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masya89 [10]
3 years ago
10

Cual sera el caudal que lleva un rio cuando se desplaza a 200 litros cada 40 segundos

Physics
1 answer:
alisha [4.7K]3 years ago
4 0

Answer:

Q = 5 L/s

Explanation:

To find the flow you use the following formula (para calcular el caudal usted utiliza la siguiente formula):

Q=\frac{V}{t}

V: Volume (volumen) = 200L

t: time (tiempo) = 40 s

you replace the values of the parameters to calculate Q (usted reemplaza los valores de los parámteros V y t para calcular el caudal):

Q=\frac{200L}{40s}=5\frac{L}{s}

Hence, the flow is 5 L/s (por lo tanto, el caudal es de 5L/s)

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A piston-cylinder device initially contains 0.08 m3 of nitrogen gas at 150 kPa and 200°C. The nitrogen is now expanded to a pres
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Answer:

V_2 = 0.125 m^3

Work done =  = 5 kJ

Explanation:

Given data:

volume of nitrogen v_1 = 0.08 m^3

P_1 = 150 kPa

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P_2 = 80 kPa

Polytropic exponent n = 1.4

\frac{T_2}{T_1} = [\frac{P_2}{P_1}]^{\frac{n-1}{n}

putting all value

\frac{T_2}{473} = [\frac{80}{150}]^{\frac{1.4-1}{1.4}

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polytropic process is given as

P_1 V_1^n = P_2 V_2^n

150\times 0.08^{1.4} = 80 \times V_2^{1.4}

V_2 = 0.125 m^3

work done = \frac{P_1 V_1 -P_2 V_2}{n-1}

= \frac{150 \times 0.8 - 80 \times 0.125}{1.4-1}

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Here, mA = 3.65 kg and mB = 7.05 kg. The string connecting the two objects is of negligible mass and the pulley is frictionless.
mr_godi [17]

Answer:

The magintude of the acceleration for both objects is 3.11m/s^2

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we can take the acceleration going up as positive for reference purposes.

for mA let's suppose that is ascending so:

T_A-m_A*g=m_A*a

and for mB (descending):

T_B-m_B*g=-m_B*a

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because the two boxes has the same acceleration because they are attached together:

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So the magintude of the acceleration for both objects is 3.11m/s^2

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