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Darya [45]
3 years ago
13

sonic is sliding down a frictionless 15m tall hill. He starts at the top with a velocity of 10m/s. At the bottom of the hill he

hits a spring and compresses IT AT .89 m What is the maximum velocity of Sonic and what is the spring constant of the spring?

Physics
1 answer:
podryga [215]3 years ago
5 0

Answer:

The maximum speed of sonic at the bottom of the hill is equal to 19.85m/s and the spring constant of the spring is equal to (497.4xmass of sonic) N/m

Energy approach has been used to sole the problem.

The points of interest for the analysis of the problem are point 1 the top of the hill and point 2 the bottom of the hill just before hitting the spring

The maximum velocity of sonic is independent of the his mass or the geometry. It is only depends on the vertical distance involved

Explanation:

The step by step solution to the problem can be found in the attachment below. The principle of energy conservation has been applied to solve the problem. This means that if energy disappears in one form it will appear in another.

As in this problem, the potential and kinetic energy at the top of the hill were converted to only kinetic energy at the bottom of the hill. This kinetic energy too got converted into elastic potential energy .

x = compression of the spring = 0.89

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Answer:

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Explanation:

It is given that,

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Energy of a gas molecules is given by :

\Delta E =\dfrac{3}{2}NkT

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3 years ago
In a Broadway performance, an 84.5-kg actor swings from a R = 4.30-m-long cable that is horizontal when he starts. At the bottom
Arada [10]

Answer:

1.57772 m

Explanation:

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As the energy of the system is conserved

\frac{1}{2}MV^2=Mgh_i\\\Rightarrow V=\sqrt{2gh_i}\\\Rightarrow V=\sqrt{2\times 9.81\times 4.3}\\\Rightarrow V=9.18509\ m/s

As the linear momentum is conserved

MV=(m+M)v\\\Rightarrow v=\frac{MV}{m+M}\\\Rightarrow V=\frac{84.5\times 9.18509}{84.5+55}\\\Rightarrow v=5.56372\ m/s

Applying conservation of energy again

\frac{1}{2}(m+M)v^2=(m+M)gh_f\\\Rightarrow h_f=\frac{v^2}{2g}\\\Rightarrow h_f=\frac{5.56372^2}{2\times 9.81}\\\Rightarrow h_f=1.57772\ m

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P = 4.1 kg × 9.8 m/s² × 4.5 m

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