physics p1 help

[rocker]

currently i am doing physics p1 and i have a lot of problem in it .I do't have redspot of physcis p1.Eventhough i have Ms i do't know that is answer for some question.Can you guys help me in the following question with possible solution and concept.
1 .06/n/11,14,34
2. 06/m/9,27,35
3. 07/m/10,23
4. 08/m/13,14,16
5. 09/m/15,18

here 06/n/11,12 means question of 2006 nov question 11 and 12








thanks

 

[beacon_of_light]

Nov06: Q11: Just memorize this formula...and solve out every similar qstn .

v1 = [(m1-m2)/(m1+m2) x u1] - [ (2m2)/(m1+m2) x u2)
Since u2 = 0
let mass be 1kg, v1=(1-1/1+1) x v = 0m/s

q14: a couple is produced. moment of couple = r x F = 2r x F = 2Fr

q34: 0 would be minimum voltage by first taking 0 ohms for potentiometer. For maximum voltage,

(20/40) x 12 = 6V

 

[beacon_of_light]

June06

q27: d = 1/N ( d is slit separation while N is no.of lines per metre)
if N increases, d decreses.

dsin theta = n x lambda
d decreases, n also decreases. and angle b/w 1st and 2nd orders increases. Just remeber proportionality ...

q35: total resistance = 30x10/30+10 = 7.5 and thatz b/w 1-10....

 

[beacon_of_light]

june07
Q10: The diagram shows momentum of the ball , so F= p2 - p1 / t2-t1 and this is the force acting by the ball...

So force acting "on" the ball is negative of this force...that is p1-p2/t2-t1

q23: at P, partical possesses PE so speed would be minimum. distance at Q is zero not displacement, at R the enrgy is entirely potential. SO accerelation at S would be maximum.....

 

[beacon_of_light]

m08:
q13: Since the system is in equilibrium, so a closed polygon(here a closed triangle)must be formed..and this is only formed in A.

q14: Taking moments about the point where two walls meet,

clockwise moments = anticlockwise moments,
Fxh + W x a = Wx 2a

q16: As the positive charge moves along the field, it gains kinetic energy and loses potential energy. work done on charge= lose of potential energy = F x s so potential energy decreases by Fs.

 

[beacon_of_light]

m09:
q15: when the level of water equals 1/2 h in both tanks, so the mass of water in each tank also lowers by 1/2m ......

Loss of GPE = mgh = 1/2m x g x 1/2h

q18: lets consider the situation after the increase in pressure. in the right limb, the liquid rises by a dis "h" and in the left limb the liquid lowers by a dis "h". So in the final position, increases presurre in the left limb will be the same in the right limb at the same vertical level. i.e p x g x 2h = 2pgh

 

[rocker]

thank u very very much for your answers,but i have flowing question regarding that:
1. 06n/34 in quesiont there is given each potentiometer 20 ohm .can u use 0 ohm.
2. 06/m/35 calulating that rough data gives 40 ohm how 7.5,how?

can you solve me question of 2004 nov 5,17 and 37.

 

[beacon_of_light]

may06:35 three resistors are in series and current through all of them will be same! while 10 and the resulting will be parallele as current divides through these ressitors.......

nov04:

Q5: time period of wave = 1/50 = 0.02s or 20ms. 8cm represent 20ms........1cm represents 20/8 = 2.5ms

q17: P= F x v

v=P/F = 3000/180= 16.67 s

time = dis/speed = 4000/16.67 = 240s

q37: V1 would be greater than V2 since combined resistance of 6 and 3 ohms is greater than 2 nd 2 ohms. and I1 will also be greater than I2 coz same amount of current divides b/w 6 and 3 ohms as welll as b/w 2 and 2ohms. using R is inversely proportional to current, same amount of current current will equallly divide in 2 and 2 ohms resistors while for 6 and 3ohms, cureent in 3 ohms will be large as compared to 6ohms. as a result I1>I2