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Physics: Post your doubts here!

Thought blocker

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Student12 said:
Doubts in waves :Question 25 (B)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_1.pdf
Question 24 (B)http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w05_qp_1.pdf
Click to expand...
I = kA²
k=I/A² (equation 1)
at other position of meter that is equation 2 :
2I=k(x²)
2I= (I/A²) * x²
( 2I * A² ) / I = x²
Taking √ both the sides, x = √ 2 * A

24)
I α a^2 and I α f^2.

Rather than doing all the math to do this, compare the amplitude and frequency of the waves and use the formula to figure out this stuff:

If Q's amplitude is twice as much, the intensity will be four times as much.
If Q's frequency is half that of P, the intensity will be one-fourth.

Net change = 0, so the intensity remains the same.
 
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aneeq247 said:
Nov 13 paper 11 question number 8 please
Click to expand...
Let's do the math behind this - suppose the distance traveled with both speeds is denoted by "s".
Suppose the time taken at 600 kmph = t(1).

Then, since Speed = Distance/Time, we can write

600 = s/t(1)
So that
t(1) = s/600

Similarly, let's do the same for the other speed. The distance traveled is still "s", the speed = 400 kmph, and the time taken to traverse this distance = t(2). So:

400 = s/t(2)
So that
t(2) = s/400

The average speed is the total distance traveled dividing by the time taken (the average velocity is different - it is equal to total displacement divided by time taken).
= (Total Distance traveled)/(Time taken to cover that distance).

Since the plane travels a distance "s" in one direction and returns the same distance "s", the total distance traveled = s+s = 2s.

The time taken = t(1) + t(2) = s/400 + s/600 = 3s/1200 + 2s/1200 = 5s/1200 = s/240

Therefore, the average speed = 2s/(s/240) = 480 kmph = C.
-sagar.
 
Asad Moosvi

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Thought blocker said:
I = kA²
k=I/A² (equation 1)
at other position of meter that is equation 2 :
2I=k(x²)
2I= (I/A²) * x²
( 2I * A² ) / I = x²
Taking √ both the sides, x = √ 2 * A

24)
I α a^2 and I α f^2.

Rather than doing all the math to do this, compare the amplitude and frequency of the waves and use the formula to figure out this stuff:

If Q's amplitude is twice as much, the intensity will be four times as much.
If Q's frequency is half that of P, the intensity will be one-fourth.

Net change = 0, so the intensity remains the same.
Click to expand...


Hey, can you please explain question 22 to me?
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf
 
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Wolfgangs said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_11.pdf

Q5, Q12 and Q18. Please, do help me in these.
Click to expand...
5)
We know S = d / t = 40 / 2.5 = 16 m/s
Uncertainty in speed = (uncertainty in distance + uncertainty in time) x speed
--------"--------------- = ( (0.1/40 ) + (0.05/2.5) ) x 16 = 0.36 ≈ 0.4 (1dp) = C

12)
Consider barrel :
Use F = ma
That is mg - T = ma (T = tension): 120g - T = 120a
Make T the subject of formula so, later we can equate it in other equation or you can solve it by simultaneous equation method also.
But I personally go for Equate so, T = 120g - 120a

Consider stake :
Again use F = ma
so T = 80g + 80a

Equate it to get acceleration :
120g-120a=80g+80a
a=0.5g

Now we use V² - U² = 2as
so V² = 2 x 0.5 (10) x 9
V = 6m/s

18)
Efficiency is given by (Useful Output Work)/(Total Input Energy) =
(Useful Output Work per Second)/(Total Input Energy per Second)
The useful output work per second = useful power output = 150 * 10^3 Joules Per Second.
Therefore, Efficiency = (150 * 10^3 Joules/Second)/(Total Input Energy per second)
Every hour, 20 liters of fuel are consumed. Therefore, every second, 20/(60 * 60) liters of fuel are consumed.
The energy in these liters is equal to [20/(60 * 60)] * 40 * 10^6 Joules.
Therefore, the efficiency is equal to
(150 * 10^3)/[20/(60 * 60)] * 40 * 10^6 = (150 * 10^3 * 60 * 60)/(20 * 40 * 10^6) = B.
 
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