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Physics: Post your doubts here!
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Re: Physics Help here! Stuck somewhere?? Ask here! 
If y = W cos 60* =W^2/y , which doesn't make any sense.
Shouldn't y = W/ cos 60*?
Where am I wrong? :S
According to the rules of trigonometry, cos 60* = adj/hyp = W/yunique840 said:
If y = W cos 60* =W^2/y , which doesn't make any sense.
Shouldn't y = W/ cos 60*?
Where am I wrong? :S
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Re: Physics Help here! Stuck somewhere?? Ask here!
Who says that W is adjacent? If you observe, W is actually the hypotenuse. The y and x components of W are actually perpendicular and parallel to the inclined plane, respectively. Whenever you want to find the component of a given slanted force you have to make it the HYPOTENUSE. The components here are with respect to the surface.
Who says that W is adjacent? If you observe, W is actually the hypotenuse. The y and x components of W are actually perpendicular and parallel to the inclined plane, respectively. Whenever you want to find the component of a given slanted force you have to make it the HYPOTENUSE. The components here are with respect to the surface.
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Re: Physics Help here! Stuck somewhere?? Ask here!
This is exactly what I needed to know. Thank you.hamidali391 said:
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Re: Physics Help here! Stuck somewhere?? Ask here!
http://xtremepapers.net/CIE/index.php?d ... 5_qp_1.pdf
Hey everyone,
I need help on question 7 and 11 from october november 2005, Questions 7 and 11.. Thanks in advance :good: :good:
http://xtremepapers.net/CIE/index.php?d ... 5_qp_1.pdf
Hey everyone,
I need help on question 7 and 11 from october november 2005, Questions 7 and 11.. Thanks in advance
:good: :good:
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Re: Physics Help here! Stuck somewhere?? Ask here!
11. the only force acting on the ball is gravity which is downwards so xv is the direction.
11. the only force acting on the ball is gravity which is downwards so xv is the direction.
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Re: Physics Help here! Stuck somewhere?? Ask here!
Guys another question popped up.
A ball is thrown horizontally in still air from the top of a very tall building. The ball is affected by air resistance.
What happens to the horizontal and to the vertical components of the ball’s velocity?
It is from november 2010 paper 13, question 11. please help!!
Guys another question popped up.
A ball is thrown horizontally in still air from the top of a very tall building. The ball is affected by air resistance.
What happens to the horizontal and to the vertical components of the ball’s velocity?
It is from november 2010 paper 13, question 11. please help!!
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11. the only force acting on the ball is gravity which is downwards so xv is the direction.
Btw i just copied it from the guy who answered my question, so thank him ^^
a021011 said:
11. the only force acting on the ball is gravity which is downwards so xv is the direction.
Btw i just copied it from the guy who answered my question, so thank him ^^
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Re: Physics Help here! Stuck somewhere?? Ask here!
11. the horizontal component of velocity decreases to zero as air resistance acting so it will reach finally towards zero velocity. Vertical component increases to a constant value towards the Earth. ball falling in the positive direction downwards.
i can't explain in a clear way as my concept on this is also weak.
babycry said:
11. the horizontal component of velocity decreases to zero as air resistance acting so it will reach finally towards zero velocity. Vertical component increases to a constant value towards the Earth. ball falling in the positive direction downwards.
i can't explain in a clear way as my concept on this is also weak.
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Re: Physics Help here! Stuck somewhere?? Ask here!
The horizontal velocity will decrease to Zero and the veritcal velocity will increase until it reaches a constant value. This question involves the concept of terminal velocity. The vertical velocity will increase initially due to the object's weight by as it is countered by drag the velocity will reach a constant value when the object is in vertical equilibrium. Whereas, the horizonal velocity will decrease to zero because of the consistent air resistance.
babycry said:
The horizontal velocity will decrease to Zero and the veritcal velocity will increase until it reaches a constant value. This question involves the concept of terminal velocity. The vertical velocity will increase initially due to the object's weight by as it is countered by drag the velocity will reach a constant value when the object is in vertical equilibrium. Whereas, the horizonal velocity will decrease to zero because of the consistent air resistance.
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Re: Physics Help here! Stuck somewhere?? Ask here!
http://www.xtremepapers.com/CIE/Interna ... _qp_11.pdf
Q.9 , 22, 34
plz anyone...
http://www.xtremepapers.com/CIE/Interna ... _qp_11.pdf
Q.9 , 22, 34
plz anyone...
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2as=v^2-u^2
2ax=0^2-u^2
a=-u^2/(2x)
2as=v^2-u^2
2(-u^2/(2x))=0^2-(1.20u)^2
-(u^2/x)s=-1.44u^2 U^2 and -ve sign cancel
s=1.44x
Q22)
I did this question before in the phy thread, so here is it again.
