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Physics: Post your doubts here!

S

Snowysangel

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Hadi Murtaza said:
24.
Intensity ∝ (Amplitude)²
Intensity ∝ (Frequency)²
Intensity = (Amplitude)² × (Frequency)²

Amplitude of wave P = x₀
Frequncy of wave P = F
So intensity of wave P = x₀² * F² = (Fx₀)²

Amplitude of wave Q = 2x₀
Frequency of wave Q = F/2 , since only half a wave passes at time t₀
So intensity of wave Q = (2x₀)² * (F/2)² = 4x₀² × F²/4 = F²x₀² = (Fx₀)²

Intensity of both is (Fx₀)² so intensity of Q will also be I₀
Click to expand...
How do we know that amplitude and frequency are directly proportional? I means re we supposed to know that?
 
Hadi Murtaza

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mehria

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sagar65265 said:
Q15)
View attachment 44836

Since the ball is balanced and in equilibrium, we can say that the net force on the ball is equal to zero. We can also split this net force into two components, one in the up-down (vertical) direction and another in the left-right (horizontal) direction. Let's also say that the upwards direction is positive and the leftward direction is negative.

In the vertical direction:

i) T has a component of T * cos(30) = (√3/2) * T = √3T/2 = (+√3T/2) Newtons
ii) W has a component of 0.15 Newtons downwards, so it has a component of (-0.15) Newtons.
iii)The wind force is horizontal, so it has no component.

Summing these up should give us zero, so

(+√3T/2) + (-0.15) = 0
√3T/2 = 0.15
√3T = 0.3

T = 0.1732 Newtons

That is the magnitude of the Tension force. Now let's do the horizontal direction:

i) T has a component of T * sin(30) in the rightward direction = T * (1/2) = T/2. Since this is to the right, the sign is negative, i.e. (-T/2) Newtons.
ii) The weight has no horizontal component in this situation.
iii) The Wind force has a component F(air) to the left, and only the left. Therefore, it has a component of (+F(air)) Newtons.

Summing these up should give zero, so

(+F(air)) + (-T/2) = 0
F(air) = T/2 = 0.1732/2 = 0.866 Newtons = 0.87 Newtons = B.

23)

When the force W is applied on R, the spring R gets the full load of the force. In other words, the force extending R is equal to the force W. Therefore, the extension of spring R is equal to (using the formula |F| = kx):

x = |F|/k = W/k = W/3k (since the spring constant of R is 3k).

Now, the spring R will exert the same force at both ends - it will exert a force W on the lower end, and it will exert a force W on the upper end, i.e. on the bar.
For the bar to be in equilibrium, the force on it have to balance out, so

(Force from spring P) + (Force from spring Q) = (Force from spring R) (The bar has negligible weight)

For moments to be zero about the center of mass, the (Force from spring P) has to be equal to the (Force from spring Q). Therefore,

2(Force from spring P) = 2(Force from spring Q) = W

Therefore, the (Force from spring P) = W/2 = (Force from spring Q).

The extension of spring P if it has to exert a force of W/2 is equal to

x = |F|/k = (W/2)/k = W/2k

The extension of spring Q is the same, so that the system is stable and balanced.

Therefore, the overall extension is

W/2k + W/3k = 3W/6k + 2W/6k = 5W/6k meters = A.
If you still have a doubt concerning either of these questions, just post it on the forums and i'll see if I can answer.

Hope this helped!
Good Luck for all your exams!
Click to expand...
thank u ^_^
 
Thought blocker

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Ahmed H. Al-Neel said:
questions 10,17,33,35,36 please!
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_13.pdf
Click to expand...
sadiaali said:
Please q16 How is D is correct when the sum of forces anti clockwise & clockwise are not equal. Oh i dont understand please anyone explain me.
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_13.pdf
Click to expand...
kitkat <3 :p said:
please tag me if u get the answers :eek:ops;
Click to expand...
I had already uploaded this paper . _ .
w12_qp_13
35)
10425805_770422926302664_410683135_n.jpg

Here where ever you see the junctions, the current divides, Try all the options and check on which two of the ammeter has same currents passing.
So in D option, only one loop is formed so every where in the circuit the current would be same. Hence D.
 
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hamzashariq

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Ahmed H. Al-Neel

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Thought blocker said:
I had already uploaded this paper . _ .
w12_qp_13
35)
10425805_770422926302664_410683135_n.jpg

Here where ever you see the junctions, the current divides, Try all the options and check on which two of the ammeter has same currents passing.
So in D option, only one loop is formed so every where in the circuit the current would be same. Hence D.
Click to expand...
i didnt really understand .. why not B for example? B has only one loop
 
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