This question sounds like it came after some activity where
some forces were observed. Since we were not there, and
we don't know what the activity was, we don't know what forces
were observed, and we have no clue to how they might be related
to the motion of the Earth around the sun.
Answer:
f= 4,186 10² Hz
Explanation:
El sistema descrito es un pendulo de torsión que oscila con con velocidad angular, que esta dada por
w = √ k/I
donde ka es constante de torsion de hilo e I es el momento de inercia del disco
El momento de inercia de indican que giran un eje que pasa por enronqueces
I= ½ M R2
reduzcamos las cantidades al sistema SI
R= 1,4 cm = 0,014 m
M= 430 g = 0,430 kg
substituimos
w= √ (2 k/M R2)
calculemos
w = RA ( 2 370 / (0,430 0,014 2)
w = 2,963 103 rad/s
la velocidad angular esta relacionada con la frecuencia por
w =2pi f
f= w/2π
f= 2,963 10³/ (2π)
f= 4,186 10² Hz
I need more details like r u reading from something
You are currently converting Distance and Length units from Centimeters to Feet 321 Centimeters (cm) = 10.5315 Feet (ft) This is a hard one but see if this helps if not let me now and i can try again..
Answer:
1) 0.43 meters per second
2) 0.21 meters per second
3) 1.02 
4) 0.66 seconds
Explanation:
part 1
By conservation of energy, the maximum kinetic energy (K) of the block is at equilibrium point where the potential energy is zero. So, at the equilibrium kinetic energy is equal to maximum potential energy (U):


With m the mass, v the speed, k the spring constant and xmax the maximum position respect equilibrium position. Solving for v

part 2
Again by conservation of energy we have kinetic energy equal potential energy:


part 3
Acceleration can be find using Newton's second law:

with F the force, m the mass and a the acceleration, but elastic force is -kx, so:


part 4
The period of an oscillator is the time it takes going from one extreme to the other one, that is going form 4.5 cm to -4.5 cm respect the equilibrium position. That period is:

So between 0 and 4.5 cm we have half a period:
