I’m pretty sure you just do it for 1m3 area, because the intensity is the same per m3, so if you have a cylinder surface area with 1m3 and depth of 5m, mass of water is 5Kg, you know the intensity per m2 so calculate power, then power = energy because it was for second so pit that into E=mcdeltat I hope so anyway that’s what I did haha

that makes sense

Reply 44

uni12354

6

Original post by sxx1237

What did people get for the normal force on the probe

As in does A or B have a greater normal force?

I got B as A has a smaller radius of orbit so acceleration is lower meaning centripetal force lower and as weight is same, it must be a decrease in contact force.

Reply 45

Intramoon

10

Original post by Matthew3868

I’m pretty sure you just do it for 1m3 area, because the intensity is the same per m3, so if you have a cylinder surface area with 1m3 and depth of 5m, mass of water is 5Kg, you know the intensity per m2 so calculate power, then power = energy because it was for second so pit that into E=mcdeltat I hope so anyway that’s what I did haha

Yes same ! I left the question out to do at the end because I was confused as no radius was given so I just made it easier for myself by saying assume area = 1m2 lol but at the same time I wrote ‘no radius given??????’ Lmaooo

Reply 46

uni12354

6

Original post by Intramoon

Yes same ! I left the question out to do at the end because I was confused as no radius was given so I just made it easier for myself by saying assume area = 1m2 lol but at the same time I wrote ‘no radius given??????’ Lmaooo

so volume of a cylinder is 5pi r^2.

density=mass/volume
so mass =1000*5pir^2

E=P*t=IA*1= I *pir^2
cant remember what I was

deltaT= E/mc = I pir^2/ (1000*5pir^2*c)
so pi r^2 cancels out
giving change in T = I/5000c

Reply 47

ComPav

1

Original post by uni12354

As in does A or B have a greater normal force?

I got B as A has a smaller radius of orbit so acceleration is lower meaning centripetal force lower and as weight is same, it must be a decrease in contact force.

F=W-R so increase in contact but i said the same for the rest

Reply 48

randommath123

6

Original post by uni12354

As in does A or B have a greater normal force?

I got B as A has a smaller radius of orbit so acceleration is lower meaning centripetal force lower and as weight is same, it must be a decrease in contact force.

I put A because at the top is directly above the axis of rotation (I.e. there’s no circular motion so not centripetal acceleration) whereas at B, there is a resultant force towards the centre because of the centripetal force/acceleration?

Reply 49

Hychro617

7

For the temperature change you Let E=mcdelta(t) but as E=P/t and P= IA you get PIA/T, and as m=density*volume PIA/T=pVcdelta(t) but V = Ah so the areas cancel out and PIA/pVc =Delta(T)

Reply 50

fruitloops87

6

Original post by randommath123

I put A because at the top is directly above the axis of rotation (I.e. there’s no circular motion so not centripetal acceleration) whereas at B, there is a resultant force towards the centre because of the centripetal force/acceleration?

this is what i got, centripetal acceleration acts towards centre so A has greater contact force as force=upthrust - acceleration

(edited 1 year ago)

Reply 51

Manny050

8

Original post by uni12354

As in does A or B have a greater normal force?

I got B as A has a smaller radius of orbit so acceleration is lower meaning centripetal force lower and as weight is same, it must be a decrease in contact force.

I was gonna put that but I wasn’t sure if velocity would also be smaller because of that. I thought they would cancel out

Reply 52

Zakmb

3

Original post by uni12354

so volume of a cylinder is 5pi r^2.

density=mass/volume
so mass =1000*5pir^2

E=P*t=IA*1= I *pir^2
cant remember what I was

deltaT= E/mc = I pir^2/ (1000*5pir^2*c)
so pi r^2 cancels out
giving change in T = I/5000c

Yeh I did that, think I got something x 10^-13

Reply 53

Mo_77377

1

Original post by Zakmb

Yeh I did that, think I got something x 10^-13

That's what I got, that's good

Reply 54

lucaswilson

1

Original post by randommath123

I put A because at the top is directly above the axis of rotation (I.e. there’s no circular motion so not centripetal acceleration) whereas at B, there is a resultant force towards the centre because of the centripetal force/acceleration?

That's what I put, I put that there is no circular motion of A so therefore there is no outward motion of the probe... so the component of weight acts vertically down. Something like that anyway, I just waffled hoping to get marks because I wasnt sure how to word it

Reply 55

tatatatata

6

I put probe B as normal contact force at B equals centripetal force and at A normal contact force=W-centripetal force as centripetal force is the resultant force so F=W-R so B has greater normal force