Physics: Post your doubts here!

taffycandy

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charges 5.0*10^-7 C,-2.5*10^-7C and 1.0*10^-7 C respectively are held fixed at three corners A,B,C of an equilateral triangle of side 5.0 cm .find the electric force on the charge at C due to the rest two.
 
sudeep1097

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Faaiz Haque said:
Physics questions - http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_1.pdf

9,10,13,14,18,19,20,22,24,25,27,29,32,33,35,37,39
Click to expand...
9 (B) change in momentum = m v1 -m v2 = 0.1 x 20 - 0.1 x 9 (-30) (since direction opposite velocity sign changes into -ve for 30 m/s)= 5
10 (C) momentum is conserved. for x momentum (p) = m x 2v. for y momentum (p) = 2m x v. I supposed v so p is equal for both.
now K.E = 1/2 x m x v^2. Finally ratio would be 2/1. Note: For K.E of X use v as 2v and m as m and for K.E of Y use v as v and m as 2m.
14 (C) power = force x velocity. Substitute F = P/V to get C
18 (D) pressure = hdg. But here h is 2h since on one side liquid rises by h and another side liquid falls by h.
22 (B) diffraction phenomenon is described as the apparent bending of waves around small obstacles and the spreading out of waves past small openings
27 (A) electric field is directed from positive to negative between charged plates and since that is from q to p. Q is positive.
29 (D) Workdone = force(F) x displacement(x) hence, x = W/f
Here W = K.E of electron = 1/2 x m x v^2 and F = qE = eE where e is cahrge of electron and E is electric field strength.
32 (A) emf measures the electrical energy gained by each coulomb of charge that passes through the power supply
33(B)
emf is 3 - 1.2 = 1.8 V
total resistance = 1/R = (1/9)+(1/18) hence R = 6 ohm
we have V = IR i.e. 1.8 = I x 6 hence Total I = 0.3 A But in parallel connection of resistors I is divided.
Also I1 ∝ 1/R1 and I2 ∝ 1/R2 where R1 = 9 ohm and R2 = 18 ohm
Dividing both equations we can remove ∝ sign so
I1/I2 = 18/9 hence I2 that is current entering 18 ohm = I1/2
By kirchoff's first law I1+I2 = 0.3A(total current) i.e I1 + (I1/2) = 0.3A
Hence I1 that is the required answer is 0.2A

35(D) In semiconductors such as LDR and thermistor as loght intensity or temperature increases resistance decrease so when both are increased
resistance in variable resistors has decreased so how does this affect the p.d. across fixed resistor is explained by kirchoffs second law.
sum of potential difference across resistor in series is equal to the emf. Hence when resistance decreases p.d. across variable resistor decreases as V∝R so to compensate for this decrease p.d. across fixed resistor increases.

37 (B) count rate means amount of nuclei decaying. If this is random it implies random nature of decay. Spontaneous means that radioactivity is not affected by external factors such as pressure etc.

39 (C) Beta particle emission means electron emission. But this electron comes from the nucleus as one neutron breaks into one proton and electron and electron is emitted. Beta particle emission thus results in increase of proton by 1 and decrease of neutron by 1 but nucleon number is unaffected hence proton number must have been 19 and nucleon number same before the emission.
 
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sudeep1097

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usama321 said:
The vertical components of both H and T are cancelled by the upwards vertical component of T. Furthermore, T balances out the horizontal component of H too. Thus T has to be the largest of the three. C is the only option.
Click to expand...
You mean vertical component of H and W are cancelled by upwards vertical componenet of T
 
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Faaiz Haque

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sudeep1097 said:
9 (B) change in momentum = m v1 -m v2 = 0.1 x 20 - 0.1 x 9 (-30) (since direction opposite velocity sign changes into -ve for 30 m/s)= 5
Click to expand...
Yo sundeep thanks so much, understood everything except Q14, q29(why is it 2Ee and not Ee)
For q33 .. i get how you got 0.3 A but dont understand next part, can you explain again? or thorougly

Someone help with this please ^
 
sudeep1097

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Faaiz Haque said:
Yo sundeep thanks so much, understood everything except Q14, q29(why is it 2Ee and not Ee)
For q33 .. i get how you got 0.3 A but dont understand next part, can you explain again? or thorougly

Someone help with this please ^
Click to expand...
In 14 we know that power = force x velocity. Change the question so you get the equation with k on one side and other quantities on other side now substitute force = power/velocity in the new equation to get the answer clear??

In 29 K.E = 1/2 x m x v^2 and Force = eE. Substitute these in the equation displacement = workdone/force. You get x = ((mv^2)/2)/eE
Hence, x = (mv^2)/2eE . The 2 comes from the equation for K.E

In 33 we can divide the equations with ∝ to remove it.
Eg: Suppose you have I1 ∝ S1 and I2 ∝ S2 then I1=k(constant) x S1 and I2 = k(constant) x S2. Now divide these two new equations you can cancel out k(constant) as they are same in both. Then looking back at 33 i just skipped supposing a constant and directly divided first equation by the second one and note I is inversely proportional with R so take care while dividing and obtain I2 in terms of I1 and use kirchoff's first law. That is sum of current entering a junction is equal to the sum of currents leaving it. I hope it's clear and by the way my name is Sudeep not sundeep :)
 
snowbrood

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sudeep1097 said:
Hello! Can anyone make it clear to me why ultrasound is pulsed and not continuous in ultrasound scan??
Click to expand...
well it can be dangerous for a human being pulsed gives short bursts of ultrasound energy this is relatively safer than a continuous.
u can touch a hot metal for short time intervals but if i ask u to hold that metal for 10min u can imagine what would happen
 
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Faaiz Haque

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sudeep1097 said:
In 14 we know that power = force x velocity. Change the question so you get the equation with k on one side and other quantities on other side now substitute force = power/velocity in the new equation to get the answer clear??

In 29 K.E = 1/2 x m x v^2 and Force = eE. Substitute these in the equation displacement = workdone/force. You get x = ((mv^2)/2)/eE
Hence, x = (mv^2)/2eE . The 2 comes from the equation for K.E

In 33 we can divide the equations with ∝ to remove it.
Eg: Suppose you have I1 ∝ S1 and I2 ∝ S2 then I1=k(constant) x S1 and I2 = k(constant) x S2. Now divide these two new equations you can cancel out k(constant) as they are same in both. Then looking back at 33 i just skipped supposing a constant and directly divided first equation by the second one and note I is inversely proportional with R so take care while dividing and obtain I2 in terms of I1 and use kirchoff's first law. That is sum of current entering a junction is equal to the sum of currents leaving it. I hope it's clear and by the way my name is Sudeep not sundeep :)
Click to expand...
Thansk a lot Sudeep.
Btw i'm not getting Q33 at all, the other two I get, I guess I had trouble in this chapter, cause I don't remember krichoff's laws.
I guess i'll ask my teacher about this one, thanks though
 
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