Physics: Post your doubts here!

[nodyed]

A copper wire of cross-sectional area 2.0mm2
carries a current of 10A.
How many electrons pass through a given cross-section of the wire in one second?
A 1.0 x 101 B 5.0 x 106 C 6.3 x 1019 D 3.1 x 1025 how c?

 

[agha saad]

appearing in A's level physics sup...
I NEED HELP IN FINDING UNCERTAINITY.... -_- en physics sux

 

[nodyed]

q36 how?
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s05_qp_1.pdf

 

[agha saad]

A copper wire of cross-sectional area 2.0mm2
carries a current of 10A.
How many electrons pass through a given cross-section of the wire in one second?
A 1.0 x 101 B 5.0 x 106 C 6.3 x 1019 D 3.1 x 1025 how c?

ITS SIMPLE
Q=ne
"n"denotes number of electrons
"e"=1.6x10^-19
sup
Q=IT
substitute both equations
IT=ne
or
I/e=n/t
10/1.6x10^-19
ans:C

 

[gary221]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_4.pdf
Question 1 part c, anyone?

U knw tht change in P.E. = change in K.E.
So, using the formulas from b i) n ii)
GMm/ 6R = (2.09 * 10^6)m
the m cancel each other out
So, substituting the values,
[(6.67 *10^11) * M]/(6 *3.40 * 10^6) = 2.09 * 10^6
So, M = 6.39 * 10^23 kg

 

[agha saad]

explain pls!

"a" is answer

 

[sagar65265]

q36 how?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_1.pdf

Point 1: Since the two resistors R and 2R are in parallel, they will have the same potential difference across them - therefore, V2 = V3
Point 2: The potential difference read by voltmeter V will be equal to the batteries' EMF since the battery is assumed to be ideal and thus has no internal resistance.
Point 3: The potential drop across the terminals of the battery (you can confirm this using Kirchoff's second law) is equal to the sum of the potential drop across the two parallel resistors (R and 2R) and the drop across the R resistor near the battery.
Point 4: These values are given by :
i) Resistor R and 2R - reading on V2 (=reading on V3)
ii) Resistor R - reading on V1
Therefore,
V = V1 + V2 or V = V1 + V3 (Since V3 = V2)
rearranging the second formula, V - V1 = V3 ===> D

Hope that helped!

Good luck for your exams!

 

[sagar65265]

explain pls!

The potential difference between two points is defined as the amount of energy transferred per unit charge as it passes from one point to the other; This means that:

V = W/Q (energy change per unit charge)
and Q = It (Current x time)
so V = W/It
20 = 12/15I
15I = 0.6
I = 0.04 A = Option A.

Hope it helped!

Good Luck for your exams!

 

[sagar65265]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_13.pdf
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_ms_13.pdf
plz explain q11,15,16,18,20,29,35
I wud be thankful to any1 who provides a good explanation
Thank you

Q11: Main idea here - Conservation of Momentum
For this one, take the initial momentum before the collision occurs. Momentum = mv and since momentum can be positive or negative, we have to take either the right direction (positive x - axis) or the left one (negative x - axis) to be positive; in this case, let us take the right direction to be positive. Then:

Total Momentum before collision = 2mu - mu = mu
Therefore, total momentum after collision = mu.
Option A : momentum after = -2mu/3 + 5mu/3 = 3mu/3 = mu so this situation follows conservation of momentum as it is the same before and after the collision.
Option B : momentum after = -2mu/6 + 2mu/3 = mu/3 so this is the answer, as here, momentum before is not equal to momentum after, thus momentum is not conserved. To confirm,
Option C : momentum after = 2mu/6 + 2mu/3 = 3mu/3 = mu so this situation follows conservation of momentum.
Option D : momentum after = (2m + m)(u/3) = 3mu/3 = mu so momentum is conserved here too.
Therefore B is the answer.

Q 16: Zero resultant force implies that the sum of forces pushing up on the object = the sum of the forces pushing down on the object. This requirement isn't fulfilled in situation A, since 50 + 30 is not equal to 90. However, this applies to all the other options, so A is eliminated directly.
the second requirement is zero resultant torque, therefore, the net moments of all the forces about ANY point is equal to zero.
Taking moments about the central force (the downwards force in this question):
Option B: 36 x 0.5 - 70 x 30 is equal to -3, so the net torque is not equal to zero, so the answer is not B.
Option C: 28 x 0.5 - 35 x 0.3 is equal to 3.5, so the net torque is still not zero. Since the answer is not A, B or C, the answer should be D. To confirm:
Option D: 42 x 0.5 - 70 x 0.3 is equal to 0, so the net torque is zero and the answer is D.

Q 18 : Taking conservation of energy into account, the sum of the kinetic energy and the potential energy in the beginning before the car leaves the motorway should be equal to the sum of the kinetic and potential energy at the end of the exit road.

