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3 years ago

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]3 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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Trey races his bicycle for 192m. A wheel of his bicycle turns 48 times as the bicycle travels this distance. What is the diamete

chubhunter [2.5K]

Answer:

1.27 m

Explanation:

Distance = 192 m

number of rotations = 48

Distance traveled in one rotation = 2 x π x r

Where, r be the radius of wheel.

so, distance traveled in 48 rotations = 48 x 2 x 3.14 x r

It is equal to the distance traveled.

192 = 48 x 2 x 3.14 x r

r = 0.637 m

diameter of wheel = 2 x radius of wheel = 2 x 0.637 = 1.27 m

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2 years ago

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Two point charges, Q1 and Q2, are separated by a distance R. If the magnitudes of both charges are tripled and their separation

stira [4]

Answer:

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Explanation:

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2 years ago

What do you mean by pressure

noname [10]

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3 years ago

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What is the total entropy change if a 0.280 efficiency engine takes 3.78E3 J of heat from a 3.50E2 degC reservoir and exhausts i

babymother [125]

Answer:

$\Delta S=1.69J/K$

Explanation:

We know,

$\eta=1-\frac{T_2}{T_1}=1-\frac{Q_2}{Q_1}$ ..............(1)

where,

η = Efficiency of the engine

T₁ = Initial Temperature

T₂ = Final Temperature

Q₁ = Heat available initially

Q₂ = Heat after reaching the temperature T₂

Given:

η =0.280

T₁ = 3.50×10² °C = 350°C = 350+273 = 623K

Q₁ = 3.78 × 10³ J

Substituting the values in the equation (1) we get

$0.28=1-\frac{Q_2}{3.78\times 10^{3}}$

$\frac{Q_2}{3.78\times 10^3}=0.72$

$Q_2=3.78\times 10^3\times0.72$

⇒ $Q_2 =2.721\times 10^3 J$

Now,

The entropy change ( $\Delta S$ ) is given as:

$\Delta S=\frac{\Delta Q}{T_1}$

$\Delta S=\frac{Q_1-Q_2}{T_1}$

substituting the values in the above equation we get

$\Delta S=\frac{3.78\times 10^{3}-2.721\times 10^3 J}{623K}$

$\Delta S=1.69J/K$

7 0

3 years ago

I attempted to answer and got 0m, please explain how to get to the answer.

mafiozo [28]

The maximum height to which the ball attain before falling back down is 1147.96 m

<h3>Data obtained from the question</h3>

The following data were obtained from the question:

Initial velocity (u) = 150 m/s
Final velocity (v) = 0 m/s (at maximum height)
Acceleration due to gravity (g) = 9.8 m/s²
Maximum height (h) =?

<h3>How to determine the maximum height </h3>

The maximum height reached by the ball can be obtained as illustrated below:

v² = u² – 2gh (since the ball is going against gravity)

0² = 150² – (2 × 9.8 × h)

0 = 22500 – 19.6h

Collect like terms

0 – 22500 = –19.6h

–22500 = –19.6h

Divide both side by –19.6

h = –22500 / –19.6

h = 1147.96 m

Thus, the maximum height reached by the ball is 1147.96 m

Learn more about motion under gravity:

brainly.com/question/22719691

#SPJ1

5 0

1 year ago