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Physics: Post your doubts here!

Faizan Muhammmad

Faizan Muhammmad

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View attachment 34015
Circuit as shown in figure :)

Switch closed what happens?
Since capacitor is charging for some time all current falls on it but in the end does the bulb light up?
And does it light up continuosly or the capacitor discharges in process and hence the bulb blinks ?

Thanks in advance :)
 
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BreakingBad

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GlitteringLights said:
Can anyone plz explain ques 4 part (c) please? thank you
http://olevel.sourceforge.net/papers/9702/9702_s13_qp_43.pdf
Click to expand...
Use the equation in the previous part to calculate the speed 'required' to reach the stationary sphere.

v = (2qV/m)^1/2

V is given as 470V, q/m (specific charge) is given as 9.58 * 10^7

Hence,

v = (2 * 9.58 * 10^7 * 470)^1/2 = 3*10^5

The 'required' speed to reach the sphere is 3*10^5. However, the speed with which the charge is traveling is only 2.5*10^5. Therefore, we can say that the charge will be deflected (both charges are positive) before it can reach the sphere.
 
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BreakingBad

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Hassan Ali Abid said:
Can anyone please ans me why the magnitute of induced emf increases in the coil when it is rotated when it is places beside a current carring wire ? Do more magnetism flux linkage are cut when coil is rotated ?
Click to expand...
In comparison to what? Try being a little clearer. I might be able to help you then.
 
GlitteringLights

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BreakingBad said:
Use the equation in the previous part to calculate the speed 'required' to reach the stationary sphere.

v = (2qV/m)^1/2

V is given as 470V, q/m (specific charge) is given as 9.58 * 10^7

Hence,

v = (2 * 9.58 * 10^7 * 470)^1/2 = 3*10^5

The 'required' speed to reach the sphere is 3*10^5. However, the speed with which the charge is traveling is only 2.5*10^5. Therefore, we can say that the charge will be deflected (both charges are positive) before it can reach the sphere.
Click to expand...

thank you! :)
 
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BreakingBad

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1357911 said:
@Kim Shaw
Can u please help me with the following questions.
http://olevel.sourceforge.net/papers/9702/9702_s13_qp_11.pdf
Q-18
Q-17
Q-16
Q-11
Q-8
id be really grateful
Click to expand...
Q8
The time when the express train crosses the goods train will be the time when both of them have traveled the same distance.
Distance traveled by goods train = speed * time = 10t
Distance traveled by express train = ut + 1/2at^2 = 1/2at^2

u = 0 (express train starts from rest) and t cancels:

1/2at = 10
at = 20 (acceleration = 0.5)
t = 40s

Answer is D

Q11:

The collision is inelastic. Masses are added on the RHS:

m1v1 + m2v2 = (m1 +m2)v
v2 = 0, so:

2 * 3 = (2 + 1) v
v = 2
KE lost = Initial KE - Final KE = 0.5m1v1^2 - 0.5(m1 + m2)v^1/2
= 0.5*2*3^2 - 0.5*3*2^2 = 3 J

Answer is D

Q16:

Since the tension T in the cable alone is countering the other forces to keep the trapdoor in equilibrium, it is the largest. The only option with T being maximum is C, so the answer is C.

Q17:

v = (2gh)^1/2
HEIGHT IS 0.72 NOT 0.8! Measure from the base of the ball. Diameter is 0.08. We have to subtract that.

Use 1/2mv^2 = initial KE to calculate mass.

u = (2gh)^1/2
Again, height is 0.45-0.08 m.
KE after impact = 1/2mu^2 = 0.39 J

Answer is C

Q18:

P = E/t
P = (1/2mv^2)/t [m = density p * volume V]
P = (1/2pVv^2)/t [Vol = Area A * length s]
P = (1/2pAlv^2)/t [l/t = speed v]
P = 1/2Alv^3

Substitute values, Answer is B.
 
1357911

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BreakingBad said:
Q8
The time when the express train crosses the goods train will be the time when both of them have traveled the same distance.
Distance traveled by goods train = speed * time = 10t
Distance traveled by express train = ut + 1/2at^2 = 1/2at^2

u = 0 (express train starts from rest) and t cancels:

1/2at = 10
at = 20 (acceleration = 0.5)
t = 40s

Answer is D

Q11:

The collision is inelastic. Masses are added on the RHS:

m1v1 + m2v2 = (m1 +m2)v
v2 = 0, so:

2 * 3 = (2 + 1) v
v = 2
KE lost = Initial KE - Final KE = 0.5m1v1^2 - 0.5(m1 + m2)v^1/2
= 0.5*2*3^2 - 0.5*3*2^2 = 3 J

Answer is D

Q16:

Since the tension T in the cable alone is countering the other forces to keep the trapdoor in equilibrium, it is the largest. The only option with T being maximum is C, so the answer is C.

Q17:

v = (2gh)^1/2
HEIGHT IS 0.72 NOT 0.8! Measure from the base of the ball. Diameter is 0.08. We have to subtract that.

Use 1/2mv^2 = initial KE to calculate mass.

u = (2gh)^1/2
Again, height is 0.45-0.08 m.
KE after impact = 1/2mu^2 = 0.39 J

Answer is C

Q18:

P = E/t
P = (1/2mv^2)/t [m = density p * volume V]
P = (1/2pVv^2)/t [Vol = Area A * length s]
P = (1/2pAlv^2)/t [l/t = speed v]
P = 1/2Alv^3

Substitute values, Answer is B.
Click to expand...
Thank you.
ummm. Question number 35 too please.
 
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BreakingBad

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Faizan Muhammmad said:
View attachment 34015
Circuit as shown in figure :)

Switch closed what happens?
Since capacitor is charging for some time all current falls on it but in the end does the bulb light up?
And does it light up continuosly or the capacitor discharges in process and hence the bulb blinks ?

Thanks in advance :)
Click to expand...
Since the capacitor and the bulb are parallel, the current splits and hence, the bulb lights up independent of the charging of the capacitor. When the capacitor is fully charged, the bulb lights up the brightest as current through it is maximum. I think the point of such a circuit would be that even if the switch is opened again, the bulb will remain on as the capacitor will prove current.

This is just my theory so you might wanna confirm it.
 
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BreakingBad

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1357911 said:
Thank you.
ummm. Question number 35 too please.
Click to expand...
Total resistance of upper route = 2 kohm
Total resistance of the 2nd route = 2.25 kohm

Find I in each route, I=V/R
I in route 1 = 1.5 mA
I in route 2 = 1.333333333 mA

Calculate resistance ONLY TILL voltmeter.
R in upper route = 1000
R in lower route = 1500

Calculate difference in voltages = IRl - IRu = (1.333333333 * 1.5) - (1.5 * 1) = 0.5

Answer is B.
 
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ZaqZainab said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_qp_1.pdf
question 3 i suck at forces i could not attend the class at school so can any one teach me the AS way?? Please
Click to expand...
The answer is B. It's simple, you don't need much of vectors knowledge in this question.
If you draw the parallel of one of the forces, to get two sides of a triangle, you will see that the angle between them is 60. Since those sides are equal, their opposite angles will also be equal i.e., 60. Since all the angles are equal, the forces are also equal, 10N each.
 
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