Physics: Post your doubts here!

[unique840]

Guys can you help me with Q5 c i in june 2004 paper 4? the ratio is coming as half with me not 2 , any ideas why is it 2?

power dissipation by direct current will be (Irms^2)*R where Irms will be Io
power dissipation by sinusoidal current will be (Irms^2)*R where Irms will be Io/(square root of 2)
[(Io^2) * R] / {[(Io/sqr root of 2)^2]*R}
Io^2*R will be cancelled
inverse of (2) will be left
inverse of 2 will be 2
ratio = 2

 

[Karimgenena]

power dissipation by direct current will be (Irms^2)*R where Irms will be Io
power dissipation by sinusoidal current will be (Irms^2)*R where Irms will be Io/(square root of 2)
[(Io^2) * R] / {[(Io/sqr root of 2)^2]*R}
Io^2*R will be cancelled
inverse of (2) will be left
inverse of 2 will be 2
ratio = 2

thanks bro, I appreciate your reply

 

[unique840]

thanks bro, I appreciate your reply

ur welcum but im not bro. m a grl

 

[aliya_zad]

how can u increase the amount of damping in a spring system??any examples..

 

[rz123]

how can u increase the amount of damping in a spring system??any examples..

http://www.myphysicslab.com/spring1.html
http://en.wikipedia.org/wiki/Vibration

 

[aliya_zad]

thank u!

 

[rz123]

welcome (bdw i simply googled it )

 

[CaptainDanger]

b and c (ii) and (iii) anyone? The answers are at the end of the image... Thanks is advance!!!

 

[Nibz]

For b:
When the phase difference is even multiple of pi, the waves are said to be 'In phase'. For b(i) the phase difference is zero, i.e, even. (0, 2 , 4, 6 etc)
When waves are 'in-phase' the interference is constructive.
So, simply add the amplitudes.
A = 3 + 1
= 4A
Frequency is constant. So, I = (A)^2 => 16I

For b(ii) the phase difference is pi (1pi), i.e, odd multiple. Waves are then 'out of phase', hence, destructive interference.
Subtract the amplitudes.
A = 3 - 1
= 2A
I = (2)^2
= 4I

 

[Nibz]

The c part involves simple calculations.

(ii) Mean intensity of the emitted pulse is I = (P) / (4 x pi x d^2)
= (2 x 10^6)/(4 x pi x (50,000)^2) = 6 x 10^-5 W/m^2

(iii) d= 50km ks = 1m^2

I = (2 x 10^6) (1) / 16 x pi^2 x (50,000)^4 = 2 x 10^-15 W/m^2

 

[CaptainDanger]

The c part involves simple calculations.

(ii) Mean intensity of the emitted pulse is I = (P) / (4 x pi x d^2)
= (2 x 10^6)/(4 x pi x (50,000)^2) = 6 x 10^-5 W/m^2

(iii) d= 50km ks = 1m^2

I = (2 x 10^6) (1) / 16 x pi^2 x (50,000)^4 = 2 x 10^-15 W/m^2

Oh right... Thanks... Should have read that part AGAIN...

 

[Nibz]

You're welcome. Yes, take care next time :]

 

[JD REBORN]

q)35 of w02 p1,Q)35 of s04 p1,Q)37 of s07) p1,Q)32 and Q)37 of w08 p1,q)34 and q)36 of s11 p1.

 

[darknessinme]

q)35 of w02 p1,Q)35 of s04 p1,Q)37 of s07) p1,Q)32 and Q)37 of w08 p1,q)34 and q)36 of s11 p1.

35 of w02 p1: The resistance wire and variable resistor share the pd of the battery at the beginning. If the resistance of the variable resistor increases while other things remain the same, the variable resistor will take a greater share of the pd so voltage across XY decreases.
For the second part, the voltage across XN is equal to cell's pd at the bottom(zero deflection). Since the voltage across XY decreased after the variable resistor was used, the contact moves closer to Y to achieve the same voltage as the cell once again(zero deflection). This is because voltage is proportional to the length of the wire (resistance is constant). Answer D.

35 of s04 p1: Similar to the first part of the previous question. The the total resistance of the circuit decreases, so there must be an increase in current (V=IR). Resistance of the variable resistor decreases, so it takes a smaller share of the voltage of the battery, so the voltmeter reading must increase. Answer D.

 

[JD REBORN]

Explain Q6)b with drawing plzz

 

[JD REBORN]

Can anyone plz explain how to draw a wave which has a phase difference of 60 degrees or 90 degrees from the original wave?

 

[Khan_971]

I did see the mark scheme of this, but I really didn't understand exactly.
I have doubt in part (a)

 

[Doctor Nemo]

I did see the mark scheme of this, but I really didn't understand exactly.
I have doubt in part (a)

The starting momentum of ball A at right angles to its motion is zero. (The y-component is zero). Therefore after the collision y-components of the momentum of the two balls must be zero. Since the y-component of the momentum of ball A after the collision is 2.6*1.2*sin(30) then the momentum of ball B must equal the negative of this value so you can calculate the y-component of its velocity as being equal to -1.3.

You can now calculate the angle that the velocity vector of B makes with the x-axis, this is 60 degrees.

Knowing the angles you can calculate the x-components of the velocities and the total momentum in the forward direction.

This is 1.2 (2.6* cos(30)+1.5*cos(60)) =1.2 *3.0 so that momentum before the collision = momentum after collision and therefore it is conserved in the collision.

 

[unique840]

q)35 of w02 p1,Q)35 of s04 p1,Q)37 of s07) p1,Q)32 and Q)37 of w08 p1,q)34 and q)36 of s11 p1.

q 37 of june 07:
resistance = resistivity * length / area
resistivity for both p and q is same as the material is same
length is also same for both
area = volume/height(or length)
area will also be same cox volume and length of both p and q is same
so resistance will be same

 

[unique840]

q)35 of w02 p1,Q)35 of s04 p1,Q)37 of s07) p1,Q)32 and Q)37 of w08 p1,q)34 and q)36 of s11 p1.

nov 08 q32
we will take all these in parallel. means there are 6 copper wires and 1 steel wire all in parallel with each other.
resistance will be =
1/R = (1/10)*6 + (1/100)
R = 1.6 ohms