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Kindly solve Jun 2011 IGCSE Physics (0625) Paper 31's Q2 (b) and (c).
A detailed answer is required. Thanks in advance.
A detailed answer is required. Thanks in advance.
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Physics Help needed urgently
- Thread starter farzansaeed07
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will try my best to help u bt if u plz attach the paper
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2.b) Power=W/t
therefore W=Pt
=(528)(5)
=2640J
If Work=(F)(displacement)
2640= (800)(0.3x) remember total displacement is 0.3m between each rung multiplied by the number of rungs he climbs
3.3=0.3x
11=x
c) Efficiency=Useful energy transferred/Total energy supplied N.B. useful energy= (800)(20)(0.3)=4800J
15%= 4800/Total energy supplied
Total energy supplied=(4800)/(15%)
=32000J
Hope this helps.
therefore W=Pt
=(528)(5)
=2640J
If Work=(F)(displacement)
2640= (800)(0.3x) remember total displacement is 0.3m between each rung multiplied by the number of rungs he climbs
3.3=0.3x
11=x
c) Efficiency=Useful energy transferred/Total energy supplied N.B. useful energy= (800)(20)(0.3)=4800J
15%= 4800/Total energy supplied
Total energy supplied=(4800)/(15%)
=32000J
Hope this helps.
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i would answer it, but the guy above me did. i got these two right on the actual 2011 exam!