When the oscillator is at maximum extension, we know all of its energy is in Potential Energy, so if the total oscillation energy is 4.1 J, we know that at maximum displacement of 0.2 m, that
<span>energy = 1/2 kA^2 where A= 0.2 m </span>
<span>k= 2E / A^2 = 2*4.1 J /0.2^2=200 N/m </span>
<span>the frequency of oscillation is (1/2pi) sqrt[k/m] </span>
<span>knowing k and m, we can substitute values and find frequency</span>
Answer:
F = 32.28 N
Explanation:
For this exercise we must use the rotational equilibrium relation
Σ τ = 0
In the initial configuration it is in equilibrium, for which all the torque and forces are compensated. By the time the payment lands on the bar, we assume that the counter-clockwise turns are positive.
W_bird L / 2 - F_left 0.595 - F_right 0.595 = 0
we assume that the magnitude of the forces applied by the hands is the same
F_left = F_right = F
W_bird L / 2 - 2 F 0.595 = 0
F = 
we calculate
F = 0.560 9.8 14.0 /2.38
F = 32.28 N
Answer:
The greatest force of gravity on the ball will occur at the point when the ball is near to hit the ground
Explanation:
We know that the earth's center attracts everything towards its center with an acceleration of 9.8 m/s² so it simply means that the change in velocity must occur to produce acceleration. When the ball comes towards the earth, its speed continuously increases and it is at maximum level when it is about to hit the ground so this is the point where gravitational force is maximum.
I hope this helps ^_^
