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Alenkasestr [34]
2 years ago
10

A traves de una manguera de 1 in de diámetro fluye gasolina con una velocidad media de 5ft/s ¿cuál es el gasto?

Physics
1 answer:
jonny [76]2 years ago
6 0

Answer:

El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.

Explanation:

El gasto es el flujo volumétrico de gasolina (Q), medido en pies cúbicos por segundo, que sale de la manguera. Asumiendo que la velocidad de salida es constante, tenemos que el gasto a través de la manguera es:

Q = \frac{\pi}{4}\cdot D^{2}\cdot v (1)

Donde:

D - Diámetro de la manguera, medido en pies.

v - Velocidad medida de salida, medida en pies por segundo.

Si sabemos que D = \frac{1}{12}\,ft y v = 5\,\frac{ft}{s }, entonces el gasto de gasolina es:

Q = \frac{\pi}{4}\cdot \left(\frac{1}{12}\,ft \right)^{2} \cdot \left(5\,\frac{ft}{s} \right)

Q \approx 0.0273\,\frac{ft^{3}}{s}

El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.

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Answer:

<h2>44 m/s</h2>

Explanation:

In this problem we are expected to calculate the velocity of Georges movements.

Given data

Total distance covered by  George= 850+250= 1100 meters

Time taken  by  George to cover the total distance= 25 seconds

We know that velocity is, v= distance/ time

Therefore substituting our data into the expression for velocity we have

v= 1100/ 25= 44 m/s

Hence the velocity in m/s is 44

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According to Newton’s second law of motion what is force equals to
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Answer:force equals to rate of change of momentum

Explanation:

F=force

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3 years ago
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2 years ago
A vertical block-spring system on earth has a period of 6.0 s. What is the period of this same system on the moon where the acce
Vera_Pavlovna [14]

Answer:

D) 15s

Explanation:

let Te be the period of the block-spring system on earth and Tm be the period of the same system on the moon.let g1 be the gravitational acceleration on earth and g2 be the gravitational acceleration on the moon.

the period of a pendulum is given by:

T = 2π√(L/g)

so on earth:

Te = 2π√(L/g1)

     =  6s

on the moon;

Tm = 2π√(L/g2)

since g2 = 1/6 g1 then:

Tm = 2π√(L/(1/6×g1))

      = √(6)×2π√(L/(g1))

and 2π√(L/(g1)) = Te = 6s

Tm = (√(6))×6 = 14.7s ≈ 15s

Therefore, the period of the block-spring system on the moon is 15s.

5 0
3 years ago
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Blababa [14]

Answer:

831.4 J

Explanation:

Info given:

mass adult: 82.5kg

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For adult:

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Difference in PE: 1011.65 J - 180.25 J = 831.4 J

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