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zalisa [80]
3 years ago
8

A tuning fork vibrates at 15,660 oscillations every minute. What is the period (in seconds) of one back and forth vibration of t

he tuning fork?
Physics
1 answer:
Ierofanga [76]3 years ago
4 0

<u>We are given:</u>

The tuning fork vibrates at 15660 oscillations per minute

<u>Period of one back-and forth movement:</u>

the given data can be rewritten as:

1 minute / 15660 oscillations

60 seconds / 15660 oscillations          <em>(1 minute  = 60 seconds)</em>

<em>dividing the values</em>

0.0038 seconds / Oscillation

Therefore, one back and forth vibration takes 0.0038 seconds

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lisov135 [29]

Answer:

(a) 70cm³

(b) 805 grams

Explanation:

(a) V = L×B×H

= 7cm×5cm×2cm

= 35cm×2cm

= 70cm³

(b) Mass = Volume × Density

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3 0
3 years ago
¿Cuantos metros recorre una motocicleta en un segundo si circula a una velocidad de 90km/h?
Zina [86]

Answer:

La motocicleta recorre 25 metros en 1 segundo si circula a una velocidad de 90 km/h

Explanation:

La velocidad es una magnitud que expresa el desplazamiento que realiza un objeto en una unidad determinada de tiempo, esto es, relaciona el cambio de posición (o desplazamiento) con el tiempo.

Siendo la velocidad es el espacio recorrido en un período de tiempo determinado, entonces 90 km/h indica que en 1 hora la motocicleta recorre 90 km. Entonces, siendo 1 h= 3600 segundos (1 h=60 minutos y 1 minuto=60 segundos) podes aplicar la siguiente regla de tres: si en 3600 segundos (1 hora) la motocicleta recorre 90 km, entonces en 1 segundo ¿cuánta distancia recorrerá?

distancia=\frac{1 segundo*90 km}{3600 segundos}

distancia= 0.025 km

Por otro lado, aplicas la siguiente regla de tres: si 1 km es igual a 1,000 metros, ¿0.025 km cuántos metros son?

distancia=\frac{0.025 km*1,000 metros}{1 km}

distancia= 25 metros

<u><em>La motocicleta recorre 25 metros en 1 segundo si circula a una velocidad de 90 km/h</em></u>

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