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Physics: Post your doubts here!

xtremeforums

xtremeforums

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Lyfroker said:
someone pleasee........:(
Click to expand...
Winter 05:
Q5: The easiest way to do this is to find the percentage uncertainty in each of mass, breadth, height, length. And add the percentage uncertainties...
(You always add the %uncertainty, whenever the values are being multiplied/divided)...
Sum of % uncertainties is 2.1% (2.1 is the final %uncertainty in the density)
So (x/2.5) x 100 = 2.1. Solve for X...

Q12: Taking moments about P:
The weight of 100 N downwards in the middle of the rod, so at B.
Anticlockwise moments:
(0.6 x 10) + (0.1 x 100) = 16 Nm
Clockwise moments:
(0.4 x 20) = 8 Nm
For equilibrium, the moments should be equal. So we need 16- 8 = 8Nm on the right side of the rod. Moments= force x dist.
So, 8 = 20 x (dist) = 0.4m to the right from P. Therefore D.

I'll answer the others in a while if no one has still answered them, IsA :)
 
N.M

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Lyfroker

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xtremeforums said:
Winter 05:
Q5: The easiest way to do this is to find the percentage uncertainty in each of mass, breadth, height, length. And add the percentage uncertainties...
(You always add the %uncertainty, whenever the values are being multiplied/divided)...
Sum of % uncertainties is 2.1% (2.1 is the final %uncertainty in the density)
So (x/2.5) x 100 = 2.1. Solve for X...

Q12: Taking moments about P:
The weight of 100 N downwards in the middle of the rod, so at B.
Anticlockwise moments:
(0.6 x 10) + (0.1 x 100) = 16 Nm
Clockwise moments:
(0.4 x 20) = 8 Nm
For equilibrium, the moments should be equal. So we need 16- 8 = 8Nm on the right side of the rod. Moments= force x dist.
So, 8 = 20 x (dist) = 0.4m to the right from P. Therefore D.

I'll answer the others in a while if no one has still answered them, IsA :)
Click to expand...


thnx :)
 
xtremeforums

xtremeforums

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N.M said:
see the carrier frequency is 50KHz ryt so the the period for the carrier wave is 1/50,000= 20µs
the audio frequency is 5KHz, T=200µs
so 200/20=10
so 10 peaks, amplitude modulated

i hope its clear...
Click to expand...
Thankyou!
Can you attach a picture of your graph, if it's not a problem for you?

N.M said:
What's virtual zero?
Click to expand...

The two input voltages (V+ and V-) must be almost the same, otherwise the amplifier would be saturated (I'm not quite sure why this is necessary for it to not be saturated, but this is what I've gathered from books and the markschemes). Since V+ is always connected to earth, it's voltage is 0V. Hence, V- should also be 0v... This is the virtual 0.
 
N.M

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xtremeforums said:
Thankyou!
Can you attach a picture of your graph, if it's not a problem for you?



The two input voltages (V+ and V-) must be almost the same, otherwise the amplifier would be saturated (I'm not quite sure why this is necessary for it to not be saturated, but this is what I've gathered from books and the markschemes). Since V+ is always connected to earth, it's voltage is 0V. Hence, V- should also be 0v... This is the virtual 0.
Click to expand...


image.jpg

hmm... makes sense... jazakillah khairen
 
N.M

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hela said:
no C Can u explain thanks
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yeah i had confusion b/w C n D

see, only B n C r showing full diffraction, but the problem with B is that it is too scattered, that means it diffracted with a larger angle, meaning lrger distance n therefore less intensity
so only C could be the answer
 
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