Let Force exerted by load be W
let K1 be the spring constant of wire X and K2 be the spring constant of wire Y
because wires are joint and of different length and spring constant, different ratio of forces will act on both of them.
Force on X=3/(3+1) W=3/4 W
Force on Y=1/4 W
For the 1st wire:
X:
3/4 W= K1 (1.5)
k1 = 1/2W
Y:
1/4W=K2 (1)
K2=1/4W
For the 2nd wire:
lengths of wires have changes so will the ratio of forces acting on them
X:
1/4W=K1 x1 where k1=1/2W
1/4W=1/2W x1
x1=1/2 mm
Y:
3/4W=K2 x2 where k2=1/4W
3/4W=1/4W x2
x2=3 mm
total extension=x1 + x2= 0.5 + 3= 3.5 mm
Q34)
Find total resistance
resistance on both wires= 0.005 x 800 x 2= 8 Ohms
resistance of relay=16/0.6= 26.67 Ohms
Total resistance = 34.67 Ohms
emf=34.67 x 0.6= 20.8 V
Q9rviboy said:
2as=v^2-u^2
2ax=0^2-u^2
a=-u^2/(2x)
2as=v^2-u^2
2(-u^2/(2x))=0^2-(1.20u)^2
-(u^2/x)s=-1.44u^2 U^2 and -ve sign cancel
s=1.44x
Q22)
I did this question before in the phy thread, so here is it again.
Let Force exerted by load be W
let K1 be the spring constant of wire X and K2 be the spring constant of wire Y
because wires are joint and of different length and spring constant, different ratio of forces will act on both of them.
Force on X=3/(3+1) W=3/4 W
Force on Y=1/4 W
For the 1st wire:
X:
3/4 W= K1 (1.5)
k1 = 1/2W
Y:
1/4W=K2 (1)
K2=1/4W
For the 2nd wire:
lengths of wires have changes so will the ratio of forces acting on them
X:
1/4W=K1 x1 where k1=1/2W
1/4W=1/2W x1
x1=1/2 mm
Y:
3/4W=K2 x2 where k2=1/4W
3/4W=1/4W x2
x2=3 mm
total extension=x1 + x2= 0.5 + 3= 3.5 mm
Q34)
Find total resistance
resistance on both wires= 0.005 x 800 x 2= 8 Ohms
resistance of relay=16/0.6= 26.67 Ohms
Total resistance = 34.67 Ohms
emf=34.67 x 0.6= 20.8 V
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nov 2010 question 9 paper 11
u = initial speed V = final speed = 0 distance = x
0 = u^2 + 2ax so:
2x = -u^2/a
x = -u^2/2a
x= Ku^2 where K is a constant of -1/2a
So x∞u^2
The second train has a speed, u, of 1.2 u
Since x∞u2 ,but u is now 1.2 u
x = u^2
? = (1.2 u)^ 2
When we cross multiply the answer is 1.44x.
Hope i helped :lol: :lol: :lol:
nov 2010 question 9 paper 11
u = initial speed V = final speed = 0 distance = x
0 = u^2 + 2ax so:
2x = -u^2/a
x = -u^2/2a
x= Ku^2 where K is a constant of -1/2a
So x∞u^2
The second train has a speed, u, of 1.2 u
Since x∞u2 ,but u is now 1.2 u
x = u^2
? = (1.2 u)^ 2
When we cross multiply the answer is 1.44x.
Hope i helped :lol: :lol: :lol:
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A thermometer can read to an accuracy of 0.5 C
The thermometer is used to measure a temperature rise from 40 C to 100 C
What is the percentage uncertainty in the measurement of temperature rise ?
post answr ASAP ... thanks
A thermometer can read to an accuracy of 0.5 C
The thermometer is used to measure a temperature rise from 40 C to 100 C
What is the percentage uncertainty in the measurement of temperature rise ?
post answr ASAP ... thanks
XPFMember
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Re: Physics Help here! Stuck somewhere?? Ask here!
absolute error = 0.5 / 2 = 0.25
%uncertainty = 0.25/(ur reading) * 100
Assalamoalaikum!mury456 said:A thermometer can read to an accuracy of 0.5 C
The thermometer is used to measure a temperature rise from 40 C to 100 C
What is the percentage uncertainty in the measurement of temperature rise ?
post answr ASAP ... thanks
absolute error = 0.5 / 2 = 0.25
%uncertainty = 0.25/(ur reading) * 100
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I wanted to ask which paper are we going to give, the practical one for Physics and Chemistry...? The timetable has not mentioned Paper 32... It has only 33 and 34 for the 2012 May June session...
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s10_qp_13.pdf
Question 26
Help please
Question 26
Help please
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the electron beam is directed at right angles from the observer so it will deflect into the positive end of the shaded area n deflected towards the observer at an angle of 90 degree. its kinda guess..or maybe i think the beam will be reflected after hitting the shaded area towards the observer...it may act at a specific angle nd then go back out of the plane paper...