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physics 9702 MULTICHOICE HELP URGENT PLEASE HELP

J

Jaspreet Dhaliwal

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Hi guys can u please help? :)

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_13.pdf QUESTIONS 10, 28
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf QUESTIONS 18,23,24,29
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_12.pdf QUESTIOn 26,33
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_13.pdf QUESTION 18,

Sorry for all questions but im not the smartest at physics ,rregret taking it this year :/ Please help ASAP, it will also be help for you guys too! :D
 
DumbIdiot

DumbIdiot

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s12 p13 no10
using f=ma
T-800=80a
1200-T=120a
solve and get a=2ms^-2
the man moves 9m while the barrel moves 9m (total =18m) for them to attain the same height
using V^2 = U^2 +2as (u=0 , a=2 , s =9) v---->6ms^-1

s12 p13 no28
using dsinx = n.wavelength
d = 1x10^-6 , x = 35 and n=1
wavelength = 574nm
 
DumbIdiot

DumbIdiot

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s12 p 12

n018
power input = rate of loss of ep of water ( mgh/t) =9.81 x 8 x 200
useful power output = 230 x32
power output/power input x 100 = 47%

n023
pressure due to gas = atm pressure + pressure due to liquid ( vertical height must be used)
=1x10^5 + (9.81 x 5 x 1000)
=1.5x10^5 pa

n024
workdone = area under graph
count the number of small squares, n under the graph from extension = 0 to extension 7mm
1 square = (0.5 x 10^-3 x 1)J
total wd = 0.5 x 10^-3 x 1 x n


n029
XY to p = 3 wavelength
v=d/t
v=c and d = 3 wavelength
find t
 
J

Jaspreet Dhaliwal

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DumbIdiot said:
s12 p 12

n018
power input = rate of loss of ep of water ( mgh/t) =9.81 x 8 x 200
useful power output = 230 x32
power output/power input x 100 = 47%

n023
pressure due to gas = atm pressure + pressure due to liquid ( vertical height must be used)
=1x10^5 + (9.81 x 5 x 1000)
=1.5x10^5 pa

n024
workdone = area under graph
count the number of small squares, n under the graph from extension = 0 to extension 7mm
1 square = (0.5 x 10^-3 x 1)J
total wd = 0.5 x 10^-3 x 1 x n


n029
XY to p = 3 wavelength
v=d/t
v=c and d = 3 wavelength
find t
Click to expand...
Hey thank u very much :D could u help with the other two q papers
 
O

oajisthebest

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W10/12

Question 33

We know that V=IR so I=V/R
(A) so if V=12 and R=2+10=12
THEN I=12/12=1 AMP
(B) if V=12 and R=2+8=10
THEN I=12/10=1.2 AMP
(C) if V=12 and R=2+6=8
THEN I=12/8=1.5 AMP
(D) if V=12 and R=2+4=6
THEN I=12/6=2 which is not equal to 1.8 option D is wrong


FEEL FREE TO ASK ANYTHING !
 
DumbIdiot

DumbIdiot

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w10 p12 no26

distance between two minima = 0.5wavelength
wavelength = 15 x2 = 30mm
speed of microwave = 3.0 x10^8
v=f x wavelength
f = (3.0 x 10^8)/(30 x 10^-3)
=10GHZ
 
J

Jaspreet Dhaliwal

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how
DumbIdiot said:
s12 p 12

n018
power input = rate of loss of ep of water ( mgh/t) =9.81 x 8 x 200
useful power output = 230 x32
power output/power input x 100 = 47%

n023
pressure due to gas = atm pressure + pressure due to liquid ( vertical height must be used)
=1x10^5 + (9.81 x 5 x 1000)
=1.5x10^5 pa

n024
workdone = area under graph
count the number of small squares, n under the graph from extension = 0 to extension 7mm
1 square = (0.5 x 10^-3 x 1)J
total wd = 0.5 x 10^-3 x 1 x n


n029
XY to p = 3 wavelength
v=d/t
v=c and d = 3 wavelength
find t
Click to expand...
how do u know the wavelength is 3 though?
 
J

Jaspreet Dhaliwal

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oajisthebest said:
S10/13

QUESTION 18

P=FV

SO (Total Force) x (Velocity)
(1000N+10000N) x (0.50 ms^-1)
P=5500W=5.5KW

Feel free to ask anything! :)
Click to expand...



hey mate
can u help me please my exam is tommorow and if i get good in this im hoping to get an A in physics?

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_12.pdf
Questions 14, 17,20,29,34

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_13.pdf
Questions 11,16,25,35

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_11.pdf
Question 13,16,25,32,35

i know there is about 14 questions there, but i hope you can help me please :)
 
O

oajisthebest

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W12/Paper 12

QUESTION 14

FIRST OF ALL,BOTH NEWTON METERS ARE SHOWING TENSION IN THE STRINGS AND WE HAVE TO FIND THE VERTICAL COMPONENTS OF BOTH THE FORCES.

VERTICAL COMPONENT OF 4 NEWTONS FORCE IS: COS(37) = X/4 , SO X = 4COS(37) = 3.19N
VERTICAL COMPONENT OF 3 NEWTONS FORCE IS: COS(53) = X/3 , SO X = 3COS(53) = 1.80N

NOW WE NEED TO FIND THE RESULTANT FORCE: FORWARD FORCE(WEIGHT) - BACKWARD FORCE
12-(3.19+1.80) = 7N
NOW USE F=MA TO FIND ACCELERATION

7=1.2 x A

A = 7/1.2
A = 5.83 ms^-2 = 6 ms^-2
 
M

moinul

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plz help me with this question :
what is the average KE of an athlete running at maximum speed during a 100 m race?
the answer is 4000.
 
O

oajisthebest

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moinul said:
plz help me with this question :
what is the average KE of an athlete running at maximum speed during a 100 m race?
the answer is 4000.
Click to expand...


WE KNOW THAT KE=1/2 MV^2

Average mass of an adult is about 65 kg
Average velocity of an athlete is about 12 ms^-1

SO KE=1/2 MV^2
KE= (1/2) x (60) x (12^2)
KE=4320 J WHICH IS APPROXIMATELY 4000 J
 
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