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Physics: Post your doubts here!

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ty. and this is the whole question basically. A cyclist is travelling at constant speed so why will his chemical energy not convert to kinetic energy. I know that the energy change will be kinetic to heat and sound energy but why will the chemical energy not be converted to K.e thts the question.
ok but do u have the answer or ur asking cause u NEED an answer?
 
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ok then
for the first row the power dissipated is obviously 0 because the circuit is incomplete so no electricty will flow
2nd row current will always go through the route with less resistance so power is 1.5kW as current "avoids" B
3rd current will only flow through A + C so total is 3.0kW
4th p= V^2/r since the resistors are in series voltage is shared equally since they have same electrical requirements so V is halfed while R remains constant so new voltage is (0.5v)^2 which is 0.25v^2 so P= 2*(0.25*240^2)/38.4 =750
5th 0.75kW + 1.5kW =2.25kW

hope it helped : D

can you please explain the highlighted part?
 
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well an antinode is 1/4 of a wavelengh right? so given that l=45cm the antinode is 45 cm and wavelengh is 0.45x4=1.8 m aand u calculate the frequency from v=fxwavelengh

isn't the wavelength 60cm as calculated before? :S
 
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How is P=Fv derived?
A question from pastpaper:Explain why a cyclist travelling at a constant speed is not making this trnasformation. Explain what transformations are taking place?

p= E/t e= work done W=f x s

p= (f x s)/t v=s/t

therefore p= fv
 
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can you please explain the highlighted part?

ok from the diagram you can see that A and B are in series right? adn the switch S2 is closed and in row 4 they ask you for total power across them which is 0.75W

and C is in parallel with A and B so 0.75 + 1.5

I am treating A and B as 1 resisitor
 
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because there is no equation of motion which applies to what you have

you don't have v or t and all equations require either v or t or both

okay thanks...
can these equations be applied in vacuum? places without air resistance and gravity?
 
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If the current of the bulb increases will it short. When the current will inc the resistance should decrease and it should not produce much heat energy so why does it short?
 
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