• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Physics: Post your doubts here!

Messages
56
Reaction score
1
Points
16
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 :)
 
Messages
5,874
Reaction score
4,242
Points
323
b and c (ii) and (iii) anyone? The answers are at the end of the image... Thanks is advance!!!
A Question!.jpg
 

Nibz

XPRS Moderator
Messages
4,615
Reaction score
9,574
Points
523
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

XPRS Moderator
Messages
4,615
Reaction score
9,574
Points
523
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
 
Messages
5,874
Reaction score
4,242
Points
323
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...
 
Messages
143
Reaction score
27
Points
38
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.
 

Attachments

  • 9702_w02_qp_1.pdf
    141.5 KB · Views: 5
  • 9702_s04_qp_1 (1).pdf
    271.5 KB · Views: 7
  • 9702_s07_qp_1.pdf
    189.9 KB · Views: 5
  • 9702_w08_qp_1.pdf
    198 KB · Views: 3
  • 9702_s11_qp_11.pdf
    609.5 KB · Views: 8
Messages
55
Reaction score
32
Points
28
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.
 
Messages
143
Reaction score
27
Points
38
Can anyone plz explain how to draw a wave which has a phase difference of 60 degrees or 90 degrees from the original wave?
 
Messages
176
Reaction score
104
Points
53
I did see the mark scheme of this, but I really didn't understand exactly.
I have doubt in part (a)
 

Attachments

  • doubt.pdf
    49.2 KB · Views: 17
Messages
59
Reaction score
39
Points
18
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.
 
Messages
755
Reaction score
159
Points
53
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
 
Messages
755
Reaction score
159
Points
53
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
 
Top