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Natali [406]
4 years ago
9

A bug on the surface of a pond is observed to move up and down a total vertical distance of 6.5 cm , from the lowest to the high

est point, as a wave passes. If the ripples decreaseto 4.7 cm, by what factor does thebug's maximum KE change?
Physics
1 answer:
m_a_m_a [10]4 years ago
5 0

Answer:

factor that bug maximum KE change is 0.52284

Explanation:

given data

vertical distance = 6.5 cm

ripples decrease to =  4.7 cm

solution

We apply here formula for the KE of particle that executes the simple harmonic motion that is express as

KE = (0.5) × m × A² × ω²     .................1

and kinetic energy is  directly proportional to square of the amplitude.

so

\frac{KE2}{KE1} =  \frac{A2^2}{A1^2}      .............2

\frac{KE2}{KE1} = \frac{4.7^2}{6.5^2}

\frac{KE2}{KE1} = 0.52284

so factor that bug maximum KE change is 0.52284

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If the gradient of an Extension versus Force graph was found to be 5.6, then what is the value of the spring constant? *
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The speed of light in ________ is about 230 million m/s.
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3 years ago
El motor de una licuadora gira a 3600 rpm, disminuye su velocidad angular hasta 2000 rpm realizando 120 vueltas. Calcular: a) La
allsm [11]

Answer:

a) α = -65,2 rad/s².

b) t = 2,57 s.

Explanation:

a) La aceleración angular se puede calcular usando la siguiente ecuación:

\omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \theta

En donde:

\omega_{f}: es la velocidad angular final =  2000 rpm = 209,4 rad/s

\omega_{0}: es la velocidad angular inicial = 3600 rpm = 377,0 rad/s

α: es la aceleración angular=?

θ: es el desplazamiento o número de vueltas = 120 rev = 754,0 rad

Las conversiones de unidades se hicieron sabiendo que 1 revolución = 2π radianes y que 1 minuto = 60 segundos.  

Resolviendo la ecuación (1) para α, tenemos:

\alpha = \frac{\omega_{f}^{2} - \omega_{0}^{2}}{2\theta} = \frac{(209,4 rad/s)^{2} - (377,0 rad/s)^{2}}{2*754,0 rad} = -65,2 rad/s^{2}  

Entonces, la aceleración angular es -65,2 rad/s². El signo negativo se debe a que el motor está desacelerando.  

b) El tiempo transcurrido se puede encontrar como sigue:

\omega_{f} = \omega_{0} + \alpha t

Resolviendo para t, tenemos:

t = \frac{\omega_{f} - \omega_{0}}{\alpha} = \frac{209,4 rad/s - 377,0 rad/s}{-65,3 rad/s^{2}} = 2,57 s

Por lo tanto, el tiempo transcurrido fue 2,57 s.

Espero que te sea de utilidad!

3 0
3 years ago
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