@tms, you have been helping me all throughout and instead of merely telling me the answer you gradually led me here.
I thank you so much, for I know I truly learned.
Because kx=mg, I can say m=kx/g
thus for 2pi sqrt(m/k) i can plug in 2pi sqrt(x/g)
thus 0.5=2pi (sqrt x/9.8) so x=0.062 or...
aha, sorry. k/m --> m/k was a silly mistake of my fault.
Though this is not a pendulum problem at all, out of m/k and L/g I want to use an equation that does not involve m.
I think for in order to let the little m not slip off, the maximum acceleration of the meterstick cannot be greater...
so I did- I used T=0.5s=(2pi)sqrt(L/g)
and I got the 'acceleration' is 157.91...
If I use 0.5s=(2pi)sqrt(k/m) I cannot retrieve anything because m is too small...
Homework Statement
A meterstick is clamped to a tabletop. The end of the meter stick is deflected downwards a small distance x and is released such the end of the meterstick moves up and down in simple harmonic motion. The meterstick is measured to oscillate up and down 10 times in 5.0...
Homework Statement
An object of mass m falls from rest subject to air resistance force directly proportional to the object's speed, F=-bv, where b is the proportionality constant in N/(m/s) and the negative sign indicates the direction of the force opposite to the direction of the velocity...
ohhh nvm i got it :D
since ma(X)=0.5kx^2 so a=kx/2m,
ma=kx-mg=kx/2 thus kx/2=mg
thus x=2mg/k thus 2x(0). But would I be able to think of all this proof in like a minute or so, which this competition yields per question? =.=
gneill- i took your advice and first saw that the only difference between the motions of the situation when the mass is in equilibrium and the situation of the maximum extension is that for the situation where there is maximum extension, there exists an overall acceleration due upside.
thus I...
Homework Statement
A perfectly elastic spring is attached to the ceiling and a mass m is hanging from the spring. he mass is in equilibrium when the spring is stretched a distance x(o). The mass is carefully lifted and held at rest in the position where the string is nether stretched nor...
I see, thanks guys, but what if
?
What will the tangential acceleration be at each points?
At C, it is mere g acting against the motion of the car
At A, it is g acting in favor of the motion of the car.
However, and B and D, g is not in the direction in the tangential motion- does that...
Does that mean gravity, of which according to the diagram shows that the car is moving 'vertically' at that moment, is also a part of the tangential acceleration?
Mr. Warlock- here it is
Tiny Tim- really? how would I use such equations in this problem?
*I am reposting as I previously posted this in the wrong category.
Homework Statement
Perhaps I am confused by the concept.
A toy car starts from rest at a height 4R above the ground and continue to a loop of radius R (frictionless). At a point C (height R from the ground) inside the...
Homework Statement
Perhaps I am confused by the concept.
A toy car starts from rest at a height 4R above the ground and continue to a loop of radius R (frictionless). At a point C (height R from the ground) inside the loop, what is the tangential acceleration of the toy car?
Homework...
..-_- no, shucks... come on.
This pic's from Nasa's page
As the other guy said I guess I read it wrong, maybe it was talking about the Eagle Nebula all the times.
haha