# The Naked Scientists Forum

### Author Topic: Who is up for a challenge?  (Read 37646 times)

#### Chemistry4me

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##### Who is up for a challenge?
« Reply #50 on: 12/06/2009 09:14:31 »
Alright, see if you can work this one out then.

Phugoid oscillations.
A boy is flying his radio-controlled aeroplane (mass = 3.67 kg) at a speed of 36.0 ms–1 in a straight and level flight at an altitude of 50.0 m. The plane suddenly experiences a region of turbulence, causing it to lift several metres above its original altitude. The amplitude of the vertical oscillations is 4.56 m. Calculate the maximum upward force acting on the plane during the time that the plane is oscillating.

#### erickejah

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##### Who is up for a challenge?
« Reply #51 on: 13/06/2009 03:45:19 »
F=ma
F=(3.67Kg)(9.8m/s^2)
F=35.966N

PE1=mgh
PE1=(3.67Kg)(9.8m/s^2)(50m)     PE2=(3.67Kg)(9.8m/s^2)(54.67m)
PE1=1798.3J                              PE2=1962.3J

= Fx
PE1   PE2

35.966N =  Fx
-------   ----
1798.3J   1962.3J

Fx=39.246

ΔF= Fx-F → 39.246-35.966 = 3.28N

This is my assumption without researching about Phugoid oscillations. I will try to look that theory ASAP to do it right. I wonder how different that answer will be

#### Chemistry4me

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##### Who is up for a challenge?
« Reply #52 on: 13/06/2009 03:52:12 »
I did not get 3.28 N.

#### erickejah

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##### Who is up for a challenge?
« Reply #53 on: 13/06/2009 23:27:10 »
Can you show me the path, dont I need the AoA, or the time in which it changes?

#### Chemistry4me

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##### Who is up for a challenge?
« Reply #54 on: 13/06/2009 23:36:20 »

#### erickejah

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##### Who is up for a challenge?
« Reply #55 on: 13/06/2009 23:36:55 »
kind of,,

#### Chemistry4me

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##### Who is up for a challenge?
« Reply #56 on: 13/06/2009 23:48:02 »
It look's something like this:

But the plane experiences an upward force of 3.67 x 9.8 N just to keep it from falling, therefore Fmax(total) = 38.45 N

#### erickejah

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##### Who is up for a challenge?
« Reply #57 on: 13/06/2009 23:59:04 »
nice :)

#### erickejah

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##### Who is up for a challenge?
« Reply #58 on: 14/06/2009 18:55:07 »
another one?

#### Chemistry4me

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##### Who is up for a challenge?
« Reply #59 on: 15/06/2009 03:42:42 »
John is sitting in a trolley and Sam pulls him along with a rope. John's mass is 65 kg, the trolley's mass is 11 kg. The tension force in the rope attached to the trolley is 95 N, and the rope is at an angle of 45o to the ground. There is a 35 N friction force on the trolley. Calculate the size of the trolley's acceleration.

#### erickejah

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##### Who is up for a challenge?
« Reply #60 on: 17/06/2009 04:47:22 »
0.883883 m/s^2

#### Chemistry4me

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##### Who is up for a challenge?
« Reply #61 on: 17/06/2009 11:32:27 »
Err...don't think so.

#### erickejah

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##### Who is up for a challenge?
« 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

• Neilep Level Member
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##### Who is up for a challenge?
« Reply #63 on: 18/06/2009 02:39:50 »
Right you are.

#### erickejah

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##### Who is up for a challenge?
« Reply #64 on: 18/06/2009 02:59:48 »
what is the next one?

#### Chemistry4me

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##### Who is up for a challenge?
« Reply #65 on: 18/06/2009 03:12:40 »
Haven't found it yet!

#### Chemistry4me

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##### Who is up for a challenge?
« 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.

#### erickejah

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##### Who is up for a challenge?
« 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

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##### Who is up for a challenge?
« Reply #68 on: 19/06/2009 02:35:50 »
the power in the resistor can be calculated by P=(I2)*R → 1.51899A*3.4Ω = 5.16456W

#### erickejah

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##### Who is up for a challenge?
« Reply #69 on: 19/06/2009 02:36:32 »
again... !!!!!!!!!!!!!!!!!!!!!!!!! this makes me mad.. let me fix it
___________________
ok its 130.749 mWh

« Last Edit: 19/06/2009 02:38:12 by erickejah »

#### Chemistry4me

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##### Who is up for a challenge?
« Reply #70 on: 19/06/2009 02:37:44 »
What's with all these mistakes?

#### erickejah

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##### Who is up for a challenge?
« 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...

« Last Edit: 19/06/2009 02:43:48 by erickejah »

#### Chemistry4me

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##### Who is up for a challenge?
« Reply #72 on: 19/06/2009 02:50:34 »
ok its 130.749 mWh
Um...how did you get that?

#### Chemistry4me

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##### Who is up for a challenge?
« Reply #73 on: 19/06/2009 02:52:37 »
Can I get an answer in Joules?

#### erickejah

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##### Who is up for a challenge?
« Reply #74 on: 19/06/2009 02:58:25 »
would it be 7.84492J ?

#### The Naked Scientists Forum

##### Who is up for a challenge?
« Reply #74 on: 19/06/2009 02:58:25 »