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Physics: Post your doubts here!

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hey can anyone help me with ON'06/02/1.c.ii and 2.c.iii
K.e=1/2 mv^2
=692550

power=workdone/time
=692500/8.1
85500 W
there is a resistive force acting which causes the acceleration to reduce or p=fv=mav
P and m are constant se when v increase a decrease
 
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hey can anyone help me with ON'06/02/1.c.ii and 2.c.iii
K.e=1/2 mv^2
=692550

power=workdone/time
=692500/8.1
85500 W
there is a resistive force acting which causes the acceleration to reduce or p=fv=mav
P and m are constant se when v increase a decrease
 
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okay remember that the 2 waves have the same frequency...so the wavelength remains the same, which is 30 boxes. you can use proportionality to do this.
If 15 boxes >> 180 degrees
x boxes >> 60 degrees
x = 5 boxes so that means, wave X is 5 boxes ahead of wave W

now it says wave X has 1/2 the Intensity >> 1/4 the amplitude of W.
the amplitude of W >> 10
10 x 1/4 = 2.5
amplitude of X >> 10 - 2.5 = 7.5 boxes in the vertical direction
 
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okay remember that the 2 waves have the same frequency...so the wavelength remains the same, which is 30 boxes. you can use proportionality to do this.
If 15 boxes >> 180 degrees
x boxes >> 60 degrees
x = 5 boxes so that means, wave X is 5 boxes ahead of wave W

now it says wave X has 1/2 the Intensity >> 1/4 the amplitude of W.
the amplitude of W >> 10
10 x 1/4 = 2.5
amplitude of X >> 10 - 2.5 = 7.5 boxes in the vertical direction

how the graph looks like ? oO'

Please help me for no. 6(b)(i) too :(
 
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how the graph looks like ? oO'

Please help me for no. 6(b)(i) too :(
6(b)(i)
It is given that the voltage of the main supply is 230V.And it is said in the ques that "The potential difference across the shower unit must not be
less than 225 V"....that is,the minimum voltage of the main supply is 225V
The minimum voltage of the cable=220V-225V=5V
Therefore....the P.d in the cable=5V
The current in the wire= 46A
For Max. Resistance,Voltage must be Min.
Therefore, v=5V,I=46A
Using V=IR
R=V/I
=5/46
=0.10869
=0.11 ohm
This is the max. resistance of the cable!
 
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6(b)(i)
It is given that the voltage of the main supply is 230V.And it is said in the ques that "The potential difference across the shower unit must not be
less than 225 V"....
Therefore....the P.d in the cable must be= 230V-225V
= 5V
The current in the wire= 46A
Therefore,
Using V=IR
R=V/I
=5/46
=0.10869
=0.11 ohm
This is the max. resistance of the cable!

Thanks mate :)

Would you mind helping me with this one : http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s08_qp_2.pdf

no. 4 (b) (c)
 
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(b) In this question, you must look for the sizes of the expression
for e.g, it is said that ∆p is measured in Pa,
So ∆p must be large...which is shown on the right hand-side where there is the value 2.2 × 10^9 which is Very Large
It is also said that "water is assumed to be incompressible" ..which means that the change in volume is Very Small..
which is shown by ∆V/V which is very small.
Hope u've understand dr!! ;)
(c) In this ques,we must use ∆p = hρg
It is given that the "approximation of h=10"
Using the eqn to find the actual h.
∆p = hρg
1.01 × 10^5= h x 1.08 × 10^3 x 9.81
h=9.53m
Therefore to determine the %error,
we take, ∆h / h x 100
∆h=10-9.53
=0.47m
Therefore %error= 0.47/9.53 x 100
= 4.9%
(You can also take the "h" as 10)
Hope it helps!! ;) :)
 
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