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3 years ago

How much kinetic energy does a person of mass 50kg running at 3m/s have? Provide

Physics

1 answer:

ddd [48]3 years ago

4 0

Answer:

0.225 KJ kinetic energy is used for a person of mass 50 Kg running at speed of 3 m/s.

Explanation:

Given:

mass = 50 kg

velocity = v = 3 m/s

To Find :

Kinetic energy = ?

Solution:

Kinetic energy of an object is the energy that it possesses due to its motion.

The formula for kinetic energy is given as

$\textrm{kinetic energy} = \frac{1}{2}\times mass\times velocity^{2}\\= \frac{1}{2}\times m\times v^{2}\\= \frac{1}{2}\times 50\times 3^{2}\\= 25\times 9\\= 225\ Joules.$

But we required kinetic energy in KiloJoules i.e KJ

1 K = 1000 = $10^{3}$

$\textrm{kinetic energy} = 225\ J\\= 0.225\times 1000\\= 0.225 \times 10^{3}\\= 0.225\ KJ$

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Two astronauts, each with a mass of 50 kg, are connected by a 7 m massless rope. Initially they are rotating around their center

kiruha [24]

Answer:

The angular velocity is $w_f = 1.531 \ rad/ s$

Explanation:

From the question we are told that

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Generally the radius is mathematically represented as $r_i = \frac{d_i}{2} = \frac{7}{2} = 3.5 \ m$

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Here $I_{k_1 }$ is the initial moment of inertia of the first astronauts which is equal to $I_{p_1}$ the initial moment of inertia of the second astronauts So

$I_{k_1} = I_{p_1 } = m * r_i^2$

Also $w_{k_1 }$ is the initial angular velocity of the first astronauts which is equal to $w_{p_1}$ the initial angular velocity of the second astronauts So

$w_{k_1} =w_{p_1 } = w_1$

Here $I_{k_2 }$ is the final moment of inertia of the first astronauts which is equal to $I_{p_2}$ the final moment of inertia of the second astronauts So

$I_{k_2} = I_{p_2} = m * r_f^2$

Also $w_{k_2 }$ is the final angular velocity of the first astronauts which is equal to $w_{p_2}$ the final angular velocity of the second astronauts So

$w_{k_2} =w_{p_2 } = w_2$

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=> $2 mr_i^2 w_1 = 2 mr_f^2 w_2$

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