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Ugo [173]
3 years ago
6

If you are building a clock by using a pendulum's oscillations to keep time and you find the period is a little too small (the c

lock counts seconds too quickly), what should you do to adjust the pendulum's period? O
a. Increase the length of the string
b. Increase the mass on the end of the string
c. Increase the amplitude of the swing
d. Decrease the length of the string
e. Decrease the mass on the end of the string
Physics
1 answer:
Leviafan [203]3 years ago
3 0

Time period of pendulum is given by

T = 2\pi\sqrt{\frac{L}{g}}

Here,

L is length

g is acceleration due to gravity

Therefore we can determine that the period of the pendulum is directly proportional to the square root of the length of the rope.

T \propto \sqrt{L}

Therefore, it is necessary to increase the length of the rope to increase the period so that it counts the seconds more slowly. The correct answer is A.

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Un movil viaja a 40km/h y comienza a reducir su velocidad a partir del instante t=0. Al cabo de 6 segundo se detiene completamen
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Answer:

1,85 m / s²

Explanation:

De la pregunta anterior, se obtuvieron los siguientes datos:

Velocidad inicial (u) = 40 km / h

Hora inicial (t₁) = 0

Tiempo final (t₂) = 6 s

Velocidad final (v) = 0

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A continuación, convertiremos 40 km / ha m / s. Esto se puede obtener de la siguiente manera:

1 km / h = 0,2778 m / s

Por lo tanto,

40 km / h = 40 km / h × 0,2778 m / s / 1 km / h

40 km / h = 11,11 m / s

Por tanto, 40 km / h equivalen a 11,11 m / s.

Finalmente, determinaremos la aceleración del móvil durante el período en el que desaceleró. Esto se puede obtener de la siguiente manera:

Velocidad inicial (u) = 11,11 m / s

Hora inicial (t₁) = 0

Tiempo final (t₂) = 6 s

Velocidad final (v) = 0

Aceleración (a) =?

a = (v - u) / (t₂ - t₁)

a = (0 - 11,11) / (6 - 0)

a = - 11,11 / 6

a = –1,85 m / s²

Por tanto, la aceleración del móvil durante el período en el que se ralentizó es de –1,85 m / s²

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3 years ago
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Answer:

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