Answer:
<h2>
f₀ = 158.12 Hertz</h2>
Explanation:
The fundamental frequency of the string f₀ is expressed as f₀ = V/4L where V is the speed experienced by the string.
where T is the tension in the string and 
is the density of the string
Given T = 600N and = 0.015 g/cm = 0.0015kg/m
The next is to get the length L of the string. Since the string is stretched and fixed at both ends, 200 cm apart, then the length of the string in metres is 2m.
L = 2m
Substituting the derived values into the formula f₀ = V/2L
f₀ = 632.46/2(2)
f₀ = 632.46/4
f₀ = 158.12 Hertz
Hence the fundamental frequency of the string is 158.12 Hertz
Impulse = change in momentum
The answer is 0.
The last one is correct (D)
A)Ep'=mgh=mgl(1-cosa).At the bottom of the swing Ep=0(reference level),so the potential energy as the child is just released is bigger than the potential energy at the bottom of the swing.;B)The speed of the child at the bottom of the swing-->v=√(2gh)=√[2gl(1-cosa)];C)I don't think that the tension does any work.
Answer:
well, as u can tell the top layer will always be the youngest layer aka the newest layer. The farther u go down the older the layers get. So the deeper u dig the farther back in time we see.
Explanation:
