AQA A-level Physics Paper 2 (7408/2) - 9th June 2023 [Exam Chat]

Original post by Kingemperor07

I've been stuck on these Isaac physics questions for some days now, does anyone know how to solve these:
- https://isaacphysics.org/questions/two_cs_three_rs?board=709e4d70-11ca-40da-9e63-ae6f9ce02e8d&stage=a_level
- https://isaacphysics.org/questions/electric_shapes?board=e27720fa-0362-4e87-9500-bb1f3f66375a&stage=a_level

are u still stuck on the Two C's and three R's? I think i have the solution for it lying somewhere

Reply 43

moonlight2460

8

Question, If in a graph of pressure against volume the area under the curve = wd, is that also = NkT?

Reply 44

Original post by Ak_5

https://isaacphysics.org/questions/bringing_to_the_boil?board=1b562310-1f71-4d5f-8930-7b825670eb59&stage=a_level

can someone help me with this i have no clue whats going on

Heres the explanation:

a)
- figure out how much power is wasted to surroundings using P = k(100-20) = 60.8W
- This however isn't the energy supplied by the heater, to get this use the efficiency equation: efficiency = output power / input power
- Therefore input power to heater = 60.8 / 0.7 = 86.8W

b)
- Same logic, use Pt = mL to get the power necessary to boil over, P = (m/t) * L, and you are given (m/t) and L in the question
- you should get 1130W, now do 1130 / 0.7 to get the input power needed = 1610W
- add this to 86.8W from the previous question to get the total power needed to keep the water at the boiling point + preventing the boiling over.
- Answer = 1610 + 86.8 = 1700W

Reply 45

Original post by prithu

are u still stuck on the Two C's and three R's? I think i have the solution for it lying somewhere

Yeah im still stuck on that question, my brain hurts from it lol

Reply 46

AbbyDodds

I'm struggling a bit with these questions on Isaac Physics, how would I do them?

https://isaacphysics.org/questions/phys_linking_22_q4?board=d7fab289-d7b4-46d1-8cf7-5c9ea758f66c&stage=a_level

Reply 47

Ak_5

Original post by Kingemperor07

Heres the explanation:

a)
- figure out how much power is wasted to surroundings using P = k(100-20) = 60.8W
- This however isn't the energy supplied by the heater, to get this use the efficiency equation: efficiency = output power / input power
- Therefore input power to heater = 60.8 / 0.7 = 86.8W

b)
- Same logic, use Pt = mL to get the power necessary to boil over, P = (m/t) * L, and you are given (m/t) and L in the question
- you should get 1130W, now do 1130 / 0.7 to get the input power needed = 1610W
- add this to 86.8W from the previous question to get the total power needed to keep the water at the boiling point + preventing the boiling over.
- Answer = 1610 + 86.8 = 1700W

Thank you so much, i was overcomplicating it unnecessarily

Reply 48

username5228292

3

Original post by AbbyDodds

I'm struggling a bit with these questions on Isaac Physics, how would I do them?

https://isaacphysics.org/questions/phys_linking_22_q4?board=d7fab289-d7b4-46d1-8cf7-5c9ea758f66c&stage=a_level

You have to utilise the equation Emf = BANomega x sin(omega x t) equation as it is concerned with a coil. You are trying to find out the flux density at max EMF and that is when sin(omega x t) = 1. So you are just left with Emf(max) = BANomega. You can then solve for B using the various info given.

Reply 49

sara2918

Anyone have a set of formulas that we need to know that arent on the data sheet?

Reply 50

JP22r

8

do we need to know about back emf?

Reply 51

Clocky265

Original post by JP22r

do we need to know about back emf?

Not as far as I am aware of...(?)

Reply 52

zainab.sambo

3

Original post by Abdullah05

Hey can anyone help me with this question:
Sea level is currently observed to increase at a total rate of about 3mm/year. Out of this total rate, approximately 1mm/year is due to thermal expansion of the warming sea water. This is known as steric sea level change.
Assume that the ocean heating occurs uniformly over the top 1000m of the ocean at a rate of 0.01∘Cyear−1. Calculate the power required for this ocean heating.

Assume that the Earth is a perfect sphere with radius 6371km
The oceans covers 70% of the Earth’s surface
The density of sea water is 1025kgm−3

The heat capacity of sea water C, =4.006×10^3Jkg−1K−1

This is what i did. i kinda just waffled.

Preview attachment 20230529_180236.jpg

20230529_180236.jpg
1.5 MB

Hi are you able to attach this differently please cuz the file won't open

Reply 53

Polp

12

Original post by sara2918

Anyone have a set of formulas that we need to know that arent on the data sheet?

My friend sent this a while back. Some of them you can derive but some you just have to know.

(edited 10 months ago)

Reply 54

sara2918

Life saving, thanks

Original post by Polp

My friend sent this a while back. Some of them you can derive but some you just have to know.

Reply 55

Ak_5

https://isaacphysics.org/questions/planet_with_cavity_num?board=4f2344c4-5312-4a21-b671-eeebf8f7ed36&stage=a_level

anyone able to do this question? can you post how to do it?

Reply 56

Original post by Ak_5

https://isaacphysics.org/questions/planet_with_cavity_num?board=4f2344c4-5312-4a21-b671-eeebf8f7ed36&stage=a_level

anyone able to do this question? can you post how to do it?

- You know that Graviational potential at the surface must be: V = -GM/R, where M is the mass of the planet and R is the radius of the planet.
- M can be written as pV where p is the density and V is the volume of the fictional planet.
- basically just convert the density into SI units and subtract the volume of the larger sphere from the inner sphere to get V.
- then use V = -GpV/R with all your values.

Reply 57

Clocky265

Original post by Ak_5

https://isaacphysics.org/questions/planet_with_cavity_num?board=4f2344c4-5312-4a21-b671-eeebf8f7ed36&stage=a_level

anyone able to do this question? can you post how to do it?

Not sure if the answer is right as I sometimes have difficulty with entering the answer as required to that website but I will do my best:

The formula for gravitational potential is (-) GM/r

G is a constant, we are given R and all we have to do is calculate the Mass of the planet.

First you calculate the volume with R and then subtract the volume with R/2 from it. That would be the volume. You then time the density and the volume together to get the mass.

Reply 58

Original post by moonlight2460

Question, If in a graph of pressure against volume the area under the curve = wd, is that also = NkT?

- the equation W = pV is for a change in volume given a constant pressure
- pV = NkT is for fixed V and p meaning that the area won't necessarily equal NkT.

(edited 10 months ago)

Reply 59

Ak_5