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Physics MCQs thread.

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hassam said:
how its .07 j

it wants the ADDITIONAL strain energy.
so at 6N, the extension is 0.03m, E = 0.09
then when it is extended another 0.01m, it becomes 0.04m which is 8.0N
the E = 0.16

0.16-0.09 = 0.07J
 
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QQ..The electric field at a certain distance from an isolated alpha particle is 3.0 x 10^7 NC-1
What is the force on an electron when at that distance from the alpha particle??
a. 4.8 x 10^-12
b. 9.6 x 10^-12
c. 3.0 x 10^7
d. 6.0 x 10^7
 
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melon159 said:
QQ..The electric field at a certain distance from an isolated alpha particle is 3.0 x 10^7 NC-1
What is the force on an electron when at that distance from the alpha particle??
a. 4.8 x 10^-12
b. 9.6 x 10^-12
c. 3.0 x 10^7
d. 6.0 x 10^7

F= Eq
F = 3 x 10^7 x 1.6 x 10 ^(-19)
F = A

the q is always the charge of the charge that is moving, not the charge of the electric field
 
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young modulus=(f x L)/(A x e)
2.0 x 10 ^ 11=(20/ pie d^2/4) x (L/e)
then calculate (e/l) x 100 % which is 5.1 x 10^-2
 
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can anyone tell me the answer of question number 2 of may june 2001.. apparently, the mark scheme isn't available on any site :\
 
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its this question abt forces acting on different positions of a circular disc.. they asked to find the direction of the resultant force.. sorry i cant give any direct link cause i cant find its soft copy on the net :\
 
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How is it 0! , can someones pls show me the directions in which the 3 forces must act in order to produce a resultant of 0!?
 
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Saturation said:
How is it 0! , can someones pls show me the directions in which the 3 forces must act in order to produce a resultant of 0!?
Take the resultant vector of 3 and 4 by the pythagoras theorem. The 3 and 4 N forces will be perpedicular to each other while the 5 N force will act at the vertex of it.
 
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its the voltage around any component in the circuit (the component can be bulb/thermister/resister...etc)
 
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