Physics May/June 2010 Paper 1 Query

[Chingoo]

I thought I'd ask since I'm probably not repeating, but I can't study A2 if AS is somewhat obscure. xD

In the recent exam, could someone explain to me how the following questions were solved? Exam paper and mark scheme in the attachment.

QUESTIONS 15, 16, 23, 28, 34.

Yes, I have no idea why they gave me an A except that God willed it. 

[S.M.A.T]

15)energy=force*distance
energy input=F*s
energy output=mgh=mgssin(a)

efficiency=energy output/energy input=mgssin(a)/Fs=mgsin(a)/F

16)Power=force*velocity=(1000+(1000*10))*0.5=11000*0.5=5500W

[S.M.A.T]

23)For option A

speed=frequency*wavelength
since there is no change in speed so increasing the frequency will decrease the wavelength.
Diffraction is smaller when the the wavelength decreases
So it will show a pattern close to diagram 2.

[S.M.A.T]

28)force=Electric field strength*charge
Work done=Electric field strength*charge*distance moved in the direction of the electric field lines
At every instant the direction of distance(the tangent of circle at every instant) is perpendicular to the direction of the electric field lines.So the component of distance  in the direction of electric field lines is zero.


34)the resistance of parallel connection is half of resistance of P because resistance of P,Q,R are equal.So voltage(V=IR) across parallel connection is also half of P.So power(Power=IV) dissipated in the parallel connection is half of P.That means the power across P is 8W and across parallel connection is 4W.So the power of one resistor(P or Q) in the parallel connection is 2W.

[Chingoo]

Thanks a lot! +Rep