Physics 9702 P21

[randomcod]

Answer was 27w.
Let me explain why using Kirchoff's first and second law.
First, (conservation of charge):
6 Amps flowed around the circuit. This implies, because it is parallel, each lamp would get 3A. Using I^2 x R for Power, it would be 3^2 x 3 = 27
Now with 2nd law (conservation of energy)
Voltage produced by battery : 12 V. V = E - Ir E = 12 I = 6 and r = 0.5. 12-6x0.5= 9
Using the equation V^2/R, it is 9^2 / 3 = 27w.
There you go ^^

 

[randomcod]

For the p.d. and e.m.f. I defined them in the other way, which I don't if it'll score me marks.
Ie pd= total energy given up by a unit charge as it moves from Point A to point B in a circuit
And emf = total work done by a unit charge as it goes around a complete circuit or loop.
Do you guys think this'll score me marks?

 

[randomcod]

For the radioactivity question. For the betas decay, I mentioned that the nucleus gains one proton, but didn't mention that it loses neutron. Is it okay?
And do we consider a beta particle to have a proton number of -1? (General question XP}

It is correct. It asked for nucleon and proton number. The nucleon number remained the same as the neutron 'lost' is compensated by the gained proton. Your answer is absolutely fine, or so I believe.

 

[waleed302]

Guys I think it was 7 waves because there were 6 crests and half a wavelength on either side.

 

[Ahmedlool]

Would i get marks if i said that the total emf in a closed loop is equal to the total potential difference?

 

[Wolfgangs]

Guys for gamma ray, I mentioned that it is a form of energy released during a nuclear reaction. It's wrong, right?

And for the question about why does not "W" affect Fb, I wrote that it is in equilibrium, which means there is no resultant force and so it is not affected. Is that somehow right?