A spring is hung vertically from a hook. A 500g mass is hung from the spring and the spring stretches 0.01m from equilibrium. If the mass is then pulled 0.02m MORE and released what will be the acceleration of the mass just as it is released? (show work)

See the explanation below.

Explanation:

By means of Hooke's second law, we can calculate the spring constant with the initial conditions of the problem.

where:

k = spring constant [N/m]

x = distance = 0.01 [m]

F = force or weight [N]

Weight can be calculated by means of the product of mass by gravitational acceleration.

m = mass = 500[g] = 0.5[kg]

Now the spring constant can be calculated:

We must now use the same Hooke's law to determine the new spring force when stretching it 0.03 [m]

Now we have to use Newton's second law to calculate acceleration. We must remember that Newton's Second Law tells us that the sum of forces is equal to the product of mass by acceleration.

where:

F = Fnew = 14.715[N]

m = mass = 500 [g] = 0.5 [kg]

a = acceleration [m/s²]

Now replacing:

the answer is a. the yacht traveled from the dock north to the 200-meter marker in the bay in less than 5 minutes, giving the passengers several more hours to fish.

im doing peak rn so sry if its wrong

as we know our atmosphere or air have different levels. every level has other gasses, and every gas has another permeability. when the light comes from lower permeability to higher permeability, it changes its direction. so that means that when the light comes from the moon to the earth, it changes it's direction many more times. it basically means that the shape of the light part of the moon is changed on the earth.

hope this ! this is not copied! search it up, it's free of plagiarism!