A car goes from point A to point B, five miles away and then returns to point A. The car is going 15 mph.

Based on Hooke's law, the spring constant of the the body's muscle mechanism is the ratio of force to extension, the effective mass is m/3 and the potential energy that can be stored is ke^2 / 2.

<h3>What is the spring constant?</h3>

The spring constant or stiffness constant of an elastic spring is constant which describes the extent a bit forceapplied to an elastic spring will extend it.

Spring constant, K = force/extension

Assuming, a body's muscle mechanism is a spring obeying Hooke's law, the effective mass of the spring with mass m is 1/3 of the mass of the spring = m/3

The potential energy that can be stored = ke^2 / 2

where K is spring constant and e is the extension produced.

Therefore, the spring constant of the the body's muscle mechanism is the ratio of force to extension, the effective mass is m/3 and the potential energy that can be stored is ke^2 / 2.

Learn more about Hooke's law at: brainly.com/question/12253978

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2 years ago

A rocket of mass 1000kg uses 5kg of fuel and oxygen to produce exhaust gases ejected at 500m/s calculate the increase its veloci

Vlad [161]

Answer:

Approximately $\rm 2.5\; m \cdot s^{-1}$ .

Explanation:

Let the increase in the rocket's velocity be $\Delta v$ . Let $v_0$ represent the initial velocity of the rocket. Note that for this question, the exact value of $v_0$ doesn't really matter.

The momentum of an object is equal to its mass times its velocity.

Mass of the rocket with the 5 kg of fuel: $1000$ .
Initial velocity of the rocket and the fuel: $v_0$ .
Hence the initial momentum of the rocket: $1000\,v_0$ .

Mass of the rocket without that 5 kg of fuel: $1000 - 5 = 995$ .
Final velocity of the rocket: $v_0 + \Delta v$ .
Hence the final momentum of the rocket: $995\,(v_0 + \Delta v)$ .

Mass of the 5 kg of fuel: $5$ .
Final velocity of the fuel: $v_0 - 500$ (assuming that the the 500 m/s in the question takes the rocket as its reference.)
Hence the final momentum of the fuel: $5\,(v_0 - 500)$ .