The initial energy = The final energy, so

0.5 x m x (28)^2 = 0.5 x m x (v)^2 + m x 9.81 x 22 (cancelling out the variable of mass m):
0.5 x (28)^2 = 0.5 x (v)^2 + 9.81 x 22
392 = 215.82 + 0.5 x (v)^2
v^2 = 2(392 - 215.82) = 352.36
Therefore v = sqrt(352.36) = 18.77 m/s = 19 m/s = B

Q 20 : The power output of an ideal transformer is the same as the power input to the transformer. Power = Current x Voltage so the initial power before the transformer works = 11,000 x 28 = 308,000 J/s = 308,000 W
Therefore, an ideal transformer should produce an output of power 308,000 W.
However, this case is not about an ideal transformer and thus the output power = 240 x 1200 = 288,000 J/s = 288,000 W
Since efficiency = (input power/output power) * 100,
efficiency = (288000/308000) * 100 = 93.506 = 94 % = D

Q 29 : There is a formula relating the amplitude of a wave to its intensity:

Intensity is proportional to (Amplitude)^2 so Intensity = Constant x (Amplitude)^2
Rearranging this, for any one wave, Intensity/(Amplitude)^2 = Constant.
So, I/A^2 =2I/x^2 where x is the final amplitude. Cancelling out the Variable I,
1/A^2 = 2/x^2
x^2 = 2A^2 and now, taking a square root on both sides,
x = amplitude of the wave when intensity is 2I = sqrt(2)A so the answer is B.

I'm not really sure about question 15 and I have some other work to do so I can't say much about Q 35, but if I do understand them, I'll try my best to reply ASAP. Sorry!

Hope this helps!

Good Luck for your exams!

 

[nodyed]

The potential difference between two points is defined as the amount of energy transferred per unit charge as it passes from one point to the other; This means that:

V = W/Q (energy change per unit charge)
and Q = It (Current x time)
so V = W/It
20 = 12/15I
15I = 0.6
I = 0.04 A = Option A.

Hope it helped!

Good Luck for your exams!

Ty was just messed up with formlae but now all clear cheers!

 

[Scafalon40]

Since the wave is amplitude modulated (the frequency of the carrier wave appears constant and thus unmodulated), the carrier wave's amplitude follows the envelope of the signal wave; in other words, if you connect the peaks of every wave to the next one, you should get something like a sine wave. In this case, the maximum amplitude occurs first at

t = approximately 2 microseconds - you can also estimate this to be at t = 0 microseconds since you can shift the graph to the left along the x axis without changing the general shape of the graph.

The wavelength is from one peak till the direct next peak, so this next peak (of both the carrier and the signal) occurs at

t = approximately 102 microseconds (or 100 microseconds if you have approximated 2 microseconds to 0)

so the difference is either 100 - 0 or 102 - 2, both of which are equal to 100 microseconds.
So the time period of the wave is 100 microseconds and as

frequency = 1/Time Period,

time period = 1 / (100 x 10^-6) = 10,000 Hz = 10 kHz.

Hope this helped!

(A neat little thing is that if you start with the beginning of the graph at t = 0 at the top of your webpage and scroll up till the top, you can actually see the carrier wave as it is!)

Good Luck for your exams!

Thanks! You too!

 

[nodyed]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w06_qp_1.pdf q34? how to calculate I2?

 

[syed1995]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w06_qp_1.pdf q34? how to calculate I2?

To find range .. Well either the resistance at the bottom resistor 0 or maximum..

You need to recall the formula Vo = (Ro/Ro+R1)*Vinput

The output given at the bottom = (Rp/Rp+Rf)*12 ----// Where Rf is the fixed resistor at top and Rp is the potentiometer across which the output is given.

If the bottom resistance is minimum/0 .. then the output at the bottom would be (0/0+20)*12 = 0V

If the bottom resistance is maximum/20 .. then the output at the bottom would be (20/20+20)*12 = 6V

so the range is 0-6V Answer is A

 

[syed1995]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w06_qp_1.pdf q34? how to calculate I2?

I think you wrote 34 instead 36 by mistake..

36)

Kirchoffs first law..

For I2 Look at the junction at the left .. and the two currents leaving that point. I2 is entering and 20.2 and 10.6 are leaving..

so according to first law.. then
I2 = 20.2 + 10.6
I2 = 30.8

I1 is + 0.2 .. cuz from the bottom junction towards ammeter = -0.2 .. so towards ammeter would be + 0.2 .. Answer is B

 

[Alool]

Does anyone know how to solve question 5 B in paper m/j 2009/21?

