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SOVA2 [1]
3 years ago
6

In just 0.30 s , you compress a spring (spring constant 5000 n/m ), which is initially at its equilibrium length, by 4.0 cm. par

t a what is your average power output?
Physics
1 answer:
Ierofanga [76]3 years ago
7 0
The average power output is the ratio between the work done to compress the spring, W, and the time taken, t:
P= \frac{W}{t} (1)

The work done is equal to the elastic energy stored by the compressed spring:
W=U= \frac{1}{2}kx^2
where k=5000 N/m is the spring constant and x=4.0 cm=0.04 m is the compression of the spring. If we substitute the numbers, we find:
W= \frac{1}{2}(5000 N/m)(0.04 m)^2=4 J

And now we can use eq.(1) to calculate the average power output:
P= \frac{W}{t}= \frac{4 J}{0.30 s}  =13.3 W
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Answer:

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Explanation:

La expresión tiene un error por omisión, su forma correcta queda descrita a continuación:

<em>"Una prenda de 320 gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene un radio de 40 centímetros y gira con una frecuencia de 4 hertz. Halle </em><em>a)</em><em> el período, </em><em>b) </em><em>la velocidad angular, </em><em>c) </em><em>la fuerza con la que gira la prenda y </em><em>d) </em><em>la velocidad lineal de la lavadora."</em>

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F = \frac{4\pi^{2}\cdot m \cdot r}{T^{2}} (2b)

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m - Masa de la prenda, en kilogramos.

r - Radio interior del tambor, en metros.

(m = 0.32\,kg, r = 0.4\,m, T = 0.25\,s)

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v = \frac{2\pi\cdot r}{T} (3)

(r = 0.4\,m, T = 0.25\,s)

v = \frac{2\pi\cdot (0.4\,m)}{0.25\,s}

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\omega = \frac{2\pi}{T}

(T = 0.25\,s)

\omega = \frac{2\pi}{0.25\,s}

\omega \approx 25.133\,\frac{rad}{s}

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