# National 5 Physics 2018-19Watch

Announcements
1 week ago
#41

Anyone give a hand with this? I keep getting ‘E’ with both the following equations, but that’s wrong:

p1 over T1 = p2 over T2

(pV) over T = constant
1
reply
1 week ago
#42
(Original post by Vladimir Poutine)

Anyone give a hand with this? I keep getting ‘E’ with both the following equations, but that’s wrong:

p1 over T1 = p2 over T2

(pV) over T = constant
Pressure is directly proportional to temperature in Kelvin; you're using degrees celcius. -273 degrees celcius = 0K, 0K = 273 degrees celcius

P/T = k

150/300 = k

k = 0.5

P/T = k

P/320 = 0.5

P = 0.5 x 320

P = 160kPa
Last edited by Strelzo; 1 week ago
1
reply
1 week ago
#43
Aha! I knew I was doing something stupid...

Thank you
(Original post by Strelzo)
Pressure is directly proportional to temperature in Kelvin, you're using degrees celcius. -273 degrees celcius = 0K, 0K = 273 degrees celcius

P/T = k

150/300 = k

k = 0.5

P/T = k

P/320 = 0.5

P = 0.5 x 320

P = 160kPa
0
reply
1 week ago
#44
(Original post by Vladimir Poutine)
Aha! I knew I was doing something stupid...

Thank you
No worries, tons of people in our class kept doing that and the teacher (who is a joking, amazing teacher) went mental (not in an aggressive way), and it was hilarious xD. One time he got someone to talk him through the question, the student went "Ehm, firstly P/T = k, then 200/49 = k, therefore k = 4.082", teacher went "Ah... yes, you've done the one thing I told you not to do", student went blank, teacher: "Pressure is directly proportional to Temperature IN KELVIN"; something about the way that discussion went down was so funny, and he purposely tried to have a laugh lol. You'd have had to have been there lol
0
reply
1 week ago
#45
Lol, I really should have been there I probably would have remembered Kelvin better.

I’mma stick that on a piece of A3 on my wall.
0
reply
1 week ago
#46
(Original post by Vladimir Poutine)
Lol, I really should have been there I probably would have remembered Kelvin better.

I’mma stick that on a piece of A3 on my wall.
Definitely xD, my teacher is class; really hope that I have him next year.
0
reply
1 week ago
#47
I haven't revised for physics, any advice? I got a B in the prelim without revision somehow lol
0
reply
1 week ago
#48
Anyone give a hand in this pls? Btw this is from 2007..
Posted on the TSR App. Download from Apple or Google Play
0
reply
1 week ago
#49
Do you have the answer? Just so I can double check that I’m right before I try and explain it to you and I then end up to be wrong
Posted on the TSR App. Download from Apple or Google Play
0
reply
1 week ago
#50
I don’t want to explain it and then it ends up being wrong and I just confuse you
Posted on the TSR App. Download from Apple or Google Play
0
reply
1 week ago
#51
Ya it's A
Posted on the TSR App. Download from Apple or Google Play
0
reply
1 week ago
#52
HERMION: Correct Answer is ?
Shine light on LDR its resistance drops in value, voltage divider means what?
Also sum of voltages always equals supply emf?
Think which answers?
0
reply
1 week ago
#53
(Original post by beth1709)
Do you have the answer? Just so I can double check that I’m right before I try and explain it to you and I then end up to be wrong
If you cant answer don't answer- although any relevant assistance Ohms Law or whatever is relevant might be ok!
0
reply
1 week ago
#54
Okay that’s what I thought. Basically as the brightness of the LDR increases the resistance of V1 has to decrease. This means that the voltage of V1 will also decrease. The only option which has V1’s voltage < 2.5 is A.
Posted on the TSR App. Download from Apple or Google Play
0
reply
1 week ago
#55
(Original post by tomctutor)
If you cant answer don't answer- although any relevant assistance Ohms Law or whatever is relevant might be ok!
What do you mean?
Posted on the TSR App. Download from Apple or Google Play
0
reply
1 week ago
#56
Ya that makes sense. Thanks a lot!
Posted on the TSR App. Download from Apple or Google Play
1
reply
1 week ago
#57
Also something to add is that if for example there is two components in a circuit and one of their voltages/resistances decreases/increases the other has to do the opposite in order to add up to the supply voltage. In this example V1’s voltage decreases by 0.5 (2.5-2) therefore V2’s must increase by 0.5 (2.5-3).
Posted on the TSR App. Download from Apple or Google Play
0
reply
1 week ago
#58
I hope that makes sense and I haven’t confused you
Posted on the TSR App. Download from Apple or Google Play
0
reply
1 week ago
#59
(Original post by beth1709)
What do you mean?
"Do you have the answer? Just so I can double check that I’m right before I try and explain it "
You should be able to provide the right answer or at least an explanation, suggestion, good guess - thats all
don't mean to offend- good your trying to help others.
The more you do this you will learn yourself.
Last edited by tomctutor; 1 week ago
0
reply
1 week ago
#60
Thank you!
Posted on the TSR App. Download from Apple or Google Play
0
reply
X

Write a reply...
Reply
new posts
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• Arts University Bournemouth
Art and Design Foundation Diploma Further education
Sat, 25 May '19
• SOAS University of London
Postgraduate Open Day Postgraduate
Wed, 29 May '19
• University of Exeter
Undergraduate Open Day - Penryn Campus Undergraduate
Thu, 30 May '19

### Poll

Join the discussion

#### How did your AQA GCSE Physics Paper 1 go?

Loved the paper - Feeling positive (402)
30.69%
The paper was reasonable (514)
39.24%
Not feeling great about that exam... (225)
17.18%
It was TERRIBLE (169)
12.9%

View All
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.