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Physics paper .....

http://filestore.aqa.org.uk/subjects/AQA-PHYA1-QP-JUN15.PDF

help with question

1b ii
4c and d
6b part ii and iii
6c
7c

bonus i remembered from the multiple choice which i think i worked out but is wrong....

Lift with power rating of 100w is used to lift up a crate of 10kg at 5m/s
calculate efficiency of process(it was something like that)

i did this

efficiency=usefultotalefficiency = \dfrac {useful}{total}


efficiency=12mv2100efficiency= \dfrac{\dfrac{1}{2} mv^2}{100}


i got 12%
answers were 8%
12%
50%
100%
(edited 7 years ago)
Original post by thefatone


bonus i remembered from the multiple choice which i think i worked out but is wrong....

Lift with power rating of 100w is used to lift up a crate of 10kg at 5m/s
calculate efficiency of process(it was something like that)

i did this

efficiency=usefultotalefficiency = \dfrac {useful}{total}


efficiency=12mv2100efficiency= \dfrac{\dfrac{1}{2} mv^2}{100}


i got 12%
answers were 8%
12%
50%
100%


you're trying to get efficiency by dividing an amount of kinetic energy by an amount of power... tbh alarm bells should be ringing, you should be dividing quantities with the same dimensions to get a dimensionless number

the question is about the rate at which work is being done against gravity... power is the rate at which work is done.

in one second the crate moves 5m vertically upwards...
the useful power output is mgΔh divided by the time taken (1 second)

that gets us a useful power output of 500 (or 490.5) W depending on the value you use for g

probably the motor in the question was rated at 1kW, not 100 W (because it doesn't make sense to have >100% efficiency) and that would get you to about 50% efficiency.
Reply 2
Original post by Joinedup
you're trying to get efficiency by dividing an amount of kinetic energy by an amount of power... tbh alarm bells should be ringing, you should be dividing quantities with the same dimensions to get a dimensionless number

the question is about the rate at which work is being done against gravity... power is the rate at which work is done.

in one second the crate moves 5m vertically upwards...
the useful power output is mgΔh divided by the time taken (1 second)

that gets us a useful power output of 500 (or 490.5) W depending on the value you use for g

probably the motor in the question was rated at 1kW, not 100 W (because it doesn't make sense to have >100% efficiency) and that would get you to about 50% efficiency.


Looks like the user you are replying to is banned.
:redface:
Original post by M14B
Looks like the user you are replying to is banned.
:redface:


I will be back 1 day after my C1 exam which is on the 18/5/16 :smile:

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