Qualification > Sciences
ALL CIE PHYSICS DOUBTS HERE !!!
ashish:
--- Quote from: Ghost Of Highbury~ on October 10, 2010, 06:39:44 am ---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?
--- End quote ---
if the string would no have been there then the disc would have rotated upon itself!
just like opening nuts with a Cylinder Opener Key
2) the moment produce will be equal to the couple ( the two string will have opposite but equal forces)
moment = (30/100)* 150
=45Nm
couple is perpendicular distance between the 2 forces * magnitude of one force
45=(12/100)*F
F=375N
sorry but i am getting problem while uploading pictures :(
the tension in the string would be toward the disc
ashish:
--- Quote from: ruby92 on October 09, 2010, 11:05:36 am ---thank you:)
I have two more doubts aswell atm.
From gravitaitonal fields:
The radius of the moons orbit is 3.84*10^8, and its period is 27.4 days.Use keplers law to calculate the period of the orbit of a satellite orbiting the earth just above the earths surface.
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Keplers law T^2 is direcly proportional to R^3
therefore T^2/R^3 = constant
(27.4*24*60*60)^2 / (3.84*10^8)^3 =9.8977*10^-14
let the period be T
T^2 / R^3 = 9.8977*10^-14
T^2 = (9.8977*10^-14)* (6.4*10^6)^3
T = 5093.74 sec
T=1h24min
is that right?
Ghost Of Highbury:
--- Quote from: ashish on October 10, 2010, 07:47:04 am ---if the string would no have been there then the disc would have rotated upon itself!
just like opening nuts with a Cylinder Opener Key
2) the moment produce will be equal to the couple ( the two string will have opposite but equal forces)
moment = (30/100)* 150
=45Nm
couple is perpendicular distance between the 2 forces * magnitude of one force
45=(12/100)*F
F=375N
sorry but i am getting problem while uploading pictures :(
the tension in the string would be toward the disc
--- End quote ---
k thanks a lot, + rep.
ashish:
--- Quote from: Ghost Of Highbury~ on October 10, 2010, 08:00:51 am ---k thanks a lot, + rep.
--- End quote ---
thanks mate :)
Deadly_king:
--- Quote from: ashish on October 10, 2010, 08:00:25 am ---Keplers law T^2 is direcly proportional to R^3
therefore T^2/R^3 = constant
(27.4*24*60*60)^2 / (3.84*10^8)^3 =9.8977*10^-14
let the period be T
T^2 / R^3 = 9.8977*10^-14
T^2 = (9.8977*10^-14)* (6.4*10^6)^3
T = 5093.74 sec
T=1h24min
is that right?
--- End quote ---
Good job pal :)
keep it up ;)
+rep
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