GCE 5054 PHYSICS HELP!

[Mahnoorfatima]

Please somebody explain these qs to me!! Im very worried!

qp 12 s10 -
Q no 7. finding the mass of trapdoor one.
Q no. 9. Why is it 18cm when the height risen is 9cm????
Qno 13. shouldnt it be like potential energy to kinetic eneggy as the paritutist is first at a height and THEN he falls at a const speed .. which is kinetic energy????
Please help me!
May Allah grant us all with A*s?! Ameeen! SUm Ameen!

 

[Mahnoorfatima]

ANd the same paper of theory Q no 2 part bi WHICH EFFING FORMULA DO I USE! I ve used like every single formula but the answer is worongg!!!!!!!!!!!!!!!

 

[Mahnoorfatima]

ANd the same paper of theory Q no 2 part bi WHICH EFFING FORMULA DO I USE! I ve used like every single formula but the answer is worongg!!!!!!!!!!!!!!!

 

[Mahnoorfatima]

ANYONE? :/

 

[Izfa Amir]

Paper 12
Q7. You have to use the principle of moments. M= Force (f) x Perpendicular Dist between force and pivot (s). First find the clockwise moment which is 15N x 60cm. so clockwise moment = 900Ncm. Since clockwise moment= aniticlockwise moment you can use that to find the weight of the trapdoor. S0 F=900/30 (cause the distance between line of weight and pivot is 30). So F=30N. Use W=mg to find mass. Mass is 3kg.

Q9. It is not the rise in height you are supposed to be looking for. You have to look at the difference in liquid level. Since the total distance between the 2 levels is 18 the atmospheric pressure is 18cm of liquid above atmospheric pressure

Q13. Because speed is not increasing KE will not increase. (KE=1/2 m v^2 , remember?) But as he falls his height above earth is decreasing which means that he is losing Potential Energy. So the PE will be converted to heat energy.

And the tension in the cable will be the weight of the person and the lift. So W=mg. W=580 x 10.

I hope this helps