Physics: Post your doubts here!

[Oliveme]

I think multiplying 20 times 3, would be wrong. Yes. you got the answer right, but the approach is wrong.
To calculate the resultant moment, we multiply the force and the perpendicular distance; in this case 2m for both 5N and 20N.
Now think about you how the 10N will have a resultant on the entire system as a whole. This is not just a simple bean and a pivot. One end of the system is bent at 90 degrees, and all the forces are acting on this end of the system. Now, for forces to have a resultant moment to this end, it has to act perpendicularly. And from the diagram we can see that the 10N is parallel not perpendicular. So the only two forces to have a resultant are the 5N and the 20N. The 10N is a trick, and is there to confuse you.
Now, f1d1-f2d2=0
20(2)-5(2)
40-10= 30 Nm. Hence, answer is A.



I've tried explain question 13 above.
Question 16-
The derivate of an energy/time graph will give us the power. We can see the graph is steepest from 10J to 40J. So the gradient=
(40-10)/(3-2)= 30W

Question 25-
I is directly proportional to A^2 (A square)
So 2I is proportional to x^2 (Since we need to find the corresponding amplitude, let's assume it to be x).
Now a ratio calculation-
2IA^2= Ix^2
2A^2=x^2 (I gets cancelled with each other)
so square root of 2A^2=x
The square root cancels indice of A ("the raised to the power 2"). which leaves us with "square root" of 2 * A.
I know it's complicated, but I hope you get it.


Question 33
We use the formula
R=("row"L)/A
L is directly proportional to R. and A is inversely proportional to R. It is mentioned in the question that the length is doubled. So since L is proportional to R, R will also be doubled. But they also mentioned that the volume remains constant. Now for a wire, whose length has been double, and for it's volume to remain constant, there has to be a reduction in the crossectional area of the wire; and in our case, since the length has been doubled, the coressectional area has to be halved, in order for the volume to remain constant. Now how will this increase in length, but the decrease in area, affect the resistance over all? The doubling of the length doubles the resistance, since it's directly proportional. And halving the area, will again double the resistance, since the area is inversly proportional. So in total, the resistance would be increased by a factor of 4. Hence the answer is D, 4R.

And as for question 14, I really can't do it, and need help myself.

Thank you.

 

[unique840]

june o9

question 8 prt b please!
jxt let me know that what is being asked in the question. Am unable to pick up the meanings!

we hav to xplain relation in which h is the constant of proportionality for example
energy is directly proportional to frequency.
when we equate energy and frequency we apply a constant h which is to remove the sign of proportionality therefore it is used as the constant of proportionality

 

[leadingguy]

we hav to xplain relation in which h is the constant of proportionality for example
energy is directly proportional to frequency.
when we equate energy and frequency we apply a constant h which is to remove the sign of proportionality therefore it is used as the constant of proportionality

thanx

 

[rz123]

kindly have a look.
test yourself question from chadda's book. page 542

A candidate obtains the following data between two variables t and r.
r/cm | t/s
6.2 | 4.6
12.0 | 6.0

the first value of r has an uncertainty of 0.2 cm, which is greater then the percentage uncertainty in t. Do the results show that t^2 is proportional to r ?

 

[unique840]

thanx

ur welcum

 

[Mustehssun Iqbal]

I think multiplying 20 times 3, would be wrong. Yes. you got the answer right, but the approach is wrong.
To calculate the resultant moment, we multiply the force and the perpendicular distance; in this case 2m for both 5N and 20N.
Now think about you how the 10N will have a resultant on the entire system as a whole. This is not just a simple bean and a pivot. One end of the system is bent at 90 degrees, and all the forces are acting on this end of the system. Now, for forces to have a resultant moment to this end, it has to act perpendicularly. And from the diagram we can see that the 10N is parallel not perpendicular. So the only two forces to have a resultant are the 5N and the 20N. The 10N is a trick, and is there to confuse you.
Now, f1d1-f2d2=0
20(2)-5(2)
40-10= 30 Nm. Hence, answer is A.



I've tried explain question 13 above.
Question 16-
The derivate of an energy/time graph will give us the power. We can see the graph is steepest from 10J to 40J. So the gradient=
(40-10)/(3-2)= 30W

Question 25-
I is directly proportional to A^2 (A square)
So 2I is proportional to x^2 (Since we need to find the corresponding amplitude, let's assume it to be x).
Now a ratio calculation-
2IA^2= Ix^2
2A^2=x^2 (I gets cancelled with each other)
so square root of 2A^2=x
The square root cancels indice of A ("the raised to the power 2"). which leaves us with "square root" of 2 * A.
I know it's complicated, but I hope you get it.


Question 33
We use the formula
R=("row"L)/A
L is directly proportional to R. and A is inversely proportional to R. It is mentioned in the question that the length is doubled. So since L is proportional to R, R will also be doubled. But they also mentioned that the volume remains constant. Now for a wire, whose length has been double, and for it's volume to remain constant, there has to be a reduction in the crossectional area of the wire; and in our case, since the length has been doubled, the coressectional area has to be halved, in order for the volume to remain constant. Now how will this increase in length, but the decrease in area, affect the resistance over all? The doubling of the length doubles the resistance, since it's directly proportional. And halving the area, will again double the resistance, since the area is inversly proportional. So in total, the resistance would be increased by a factor of 4. Hence the answer is D, 4R.

