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Physics: Post your doubts here!

selrey

selrey

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How do i do this question?
The density of air on the Earth decreases almost linearly with height from 1.22 kg m –3 at sea level
to 0.74 kg m –3 at an altitude of 5000 m.
Atmospheric pressure at the Earth’s surface on a particular day is 100 000 Pa. The value of g
between the Earth’s surface and an altitude of 5000 m can be considered to have a constant
value of 9.7 m s –2 .
What will be the atmospheric pressure at an altitude of 5000 m?
OPTIONS:
36000 Pa
48000 Pa
54000 Pa
59000 Pa
 
amina1300

amina1300

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selrey said:
How do i do this question?
The density of air on the Earth decreases almost linearly with height from 1.22 kg m –3 at sea level
to 0.74 kg m –3 at an altitude of 5000 m.
Atmospheric pressure at the Earth’s surface on a particular day is 100 000 Pa. The value of g
between the Earth’s surface and an altitude of 5000 m can be considered to have a constant
value of 9.7 m s –2 .
What will be the atmospheric pressure at an altitude of 5000 m?
OPTIONS:
36000 Pa
48000 Pa
54000 Pa
59000 Pa
Click to expand...
Holmes said:
View attachment 62388
M/J 2016 (13)
Explain please
Click to expand...
Im trying but all I get is 60655...Pa <_> SOMEONE SOLVE THIS QUESTION!!
 
G

GarryTheGhost

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amina1300 said:
http://papers.gceguide.com/A Levels/Physics (9702)/9702_w16_qp_12.pdf
Q4 , 10 26 ,29
Click to expand...

amina1300 said:
http://papers.gceguide.com/A Levels/Physics (9702)/9702_w16_qp_12.pdf
Q4 , 10 26 ,29
Click to expand...
question 4:
between the 2 pulses there is a difference of 3 boxes. since the time interval between the pulses is 0.006 seconds, we say that 3boxs=0.006 seconds
so 1 box=2*10^-3seconds hence answer is C
question 10:
not sure if i'm correct but here it goes.....
v^2=u^2+2as
since the vehicle decelerates at the same rate at both scenarios we can say that a=-(u^2/x) where x is the distance between red and yellow and minus because its decelerating.
if speed is increased by 20% then the new initial speed will be 1.2u
then i'll make the s the subject where s is the minimum distance required
s= (v^2-u^2)/2a
s=(0-1.44u^2)/2(u^2/x)
s=1.44x is the answer so C
question 26
ummmmmm. please explain this to me.
the timeperiod of the wave is T=1/250 which is 4*10^-3. which means that after a whole 4ms point p will go back to 1mm the questions asks us about 5ms. so basically 1ms after the full cycle of p. we can use the formula
phase difference= (change in time)/timeperiod * 360
so it gives us 90 degrees. (1*10^-3/4*10^-3)*360. so i'd guess for B. so 0mm. if you manged to tell me why because i forgot how a 90 degrees phase difference looks like hook me up yea.
question 29
i just found out that wave length= twice the length between 2 maxima. so wavelength= 1.5cm lets make that into meters shall we.
so frequency= speed/wavelengh
and so therefore frequency= 3*10^8/3*10^-2
so answer is 1*10^10 so C
 
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