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Consider the wave as displacement time graph...
where gradient is Max velocity is maximum at B and D gradient is zero so lowest velocity there...
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Physics: Post your doubts here!
- Thread starter XPFMember
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where gradient is Max velocity is maximum at B and D gradient is zero so lowest velocity there...
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31)
The total resistance of the cable is 8 ohms. The cable is 800 meters long since that's the distance between the power supply and the relay and there are 2 wires in the cable. 0.005 ohms per m means in one wire the resistance is 0.005 ohms x 800 = 4 ohms. So total resistance is 2 x 4 = 8 ohms.
Add the resistance of the wires to the resistance of the relay. Relay's resistance is R = V/I = 16/0.6 = 26.7 ohms. (26.7 + 8)
To find the output voltage = IR = 0.6 x (26.7 + 8) = 20.82 and the answer is C.
We are treating the relay and wires as one unit. The current through wires does not change and so we can plug the 0.6 amps of the relay into the V=IR formula.
It just seems like it's complicated but it really isn't. Try looking at it in a simplified way.
37)
1/30 + 1/10 = 4/30 now inverse it you get 7.5
so B
Aoa everyone.
Im REALLY stuck on this one specific type of question in all my pastpapers and they are the diffraction grating ones. I havent even gotten a single one right yet and I would EXTREMELY appreciate if someone could just explain 2 of these so I'll understand the rest myself. JazakAllah
Q. 30 from O.N/2012
Q. 27 from O.N/2010
Im REALLY stuck on this one specific type of question in all my pastpapers and they are the diffraction grating ones. I havent even gotten a single one right yet and I would EXTREMELY appreciate if someone could just explain 2 of these so I'll understand the rest myself.
JazakAllah Q. 30 from O.N/2012
Q. 27 from O.N/2010
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31)
The total resistance of the cable is 8 ohms. The cable is 800 meters long since that's the distance between the power supply and the relay and there are 2 wires in the cable. 0.005 ohms per m means in one wire the resistance is 0.005 ohms x 800 = 4 ohms. So total resistance is 2 x 4 = 8 ohms.
Add the resistance of the wires to the resistance of the relay. Relay's resistance is R = V/I = 16/0.6 = 26.7 ohms. (26.7 + 8)
To find the output voltage = IR = 0.6 x (26.7 + 8) = 20.82 and the answer is C.
We are treating the relay and wires as one unit. The current through wires does not change and so we can plug the 0.6 amps of the relay into the V=IR formula.
It just seems like it's complicated but it really isn't. Try looking at it in a simplified way.
37)
1/30 + 1/10 = 4/30 now inverse it you get 7.5
so B
In question 31, how do you know the two wires inside are in series? They could be in parallel too, the question doesn't specify.
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What is the answer ?
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Two wires P and Q are made from the same material.
Wire P is initially twice the diameter and twice the length of wire Q. The same force, applied to
each wire, causes the wires to extend elastically.
What is the ratio of the extension in P to that in Q?
(A) 1/2 B 1 C 2 D 4
Wire P is initially twice the diameter and twice the length of wire Q. The same force, applied to
each wire, causes the wires to extend elastically.
What is the ratio of the extension in P to that in Q?
(A) 1/2 B 1 C 2 D 4
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- Messages
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- Messages
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- 34,837
- Points
- 698
w12_12Aoa everyone.
Im REALLY stuck on this one specific type of question in all my pastpapers and they are the diffraction grating ones. I havent even gotten a single one right yet and I would EXTREMELY appreciate if someone could just explain 2 of these so I'll understand the rest myself.
Q. 30 from O.N/2012
Q. 27 from O.N/2010
30)
Apply d*sin thetha = n*lamida
d = 1/n where n is no. of line/m. Metre not millimetre.
Grating's got 300lines/mm. So this means it has 300*1000 lines/m
(Coz 1m = 1000mm)
d = 1/n ie 1/300,000
U wanna see all the maximas possible, so to get max. no. of maximas on one half of the screen, u put thetha = 90
lamida is given. Convert it into m from nm.
U get n = 4
This means 4 maximas on one half of the screen.
On whole (both halves) of screen, maximas are = 4*2 = 8
BUT, u also have to consider the central maxima.
So total maximas are 8+1 = 9
Central is in middle.
4 maximas above and 4 below it.
27)
Let the distance between the double slit and the screen be 1m initially.
When the distance is increased BY 2m the NEW distance is 3m.
Using the formula:
Wavelength= (fringe seperation x distance between the double slits)/ distance between the screen and the slits.
Wavelenght=600nm=600x10^-9m.
Fringe seperation= 3mm= 3x10^-3m
Distance between screen and slits= 3m (1+2)
Distance between the double slit=?
Put these values in the formula, the final answer is 6x10^-4m =0.6mm
Hence the answer is B.
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10)
look ... this question deals with relative speeds rather than just momentum.... in a perfectly elastic collision, the relative speed of approach is equal to the relative speed of separation.. but what is meant by relative speed of the 2 objects.... imagine u are moving in a car at a speed of 100 km/h beside another car moving at the same speed that is 100 km/h in the same direction.. u will feel that the other car is not moving.....also if u move at a speed of 100 km/h and u look at another car moving in opposite direction at a speed of 100 km/h u feel that the other car is moving very fast .. that is at a speed of 200 km/h......... so to find relative speed of 2 objects ...add their speeds if they are moving in opposite direction and subtract their speeds when they are moving in the same direction... but only in a perfectly elastic collision.
In this question, the relative speed of approach is equal to u1 + u2 since they are moving in opposite directions.. this is equal to the relative speed of separation which is equal to v2 - v1 since they are moving in the same direction after the collision..
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Any doubts left ?
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Can someone help me with JUN 11 P11 Qs 27?
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s11_qp_11.pdf
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s11_qp_11.pdf
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Ahahaha..... Thanks... I know about relative speeds... Add Maths RV.... Bro u have amazing concepts.... (Y)
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find d which is 1/n -->1/500 * 10^ (-3) = 2 x 10^-6.
then d sin90=n x (600 x 10^-9) = 3 then he asked for the images so it is 3 orders for one side which is 45 degrees so for the 90 degrees it is 3 x 2= 6 + the normal ray = 7 so D
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"there are two vectors, P and P' where P is initial momentum and P' is final momentum.
P=mv
P'=-mv
change in momentum is P'-P which is same as P + (-P)
just add them like this.
-mv + (-mv) = -2mv"
Haris Bin Zahid
This answer is for you from Suchal.. He is unable to post here..
P=mv
P'=-mv
change in momentum is P'-P which is same as P + (-P)
just add them like this.
-mv + (-mv) = -2mv"
Haris Bin Zahid
This answer is for you from Suchal.. He is unable to post here..