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Quantum Phenomena

https://00c07169b8f01b5b433281e611eed08ab9509441.googledrive.com/host/0B1ZiqBksUHNYaHN6VmVKTlVMWms/January%202010%20QP%20-%20Unit%201%20AQA%20Physics.pdf

question 4b part 3 how do i do it? i know i've done something wrong since the answer i got for 4b part 4 was the wavelength of light given in the question >.>

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Reply 1
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Kyx
6
Original post by thefatone
https://00c07169b8f01b5b433281e611eed08ab9509441.googledrive.com/host/0B1ZiqBksUHNYaHN6VmVKTlVMWms/January%202010%20QP%20-%20Unit%201%20AQA%20Physics.pdf

question 4b part 3 how do i do it? i know i've done something wrong since the answer i got for 4b part 4 was the wavelength of light given in the question >.>


we know that max kinetic energy = hf - phi.

Kinetic energy is 1/2 mv^2 so, kinetic energy/(1/2)m should equal v^2

Then square root the answer. Is this correct?
Reply 2
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thefatone
OP
6
Original post by Kyx
we know that max kinetic energy = hf - phi.

Kinetic energy is 1/2 mv^2 so, kinetic energy/(1/2)m should equal v^2

Then square root the answer. Is this correct?


well ffs then i didn't even know to use that -.-

i'll tag you a few more times when i get stuck
(edited 8 years ago)
Reply 3
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Kyx
6
Original post by thefatone
well ffs then i didn't even know to use that -.-


glad to help :smile:
Reply 4
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thefatone
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6
Original post by Kyx
glad to help :smile:


what about q5 part 2? i don't understand
Reply 5
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Kyx
6
Original post by thefatone
what about q5 part 2? i don't understand


Using the resistivity formula, make R the subject.

Then it would be easier to see the relationship :smile:
Reply 6
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thefatone
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6
Original post by Kyx
Using the resistivity formula, make R the subject.

Then it would be easier to see the relationship :smile:


so i saw radius is halved and length is multiplied by 4

so i got this ρ=12RA4L \rho = \dfrac{\frac{1}{2} RA}{4L}

then i said well this means that ρ=RA6L \rho= \dfrac{RA}{6L}
Reply 7
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Kyx
6
Original post by thefatone
so i saw radius is halved and length is multiplied by 4

so i got this ρ=12RA4L \rho = \dfrac{\frac{1}{2} RA}{4L}

then i said well this means that ρ=RA6L \rho= \dfrac{RA}{6L}


No. In area, the radius is squared. It is 1/4 divided by 4, which makes it 16 methinks.
Reply 8
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thefatone
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6
Original post by Kyx
No. In area, the radius is squared. It is 1/4 divided by 4, which makes it 16 methinks.


oh i see so it's 1/4 which mean the bottom is also multiplied by 4 so that's a total of 16 times .....

wait why does resistance increase? why doesn't it ... NVM

thanks for the help i'll probably tag you in again in a minute
Reply 9
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Kyx
6
Original post by thefatone
oh i see so it's 1/4 which mean the bottom is also multiplied by 4 so that's a total of 16 times .....

wait why does resistance increase? why doesn't it ... NVM

thanks for the help i'll probably tag you in again in a minute


Rearrange to make R the subject :smile:

You get 4 divided by 1/4 which equals 16 :smile:
Reply 10
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thefatone
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6
Original post by Kyx
Rearrange to make R the subject :smile:

You get 4 divided by 1/4 which equals 16 :smile:


yup thanks i got it ^-^

can you help on 5b i have no idea where to start
Reply 11
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Kyx
6
Original post by thefatone
yup thanks i got it ^-^

can you help on 5b i have no idea where to start


First of all, what equipment would you need?

Then, how would you set it up (parallel, series)?

What measurements you would make?

How you would use the measurements to find the resistivity?

How you can make these measurements as precise as possible?
Reply 12
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thefatone
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6
Original post by Kyx
First of all, what equipment would you need?

Then, how would you set it up (parallel, series)?

What measurements you would make?

How you would use the measurements to find the resistivity?

How you can make these measurements as precise as possible?


1. obviously voltmeter ammeter putty wires cells ruler can't think of anything else
2. no idea maybe series?
3. taking down values of V and I when i connect the circuit up. radius of putty, length of putty

4. R=\dfrac{V}{I}\ and\ sub in values into \rho= \dfrac{RA}{L}
5. use a measuring tool with smallest increment?
Reply 13
Avatar for Kyx
Kyx
6
Original post by thefatone
1. obviously voltmeter ammeter putty wires cells ruler can't think of anything else
2. no idea maybe series?
3. taking down values of V and I when i connect the circuit up. radius of putty, length of putty

4. R=\dfrac{V}{I}\ and\ sub in values into \rho= \dfrac{RA}{L}
5. use a measuring tool with smallest increment?


2. how would you put the voltmeter? How would you put the ammeter?

Spoiler

5. Repeat readings?
Reply 14
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thefatone
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6
Original post by Kyx
2. how would you put the voltmeter? How would you put the ammeter?

Spoiler

5. Repeat readings?


ah ok so a basic circuit

but wouldn't repeating give me accurate answers not precise ones?
Reply 15
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Kyx
6
Original post by thefatone
ah ok so a basic circuit

but wouldn't repeating give me accurate answers not precise ones?


Correct :smile:

Glad you saw that
Reply 16
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thefatone
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Original post by Kyx
Correct :smile:

Glad you saw that


my teachers especially physics and chem ramble on so much about the difference between accurate an precise is unreal so that's stayed in my head
Reply 17
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Kyx
6
Original post by thefatone
my teachers especially physics and chem ramble on so much about the difference between accurate an precise is unreal so that's stayed in my head


Probably for the best :smile:
Reply 18
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thefatone
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6
Original post by Kyx
Probably for the best :smile:


yup
Reply 19
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thefatone
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Original post by Kyx
Probably for the best :smile:


how would i do question 6a part 2?

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