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ALL CIE PHYSICS DOUBTS HERE !!!

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ruby92:
m/j 2008 paper 4 q1 b

ashish:

--- Quote from: ruby92 on October 09, 2010, 02:41:35 pm ---m/j 2008 paper 4 q1 b

--- End quote ---

the centripetal force is provided by the frictional force..

note w=omega
W=weight

Fc= Frictional force
M(w^2)r =0.72W
M(w^2)(35/100) =0.72Mg
M cancel out

w^2=(0.72*9.81)/0.35

w=4.49 rad/s


frequency of rotation =  w/2pi
                             =0.7146
in 60 sec no. of rotation = 0.7146*60
=43

ruby92:
THANK YOU :)
also could someone please help me with oct/nov 2009 paper 41 q1 bii

ashish:
ok

Gravitational force = centripetal force
(GMm)/x^2 = Mw^2 x            (w=omega)
m is eliminated on both side
GM= w^2 x^3-----------1

we know that g is Gravitational field strength on earth surface

g=GM/R^2

BY making GM subject of formula

GM = gR^2------------

now
now replace GM by gR^2 in 1

gR^2 =w^2 x^3

:D

Ghost Of Highbury:
Question Attached.

Doubt : (i) (direction)

           2) For a force F of magnitude 150N, determine the force in S1.

I didn't understand the diagram very well. How exactly will the system move if a force is applied to the end of the lever. Also, how are equal forces produced in the strings? Could someone explain please?

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