-
We need your support!
We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
Click here to Donate Now (View Announcement)
Phyics m/j 2009 inquiry PLEASE HELP :(
- Thread starter boodi196
- Start date
First of all, don't you have the mark scheme?
Rays from his hat and his feet make for 2 isoceles triangles, which can also be considered sets of identical, but mirrored right angles, both with angles of incidence/reflection a for the ray from his hat (or b from his feet). It doesn't matter how far the mirror is (the only things that change would be length/base/height of right-angled triangle and the value of angle a/b -- neither of which matter), you will get the height, h, of mirror equal to half the height of the man. See amended diagram, Fig.5.1.
Rays from his hat and his feet make for 2 isoceles triangles, which can also be considered sets of identical, but mirrored right angles, both with angles of incidence/reflection a for the ray from his hat (or b from his feet). It doesn't matter how far the mirror is (the only things that change would be length/base/height of right-angled triangle and the value of angle a/b -- neither of which matter), you will get the height, h, of mirror equal to half the height of the man. See amended diagram, Fig.5.1.
- Messages
- 5,874
- Reaction score
- 4,242
- Points
- 323
What is the paper code and number????
Not the correct question? I was just guessing at the paper you meant... So I picked the hardest question. 
Perhaps you meant the windmill question? Here goes:
5(b)i : wind speed is halved, therefore everything else being unchanged,
the mass = original mass / 2 = 7500 / 2 = 3750kg
5(b)ii : fraction of power output on second day to first day is
[1/2 x m/2 x (v/2)2] / (1/2 x m x v2)
= 1/8
Perhaps you meant the windmill question? Here goes:
5(b)i : wind speed is halved, therefore everything else being unchanged,
the mass = original mass / 2 = 7500 / 2 = 3750kg
5(b)ii : fraction of power output on second day to first day is
[1/2 x m/2 x (v/2)2] / (1/2 x m x v2)
= 1/8