Solution
distance travelled by Chris
\Delta t=\frac{1}{3600}hr.
X_{c}= [(\frac{21+0}{2})+(\frac{33+21}{2})+(\frac{55+47}{2})+(\frac{63+55}{2})+(\frac{70+63}{2})+(\frac{76+70}{2})+(\frac{82+76}{2})+(\frac{87+82}{2})+(\frac{91+87}{2})]\times\frac{1}{3600}
=\frac{579.5}{3600}=0.161miles
Kelly,
\Delta t=\frac{1}{3600}hr.
X_{k}=[(\frac{24+0}{2})+(\frac{3+24}{2})+(\frac{55+39}{2})+(\frac{62+55}{2})+(\frac{71+62}{2})+(\frac{79+71}{2})+(\frac{85+79}{2})+(\frac{85+92}{2})+(\frac{99+92}{2})+(\frac{103+99}{2})]\times\frac{1}{3600}
=\frac{657.5}{3600}
\Delta X=X_{k}-X_{C}=0.021miles
Answer:
Explanation:
From the question we are told that:
Mass 
Drop distance 
Generally the equation for Spring Constant is mathematically given by
Explanation:
For each object, the initial potential energy is converted to rotational energy and translational energy:
PE = RE + KE
mgh = ½ Iω² + ½ mv²
For the marble (a solid sphere), I = ⅖ mr².
For the basketball (a hollow sphere), I = ⅔ mr².
For the manhole cover (a solid cylinder), I = ½ mr².
For the wedding ring (a hollow cylinder), I = mr².
If we say k is the coefficient in each case:
mgh = ½ (kmr²) ω² + ½ mv²
For rolling without slipping, ωr = v:
mgh = ½ kmv² + ½ mv²
gh = ½ kv² + ½ v²
2gh = (k + 1) v²
v² = 2gh / (k + 1)
The smaller the value of k, the higher the velocity. Therefore:
marble > manhole cover > basketball > wedding ring
It would be 12hz because it
