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Deffense [45]
4 years ago
15

Un carro tiene una velocidad angular de 95 rev/min. a) ¿Cuál es la velocidad tangencial del objeto si se colocan unos contrapeso

s a una distancia de 598 mm? b) Calcule la aceleración centrípeta.
Physics
1 answer:
Maru [420]4 years ago
6 0

Answer:

v = 5.949 m / s ,      a = 59.18 m / s²

Explanation:

Linear and angular quantities are related

           v = w r

we search the magnitudes to the SI system

           w = 95 rev / min (2π rad / 1rev) (1 min / 60 s) = 9.948 rad / s

           d = 598 mm = 0.598 m

a) let's calculate the linear velocity

           v = 9.948 0.598

           v = 5.949 m / s

b) Cenripetal acceleration

            a = v² / r

            a = 5.94 2 / 0.598

            a = 59.18 m / s²

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Example 3 :
kherson [118]

The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.

<h3>How to determine the friction factor</h3>

Using the formula

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Loss When V = 1m/s

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Head loss = 0. 289 × \frac{1}{2*6. 6743 * 10^-11} × \frac{1^2}{0. 120} × \frac{1}{100}

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Head loss When V = 3m/s

Head loss = 0. 096 × \frac{1}{1. 334 *10^-10} × \frac{3^2}{0. 120} × \frac{1}{100}

Head loss = 5. 3× 10^8

Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss  when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.

Learn more about friction here:

brainly.com/question/24338873

#SPJ1

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