And as for question 14, I really can't do it, and need help myself.

But there's still a perpendicular distance from the 10 N force to the pivot. And that distance is upwards, rather than towards left. see the uploaded diagram. The dotted line there show the perpendicular distances. There is still moment of force produced by the 10 N force and the other two forces as well.

 

[omg]

we have to find time for total error of 10%. it already has an error of 2%. means we have to find the time having an error of further 8%. the value given is Ao. 2% error is included in it. further error of 8% means Ao - 8% = 0.92Ao

thanks a lot dude

 

[DragonCub]

can smby explain me the derivation of PV=NkT????

The original form of the equation for ideal gas is p V = n R T where
n = the mole number of gas
R = ideal gas constant
T = temperature in absolute scale

In derivation,
N = the number of gas molecules = n × NA (Avogadro's constant)
k = Boltzmann's constant = R ÷ NA

So you can see p V = N k T = (n NA) (R / NA) T = n R T
Hope this is helpful to you

 

[omg]

The original form of the equation for ideal gas is p V = n R T where
n = the mole number of gas
R = ideal gas constant
T = temperature in absolute scale

In derivation,
N = the number of gas molecules = n × NA (Avogadro's constant)
k = Boltzmann's constant = R ÷ NA

So you can see p V = N k T = (n NA) (R / NA) T = n R T
Hope this is helpful to you

yeahhh thankss

 

[unique840]

thanks a lot dude

ur welcum

 

[Bloodlines]

guys plz answer physics w 07 qp1 question no. 23...

 

[Bloodlines]

reply plzzz!....

 

[Arshiful]

someone pls kindly help me out, um stuck in communicating information in application chapter... donno why but i cant understand what the booklet says... i am seriously having problem digging deeper into its concepts..... someone can u pls provide me good links where i can find easy (understandable) notes or video lecture or anything which will make everything clear to me.... pls help me .....thnx in advance

 

[omg]

someone pls kindly help me out, um stuck in communicating information in application chapter... donno why but i cant understand what the booklet says... i am seriously having problem digging deeper into its concepts..... someone can u pls provide me good links where i can find easy (understandable) notes or video lecture or anything which will make everything clear to me.... pls help me .....thnx in advance

i was about to post the same qs!!!!!!!!!!!!!!!!!!!!!!!!
im stuck there toooo!!! -.-

 

[Oliveme]

Asalam-o-alaikum
please help me with these questions. Physics paper 11 may 2011 questions 14, 15, 35.
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s11_qp_11.pdf

thank you very much.

 

[leadingguy]

i was about to post the same qs!!!!!!!!!!!!!!!!!!!!!!!!
im stuck there toooo!!! -.-

someone pls kindly help me out, um stuck in communicating information in application chapter... donno why but i cant understand what the booklet says... i am seriously having problem digging deeper into its concepts..... someone can u pls provide me good links where i can find easy (understandable) notes or video lecture or anything which will make everything clear to me.... pls help me .....thnx in advance

from whch book are u studying?? do u guys have the black one ???? "internalional A/AS level physics??" chriss MEE, mike crundel, brian arnold are the authors. of the book. well apart have not studied this topic bt yes have gone through some of it till a.m and f.m itx quite typical but not that bad!

 

[unique840]

guys plz answer physics w 07 qp1 question no. 23...

speed is 8m/s which means
8m in 1 s so according to unitary method
50 m in x s 50 m cox one cycle is of 50 m
x = 6.25s
time period is 6.25
frequency is 1/6.25 = 0.16 hz
for max speed: 2*pie*a*f
for max speed we will take max amplitude.
so it will be 2*pie*2*0.16 = 2.01
max kinetic energy = 1/2 (mv^2)
1/2 (0.002*2.01^2)
= 4mJ

 

[omg]

from whch book are u studying?? do u guys have the black one ???? "internalional A/AS level physics??" chriss MEE, mike crundel, brian arnold are the authors. of the book. well apart have not studied this topic bt yes have gone through some of it till a.m and f.m itx quite typical but not that bad!

YEAH i have dat book bt idk y is it so confusing!!
maybe cuz idk the basics!
:/

 

[unique840]

Asalam-o-alaikum
please help me with these questions. Physics paper 11 may 2011 questions 14, 15, 35.
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s11_qp_11.pdf

thank you very much.

14) rebounding height is half of initial means potential energy is half after rebounding.
p.e = k.e therefore k.e is also half after the rebound
final k.e = 1/2 of initial k.e
1/2*m*v^2 = 1/2 (1/2*m*u^2)
v^2 = 1/2* u^2
taking square root on both sides
v = u/sqr rt of 2
v/u = (u/sqr rt of 2) / u
u cancels out
ratio is 1/sqr rt of 2

 

[leadingguy]

Asalam-o-alaikum
please help me with these questions. Physics paper 11 may 2011 questions 14, 15, 35.
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s11_qp_11.pdf

thank you very much.

question 15
look the box at rest having potential energr mgh = 9.81*3*2=58.9J.


now this energy will be converted to kinetic energy at the bottom of the slope.
means at bottom 1/2mv^2 = 58.9J now subtract the work done against friction while coming down! W=f.d



means W= 5*7 = 35J now subtract it from the original ans. u will get the net kinetic energy at the bottom of the slope which is 58.9-35=23.9J



now look 1/2 mv^2 =23.9 find v??

v= 4.88 round off u will get 4.9ms-1Ans