So, the ammeter would read 1.0 A, thus (C).
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Arrowshaft
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Note that dark spots occur during destructive interference, that is,
Using this, we can see that it is a second order dark spot, so
Thus, we get
You should also know from theory, that
is the definition of the path difference, hence the path difference is
I hope to hear back from @Arrowshaft in regards to his 98 in physics lmao
Arrowshaft
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I wish lol
ye 100 you mean
TheOnePheeph
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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.