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ALL CIE PHYSICS DOUBTS HERE !!!

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$!$RatJumper$!$:
Thankx for all ur help +rep

umm and could you also explain why 3(c) direction is 35 below the horizontal. isnt it just 90 degrees?

$!$RatJumper$!$:

--- Quote from: astarmathsandphysics on November 08, 2010, 09:22:18 am ---Q4 (a)(ii)- why doesnt it work when I find the area under of the triangle for the first 0.3 sec then times it by 2

4aii) It should work the way you did it. I get about 2*1/2*0.3*0.10 =0.03 by APPROXIMATING THE slope by a line


--- End quote ---

why did u use 0.1

TJ-56:

--- Quote from: $!$RatJumper$!$ on November 08, 2010, 12:32:24 pm ---Thankx for all ur help +rep

umm and could you also explain why 3(c) direction is 35 below the horizontal. isnt it just 90 degrees?

--- End quote ---
I think you just need to state that the direction is horizontal, but its not 90 degrees, its 90 degrees when its in the center (i.e. the weight acts through the pivot), but no need since they dont want the angle

$!$RatJumper$!$:
ok.. and can u explain 4(a)(ii) .. why should we use 0.1

TJ-56:

--- Quote from: $!$RatJumper$!$ on November 08, 2010, 01:04:05 pm ---ok.. and can u explain 4(a)(ii) .. why should we use 0.1

--- End quote ---
Ok i didnt understand the calculation that much, but heres what i did, ur way, calc till 0.3 seconds then multiplied the value by 2:
Divided the area into 2 triangles 2 squares and 1 trapezium (thats alot but its the only way it matched the ms value): so respectively:
triangles : 0.1*4*0.5 + 0.1*2.9*0.5
squares: .2*4 + 0.1*2
trapezium: 0.1 * (0.9+2 / 2)
Multiply the sum by 2, then by 10^-2 (in metres) gives us 0.0298 which is 0.030 rounded to 2 sf

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