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Physics: Post your doubts here!

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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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Anyone?
 

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For those of you who couldn't solve this question :
Solution;
P=10^5-p(average)gh [ Basically you take the average density]

Question 15 Candidates found this question difficult. Atmospheric pressure is caused by the weight of air above the point, so the correct method is to calculate the pressure caused by the part of the atmosphere between sea level and 5000m, and to subtract this from the pressure at sea level. Candidates answering B or C had both calculated the pressure difference correctly using the average density; those choosing C gave the correct answer because they had subtracted the pressure difference from 100000Pa.
 
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Ammeter reading = 0
P.D across junction = 0
P.D(R4) = P.D(R2 )
R4/(R3 + R4) * V = R2 /( R2 + R1 ) *V
(R2 + R1)* R4 = R2 *(R3 + R4)
R2R4 +R4R1 = R2R3 + R4R2
R4R1 = R2R3

AND = D
oh finally got the answer
Thanks a lot. :)
 
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E = (F/A) /( extension /length)
--> Stress : F/A
F = W = mg ( g is constant)
m = density * volume (density is constant as same material )
V = l * l* l ( The Load is a cube )
[The model is one tenth full-size in all linear dimensions.]
V = 1/10*1/10*1/10
V = 1/1000
When V decreases mass decreases and hence Force decreases
F = 1/1000
Area = A = l*l
= 1/10 * 1/10
= 1/100
So Stress = F/A = (1/1000)/(1/100) ---->1/10

----> Strain : e/l
Suppose extension = 1
length = 1/10
Strain = 10

E = stress/ strain = 0.1 / (10)
=> 1/100

Ration of original to model = 1/(1/100)
= 10^2 ANSWER = C

Great
no one could make me understand this question, but now I Got it.
Thanks again
 
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