Un cuerpo de m=0,5 Kg se desplaza horizontalmente con v=4m/s y luego de un lapso de tiempo se mueve con v=20 m/s. ¿cual ha sido la variación de la energÍa cinética?

Answers

Answer 1

Responder:

64 Julios

Explicación:

La energía cinética se expresa mediante la fórmula KE = 1 / 2mv² donde;

m es la masa del cuerpo

v es la velocidad del objeto

Dado que el cuerpo se mueve horizontalmente con v = 4 m / sy después de un período de tiempo se mueve con v = 20 m / s, entonces la variación en la velocidad será de 20 m / s - 4 m / s = 16 m / s.

Parámetros dados

masa del objeto m = 0,5 kg

Variación de velocidad = 16 m / s

Variación de la energía cinética = 1/2 * 0,5 * 16²

Variación de la energía cinética = 1/2 * 0,5 * 256

Variación de la energía cinética = 0,5 * 128

Variación de la energía cinética = 64 Julios


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Answer:

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Answer:

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Answers

Given :

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Answer :

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_________________________________

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Answers

Answer:

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Answers

Answer:

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Hey there!

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Answers

Answer:

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Answers

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Answer:

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Answers

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Convert 543 g to kg (Use correct number of significant figures for answer).
a
Ob
e
0.5 kg
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0.543 kg
5.0 kg
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A roadrunner at rest suddenly spots a rattlesnake slithering directly away at a constant speed of 0.75 m/s. At the moment the snake is 10. Meters from the bird, the roadrunner starts chasing it with a constant acceleration of 1.0 m/s^2. How long will it take the roadrunner to catch up to the snake?

Answers

Answer:

The roadrunner will take approximately 5.285 seconds to catch up to the rattlesnake.

Explanation:

From the statement we notice that:

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2) Initial distance between the roadrunner and rattlesnake is 10 meters. ([tex]x_{o, R} = 0\,m[/tex], [tex]x_{o,S} = 10\,m[/tex])

3) The roadrunner catches up to the snake at the end. ([tex]x_{S} = x_{R}[/tex])

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[tex]x_{S} = x_{o,S}+v_{S}\cdot t[/tex]

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[tex]x_{o, S}[/tex] - Initial position of the rattlesnake, measured in meters.

[tex]x_{S}[/tex] - Final position of the rattlesnake, measured in meters.

[tex]v_{S}[/tex] - Speed of the rattlesnake, measured in meters per second.

[tex]t[/tex] - Time, measured in seconds.

Roadrunner

[tex]x_{R} = x_{o,R} +v_{o,R}\cdot t +\frac{1}{2}\cdot a\cdot t^{2}[/tex]

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[tex]x_{o, R}[/tex] - Initial position of the roadrunner, measured in meters.

[tex]x_{R}[/tex] - Final position of the roadrunner, measured in meters.

[tex]v_{o,R}[/tex] - Initial speed of the roadrunner, measured in meters per second.

[tex]a[/tex] - Acceleration of the roadrunner, measured in meters per square second.

[tex]t[/tex] - Time, measured in seconds.

By eliminating the final positions of both creatures, we get the resulting quadratic function:

[tex]x_{o,S}+v_{S}\cdot t = x_{o,R}+v_{o,R}\cdot t +\frac{1}{2}\cdot a \cdot t^{2}[/tex]

[tex]\frac{1}{2}\cdot a \cdot t^{2} + (v_{o,R}-v_{S})\cdot t + (x_{o,R}-x_{o,S}) = 0[/tex]

If we know that [tex]a = 1\,\frac{m}{s^{2}}[/tex], [tex]v_{o, R} = 0\,\frac{m}{s}[/tex], [tex]v_{S} = 0.75\,\frac{m}{s}[/tex], [tex]x_{o, R} = 0\,m[/tex] and [tex]x_{o,S} = 10\,m[/tex], the resulting expression is:

[tex]0.5\cdot t^{2}-0.75\cdot t -10=0[/tex]

We can find its root via Quadratic Formula:

[tex]t_{1,2} = \frac{-(-0.75)\pm \sqrt{(-0.75)^{2}-4\cdot (0.5)\cdot (-10)}}{2\cdot (0.5)}[/tex]

[tex]t_{1,2} = \frac{3}{4}\pm \frac{\sqrt{329}}{4}[/tex]

Roots are [tex]t_{1} \approx 5.285\,s[/tex] and [tex]t_{2}\approx -3.785\,s[/tex], respectively. Both are valid mathematically, but only the first one is valid physically. Hence, the roadrunner will take approximately 5.285 seconds to catch up to the rattlesnake.

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