 

[hellangel1]

U knw tht change in P.E. = change in K.E.
So, using the formulas from b i) n ii)
GMm/ 6R = (2.09 * 10^6)m
the m cancel each other out
So, substituting the values,
[(6.67 *10^11) * M]/(6 *3.40 * 10^6) = 2.09 * 10^6
So, M = 6.39 * 10^23 kg

Thanku

 

[Tkp]

for the sound at m to be zero that means destructive interference.so for destructive interference the path difference must be be a odd number of multiple wavelength
and for the second conditon if the amplitudes are same so there will be destructive interference as the two waves cancel out each other
so for the first 1 it would be the path difference must be be a odd number of multiple wavelength and 2 nd one amplitude amplitude of both the waves are same at m.
b)v=f lambda.so u put these values and u get 2 value of l(convert the unit in cm).that means the minima would be within the range.now find the path differnce.find the length of s2m by pythagoras theorem which is 128 cm.so the path difference for s2m-s1m is 28 cm.now apply the formula of path difference for destructive interference
28=(n+.5)l put the value of n to o,1,2,3,4.now u will get 2 values which are within the range.so 2 minima would be detected

Does anyone know how to solve question 5 B in paper m/j 2009/21?

 

[snoonono]

Please help fast! I need explanation of intensity of a wave, how to find it, what I need to have, how to find what I need to have if not provided! Thanks!

 

[goodluckayesha]

Q11: Main idea here - Conservation of Momentum
For this one, take the initial momentum before the collision occurs. Momentum = mv and since momentum can be positive or negative, we have to take either the right direction (positive x - axis) or the left one (negative x - axis) to be positive; in this case, let us take the right direction to be positive. Then:

Total Momentum before collision = 2mu - mu = mu
Therefore, total momentum after collision = mu.
Option A : momentum after = -2mu/3 + 5mu/3 = 3mu/3 = mu so this situation follows conservation of momentum as it is the same before and after the collision.
Option B : momentum after = -2mu/6 + 2mu/3 = mu/3 so this is the answer, as here, momentum before is not equal to momentum after, thus momentum is not conserved. To confirm,
Option C : momentum after = 2mu/6 + 2mu/3 = 3mu/3 = mu so this situation follows conservation of momentum.
Option D : momentum after = (2m + m)(u/3) = 3mu/3 = mu so momentum is conserved here too.
Therefore B is the answer.

Q 16: Zero resultant force implies that the sum of forces pushing up on the object = the sum of the forces pushing down on the object. This requirement isn't fulfilled in situation A, since 50 + 30 is not equal to 90. However, this applies to all the other options, so A is eliminated directly.
the second requirement is zero resultant torque, therefore, the net moments of all the forces about ANY point is equal to zero.
Taking moments about the central force (the downwards force in this question):
Option B: 36 x 0.5 - 70 x 30 is equal to -3, so the net torque is not equal to zero, so the answer is not B.
Option C: 28 x 0.5 - 35 x 0.3 is equal to 3.5, so the net torque is still not zero. Since the answer is not A, B or C, the answer should be D. To confirm:
Option D: 42 x 0.5 - 70 x 0.3 is equal to 0, so the net torque is zero and the answer is D.

Q 18 : Taking conservation of energy into account, the sum of the kinetic energy and the potential energy in the beginning before the car leaves the motorway should be equal to the sum of the kinetic and potential energy at the end of the exit road.

The initial energy = The final energy, so

0.5 x m x (28)^2 = 0.5 x m x (v)^2 + m x 9.81 x 22 (cancelling out the variable of mass m):
0.5 x (28)^2 = 0.5 x (v)^2 + 9.81 x 22
392 = 215.82 + 0.5 x (v)^2
v^2 = 2(392 - 215.82) = 352.36
Therefore v = sqrt(352.36) = 18.77 m/s = 19 m/s = B

Q 20 : The power output of an ideal transformer is the same as the power input to the transformer. Power = Current x Voltage so the initial power before the transformer works = 11,000 x 28 = 308,000 J/s = 308,000 W
Therefore, an ideal transformer should produce an output of power 308,000 W.
However, this case is not about an ideal transformer and thus the output power = 240 x 1200 = 288,000 J/s = 288,000 W
Since efficiency = (input power/output power) * 100,
efficiency = (288000/308000) * 100 = 93.506 = 94 % = D

Q 29 : There is a formula relating the amplitude of a wave to its intensity:

Intensity is proportional to (Amplitude)^2 so Intensity = Constant x (Amplitude)^2
Rearranging this, for any one wave, Intensity/(Amplitude)^2 = Constant.
So, I/A^2 =2I/x^2 where x is the final amplitude. Cancelling out the Variable I,
1/A^2 = 2/x^2
x^2 = 2A^2 and now, taking a square root on both sides,
x = amplitude of the wave when intensity is 2I = sqrt(2)A so the answer is B.

I'm not really sure about question 15 and I have some other work to do so I can't say much about Q 35, but if I do understand them, I'll try my best to reply ASAP. Sorry!

Hope this helps!

Good Luck for your exams!

Thnq u soooo much n good luck to u too

 

[sanii94]

may/june 12 paper 42 Q5 (b) (i)
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_42.pdf

^please help with this