Qualification > Sciences

P.H.Y.S.I.C.S P2 D.O.U.B.T.S CIE

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poisonedrose:

--- Quote from: thecandydoll on June 06, 2010, 12:33:43 pm ---Anyone solve this for me  ? :D
M/J 2007-----Physics Q4d (i) and (ii)
And M/J/2007----PHYSICS q1 the whole thing!
I dont get it =(
thank youuuu!

--- End quote ---
Using an appropriate scale, draw a horizontal line representing horizontal velocity of 18m/s.I used 2cm (10 boxes)=5 m/s. And it is a straight, horizontal line because horizontal velocity does not change,.

then calculate vertical velocity at point of impact with ground by using the equations of motion with the formula v2=u2+2as. U being 0. a being 9.8 and s being 16. The vertical velocity gottn is approximately 17.7m/s which we can draw using our scale, as a vertical straight line.

We join the two lines- making a triangle and we can find the resultant velocity by the using Pythagoras theorem .. (which would be the hypotenuse) a2+b2=c2.

Once you are drawn with the vector diagram, measure the asked angle with a protractor or with trigonometry

zxcvbnm:
may 06-question 5, part (c)

sizbeauty:
m/june02
Q4(b)(part 1)
oct/nov02
Q5(b)(part1) how do u calculate the phase difference in this one  ??? ??? ??? ???
plz help me!!!!!!

nid404:

--- Quote from: zxcvbnm on June 06, 2010, 06:07:24 pm ---may 06-question 5, part (c)

--- End quote ---

check the attachment.

nid404:

--- Quote from: sizbeauty on June 06, 2010, 07:16:59 pm ---m/june02
Q4(b)(part 1)

--- End quote ---


a) v2=u2+2as
v2=2X9.8X1.6
v=5.6m/s

b) p.e lost=k.e gained

p.e lost=mgh=73/1000 X 9.8 X 1.6=1.14
k.e gained=90% of p.e lost= 1.03J

1.03=mgh
1.03=73/1000 X 9.8 X h
h=1.44 m

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