Physics O/N 2012 Paper 32 Question 2b, NEED HELP!

[Fady Soliman]

can anyone help me in solving this question, i have tried it a million times, yet could not make it like wat the marking scheme says, plz!!!

 

[igcsephysicstutor]

choose a starting point. draw the weight vertically downwards, to scale. where the arrow of that force is (i.e. at the bottom of the arrow) start and draw the second, lift force which is pointing upwards and to the left, making an angle of 30-degrees to the vertical (i.e. in the same direction as the rudder on the aeroplane diagram). the resultant force is from the starting point to the end of the lift force, and should be horizontal, going from right to the left. hmm its easier to draw than explain in words! just be aware that there is more than one way to go vector force diagrams, you can do parallelograms or triangles. both work, but sometimes one method is easier than another.

there are detailed instructions on our tutorial video here: http://igcsephysicstutor.com/tutorial-videos/general-vectors-and-resultant-forces.html

hope this helps.

 

[sj0007]

choose a starting point. draw the weight vertically downwards, to scale. where the arrow of that force is (i.e. at the bottom of the arrow) start and draw the second, lift force which is pointing upwards and to the left, making an angle of 30-degrees to the vertical (i.e. in the same direction as the rudder on the aeroplane diagram). the resultant force is from the starting point to the end of the lift force, and should be horizontal, going from right to the left. hmm its easier to draw than explain in words! just be aware that there is more than one way to go vector force diagrams, you can do parallelograms or triangles. both work, but sometimes one method is easier than another.

there are detailed instructions on our tutorial video here: http://igcsephysicstutor.com/tutorial-videos/general-vectors-and-resultant-forces.html

hope this helps.

I would be very grateful if you could help me wid dis plzzzz its really urgent..
http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Physics (0625)/0625_s12_qp_32.pdf
Q11 b
how to solve it?

 

[igcsephysicstutor]

you need to take off the background radiation, which looks to be say 12. then the radiation count starts at (52-12) approx 40.

count the time for it to halve. half would be 20 plus the background which is (20+12) = 32. on the graph, 32 is about 1.9 days.

 

[Fady Soliman]

Well, i know how to go about with the question and draw the vectors, i have did nearly every singl question correct, but for this one, i can't figuire out how to make the resultant force horizontally left, it is always with a bearing of 345 degrees from the weight? also i don't get it when in the marking scheme they say smthing about the cosine rule, wat does it have to do with this question, they didn't ask for calculations!
Can u plz draw it and post a pic, plz! i would be grateful!

 

[Fady Soliman]

oh, when i read your comment again, i find smthing that might help!
did u say that the resultant was from the beginning point to the end of the lift force, why is that? isn't it supposed to be the diagonal of the parallelogram that is formed?

 

[igcsephysicstutor]

draw first force. where first force finishes, draw second force. resultant force is from start of first force to end of secod force. it forms a triangle, not a parallelogram. in this case its a better way to solve the problem graphically.

 

[sj0007]

you need to take off the background radiation, which looks to be say 12. then the radiation count starts at (52-12) approx 40.

count the time for it to halve. half would be 20 plus the background which is (20+12) = 32. on the graph, 32 is about 1.9 days.

THANKU SOOOOO MUCH YOU ARE A LIFE SAVER.....i didn't expect a reply so soon.