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Author Topic: Who is up for a challenge?  (Read 37647 times)

Offline lightarrow

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« Reply #125 on: 07/07/2009 19:56:11 »
How can we include √(L/C) in our calculations when we are told nothing about them ?.
Not understood this one.

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We are told nothing about the value of R we can only assume a convinient value.
In my opinion the only information we can derive from the data given is the Q factor (1.25) and the phase angle of the current (53.13).
Exactly. We have 3 equations and 4 unknown. Solved them we have the parameters I wrote, from which we can only find what you said.
 

Offline Chemistry4me

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« Reply #126 on: 09/07/2009 00:40:00 »
Why does Ek = 0.5mv2 but E = mc2
How are the two equations related? What happened to the 'm'?
 

Offline syhprum

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« Reply #127 on: 09/07/2009 05:19:14 »
E=mc^2 is derived from geometric considerations and is relevent as velocities approach c E=0.5*mv^2 belongs to the world before Einstein and Lorentz and does not take into account the changes that take place in M as the velocity increases.
 

Offline lightarrow

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« Reply #128 on: 09/07/2009 13:06:20 »
Why does Ek = 0.5mv2 but E = mc2
They are both false :)

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How are the two equations related? What happened to the 'm'?
Ek = 0.5mv2 is valid only for bodies with non zero mass and at non relativistic speeds; E = mc2 is valid only at zero speed.
The correct one, valid for all speeds and with massive or non-massive bodies (in special relativity) is:

E2 = (cp)2 + (mc2)2

where p = momentum. If m = 0 (photons, gluons, ecc.) then E = cp. At zero velocity p = 0 so E = mc2.

Kinetic energy is total energy minus rest energy:

Ek = E - mc2
« Last Edit: 09/07/2009 13:08:41 by lightarrow »
 

lyner

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« Reply #129 on: 09/07/2009 18:58:14 »
How can we include √(L/C) in our calculations when we are told nothing about them ?.
Not understood this one.

Quote
We are told nothing about the value of R we can only assume a convinient value.
In my opinion the only information we can derive from the data given is the Q factor (1.25) and the phase angle of the current (53.13).
Exactly. We have 3 equations and 4 unknown. Solved them we have the parameters I wrote, from which we can only find what you said.
Yes - the circuit can be scaled up or down in Impedance, according to taste and availability of components. It's just the ratios that count.
 

Offline Chemistry4me

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« Reply #130 on: 10/07/2009 00:17:16 »
Ek = 0.5mv2 is valid only for bodies with non zero mass and at non relativistic speeds; E = mc2 is valid only at zero speed.
The correct one, valid for all speeds and with massive or non-massive bodies (in special relativity) is:

E2 = (cp)2 + (mc2)2

where p = momentum. If m = 0 (photons, gluons, ecc.) then E = cp. At zero velocity p = 0 so E = mc2.

Kinetic energy is total energy minus rest energy:

Ek = E - mc2
Ooh  :o (showing my ignorance here) That's a new one! I must really remember that.
 

Offline Chemistry4me

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« Reply #131 on: 20/07/2009 07:17:53 »
This question has me scratching my head, I can't seem to get the 'correct' answer. [:I] :-\ Maybe I'm not understanding something. It goes something like this: A uniform beam 2.0 metres long is known to weigh between 400 N and 500 N. Using a spring scale, S and a pivot block, P, an experiment was made with the beam balanced horizontally on P and S to enable the weight to be calculated. The spring scale reads 85 N when P is 0.75 m from one end, what is the weight of the beam?
 

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« Reply #131 on: 20/07/2009 07:17:53 »

 

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