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Physics: Post your doubts here!

hussamh10

hussamh10

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Tasnim Akhter said:
hey can anyone help me with ON'06/02/1.c.ii and 2.c.iii
Click to expand...
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
 
G

geek101

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bamteck said:
www.xtremepapers.com/papers/CIE/.../9702_w07_qp_2.pdf

HELP NEEDED for no. 5 (a) :(
Click to expand...

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
 
bamteck

bamteck

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geek101 said:
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
Click to expand...

how the graph looks like ? oO'

Please help me for no. 6(b)(i) too :(
 
TeKnOzOr

TeKnOzOr

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bamteck said:
how the graph looks like ? oO'

Please help me for no. 6(b)(i) too :(
Click to expand...
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!
 
bamteck

bamteck

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TeKnOzOr said:
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!
Click to expand...

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)
 
TeKnOzOr

TeKnOzOr

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bamteck said:
Thanks mate :)

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

no. 4 (b) (c)
Click to expand...

(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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caro24

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plz can someone help me..what are the experiments that i can be asked in exams for nuclear physics p2,,and how can i describe them plz plz ......i really need your help..god bless u
 
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geek101

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hussamh10 said:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_21.pdf
Q2 c part how to determine what is the pivot to be taken
bamteck
TeKnOzOr
geek101
Click to expand...

we are going to calculate moments about a point it can be any point, because the forces on A and B are unknown and the force on the midpoint (weight) is known. we will take moments around an unknown point, lets take A as the pivoting point >> to calculate the Force (tension) on B.

distance of the mid point from A = (1.5/2) - 0.3 = 0.45 m
distance of point B from A = 1.5 - 0.3 = 1.2
upward moments = downward moments
therefore, 0.45 x 2.4 = B x 1.2
B = 0.9 N
 
TeKnOzOr

TeKnOzOr

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hussamh10 said:
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_qp_23.pdf
Q1 d part can any one show it how to do it in vector triangl way not mathematically....thanx plzzz
bamteck TeKnOzOr
geek101
Click to expand...
Hehe dude...i always do it by the resolving forces methods..I actually didn't learn the vector methods..I tried to use ONLY one method for not confusing!!
Why don't you use the resolving method because its the easiest,according to me..!! ;) :)
If yo don't know this method,just tell me i'll explain to you man..Don't worry ;)
 
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