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salantis [7]
3 years ago
15

A mass on a spring A oscillates at twice the frequency of the same mass on spring B. Which statement is correct?A.The spring con

stant for B is one quarter of the spring constant for A.B.The spring constant for B is 4 times the spring constant for A.C.The spring constant for B is half of the spring constant for A.D.The spring constant for B is 1.41 times the spring constant for A.E.The spring constant for B is twice the spring constant for A.
Physics
1 answer:
Nataliya [291]3 years ago
3 0

Answer:

A.The spring constant for B is one quarter of the spring constant for A.

Explanation:

If spring A oscillates at twice the frequency of spring B, and period is frequency inverted. It means spring B has a period twice of spring A's.

T_B = 2T_A

As T = 2\pi\sqrt{\frac{m}{k}}, and the 2 springs have the same mass

2\pi\sqrt{\frac{m}{k_B}} = 2\pi\sqrt{\frac{m}{k_A}}

\sqrt{k_A} = 2\sqrt{B}

k_A = 4k_B

k_B = k_A/4

So A.The spring constant for B is one quarter of the spring constant for A. is the correct answer.

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Answer:

a) La aceleración angular es: \alpha=2\: rad/s^{2}

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Explanation:

a)

La aceleración angular se define como:

\alpha=\frac{\Delta \omega}{\Delta t}

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b)

El desplazamiento angular puede ser calculado usando la siguiente ecuación:

\theta=\theta_{i}+\omega_{i}t+\frac{1}{2}\alpha t^{2}

Aqui el angulo inicial es 0, por lo tanto.

\theta=20(5)+\frac{1}{2}(2)(5)^{2}

\theta=125\: rad

El engranaje gira 125 radianes.

c)

Lo que debemos hacer aquí es convertir radianes a revoluciones.

Recordemos que 2π rad = 1 rev

Entonces:

\theta=125\: rad \times \frac{1\: rev}{2\pi\: rad}=19.89\: rev

Por lo tanto el engranaje hara aproximadamente 20 revoluciones.

Espero te haya sido de ayuda!

6 0
3 years ago
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Answer:

13.6 cm

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sin θ = x / √(x² + 144)

Substituting:

3/4 = x / √(x² + 144)

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