Who is up for a challenge?

erickejah · « **Reply #60 on:** 17/06/2009 04:47:22 »

0.883883 m/s^2

Chemistry4me · « **Reply #61 on:** 17/06/2009 11:32:27 »

Err...don't think so.

erickejah · « **Reply #62 on:** 17/06/2009 17:24:22 »

o yeah there is some friction [ $:-\$ ]
Fx=95cos45
Fx=67.1751N

35N of friction:

67.1751N-35N= 32.1751

F=ma
32.1751N=(76Kg)a
a= 0.423357m/s^2

Chemistry4me · « **Reply #63 on:** 18/06/2009 02:39:50 »

Right you are.

erickejah · « **Reply #64 on:** 18/06/2009 02:59:48 »

what is the next one?

Chemistry4me · « **Reply #65 on:** 18/06/2009 03:12:40 »

Haven't found it yet! [

]

Chemistry4me · « **Reply #66 on:** 18/06/2009 06:11:03 »

Calculate the heat energy that is produced by the 3.40 Ω resistor in one minute when the diode is connected in the circuit.

[ Invalid Attachment ]

erickejah · « **Reply #67 on:** 19/06/2009 02:15:37 »

Since the diode is not being polarized and it will never be. the circuit becomes to be a series circuit with a total resistance of:
5.2Ω+3.4Ω = 7.9Ω

the current going trough the circuit is I=12V/7.9Ω → I = 1.51899A
the power in the resistor can be calculated by P=(I^2)*R → 1.51899A*3.4Ω = 5.16456W

then the total power dissipated in 1 minute is equal to:
Energy(watt hours)= Power (Watts)x Time (Hours)
W=(5.16456W)x(1/60)
W=86.076mWh
I think [

] [

]

Chemistry4me · « **Reply #68 on:** 19/06/2009 02:35:50 »

Quote from: erickejah on 19/06/2009 02:15:37

the power in the resistor can be calculated by P=(I²)*R → 1.51899A*3.4Ω = 5.16456W

erickejah · « **Reply #69 on:** 19/06/2009 02:36:32 »

again... !!!!!!!!!!!!!!!!!!!!!!!!! this makes me mad.. let me fix it
___________________
ok its 130.749 mWh

[

]

Chemistry4me · « **Reply #70 on:** 19/06/2009 02:37:44 »

What's with all these mistakes? [

]

erickejah · « **Reply #71 on:** 19/06/2009 02:39:25 »

i dont know i guess that i have a split brain.. lol
--------------------------------
it may be the flow affected by the used of too much Facebook...

anyhow, is this answer close to the one that u had?

Chemistry4me · « **Reply #72 on:** 19/06/2009 02:50:34 »

Quote from: erickejah on 19/06/2009 02:36:32

ok its 130.749 mWh

Um...how did you get that?

Chemistry4me · « **Reply #73 on:** 19/06/2009 02:52:37 »

Can I get an answer in Joules?

erickejah · « **Reply #74 on:** 19/06/2009 02:58:25 »

would it be 7.84492J ?

Chemistry4me · « **Reply #75 on:** 19/06/2009 03:06:53 »

P = E/t

erickejah · « **Reply #76 on:** 19/06/2009 03:08:28 »

so,
7.84492=E/60
E = 470.695J

Chemistry4me · « **Reply #77 on:** 19/06/2009 03:10:08 »

Yes. That was what I got.

erickejah · « **Reply #78 on:** 19/06/2009 03:13:09 »

thanks for baring with me.

erickejah · « **Reply #79 on:** 20/06/2009 01:20:59 »

when are u going to put the next one, I will double check my answers this time..

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