phy p2 prob!!!!

[Fatima Hasan]

can sum 1 plzz tel how to solve Q5, (b0...i know about phase diff of 90 n 180 but how will we show phase diff of 60????? :?

 

[mista.lova.lova]

u need to mention, which year!!!??

 

[beacon_of_light]

find it in terms of time period...that will be easier to plot points...

 

[uzair_manunited]

Still cannot understand anything about DYNAMICS

 

[mista.lova.lova]

shabaash....bhool jao phir aerospace engineer banna...

 

[Fatima Hasan]

ooops...sorry!! nov 02! Q5

 

[x.daydreamer]

Anyone solve this for me ?
M/J 2007-----Physics Q4d (i) and (ii)
And M/J/2007----PHYSICS q1 the whole thing!
I dont get it =(
thank youuuu!

 

[beacon_of_light]

Anyone solve this for me ?
M/J 2007-----Physics Q4d (i) and (ii)
And M/J/2007----PHYSICS q1 the whole thing!
I dont get it =(
thank youuuu!

For q4:

Using an appropriate scale, draw a horizontal line representing horizontal velocity of 18m/s.

then calculate vertical velocity at point of impact with ground. resultant velocity is 25m/s and horizontal is 18m/s...using pythagoras theorem ..calculate vertical velocity which is approximately equal to 17.3m/s.....now draw this on the graph and then draw out the resultant!!!

Once you are drawn with the vector diagram, measure the asked angle with a protractor!

For q1: As the ques says that the metre drawn( with a pointer) indicated the vol of fuel in tank...u need to calibrate it. first at 20x 10^3 cm^3, the angle read form

graph is 40 ...at the nxt 20 x10^3 interval the angle is 70...placing a protractor over the pointer such that its line coincides with the line of pointer, draw the angles( i hope u knw them) ..and then u get a caliberated scale!!!

With increasing vol of fuels, deflection of pointer reduces...at 80 cm^3 deflection is only 102...so at 1x 10^5 cm^3 deflection is an estimated value of 110.....

as the scale shows larger deflection at low fuel volumes....so it is very sensitive at low fuel volumes!!!

Hope u get that

 

[beacon_of_light]

ooops...sorry!! nov 02! Q5

Lets find 60 in terms of time in seconds. that will be easier to plot otherwise you can plot the points in terms of 1/3 pi radians...

so, let x be the time T2 lags behind T1...(and 3s is the time period of the two waves)

x/ 3 = 60/360 .....(thats a formula)....

x = 0.5s...this means every point on t2 must be 0.5s behind T1...draw it as a new curve for T2...

and on the graph 0.5s = 1cm

 

[Fairy]

Can someone plz explain Q6 of May 06 (the whole question) and also tell what basic points we should be acquainted with of this part of the topic (Stationary waves and the end correction stuff). i never get it

 

[beacon_of_light]

Before solving out this question, you need to read chapter related to formation of air columns...

well specifically speaking for this question,

If an antinode forms on the open end of tube , sound is heard..if a node forms on the open end no sound is heard...
so in fig 6.1, the wave form is shown although the question doesn't specify an antinode formation on the open end!

now in fig 6.2, we,ve to draw a second wave when second time sound is heard....draw a wave with 2nd antinode on the mouth end..and the total wave is 3/4 lambda...means 2 nodes n 2 antinodes....

32.4cm is the distance b/w 2 nodes..which is 1/2 lambda...so double this distance n find the speed...

for the last part,
since 32.4 cm is 1/2 lambda...so 1/4 lambda will be 16.2cm (that is the distance b/w node and adjacent antinode)..while 15.7 cm is not equal to 16.2cm..so antinode will be (16.2cm - 15.7 cm) = o.5cm above open end of tube in fig 6.1....................