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The speed of the incoming water is 25 cm/s and the cross-sectional area of the hose carrying the water is 2.0 cm2. 1) Determine the time required for a 25-L container to be filled with water. (Express your answer to two significant figures.)

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okpalawalter8

Answer:

The time taken is  [tex]t = 500 \ s[/tex]

Explanation:

From the question are told that

  The speed of the incoming water is  [tex]v = 25 cm/s[/tex]

  The cross-sectional area of the host pipe is  [tex]a = 2.0 cm^2[/tex]

Generally the rate at which the container is been filled is mathematically represented as

         [tex]\r V = v * a[/tex]

=>      [tex]\r V = 25 * 2[/tex]

=>      [tex]\r V = 50\ cm^3/s[/tex]

Generally [tex]2.5 L = 25000 cm^3[/tex]

Generally the time taken is  mathematically represented as

      [tex]t = \frac{25000}{50}[/tex]

=>   [tex]t = 500 \ s[/tex]

onyebuchinnaji

The time required for a 25-L container to be filled with water is 8.3 min.

The given parameters;

  • speed of the water, v = 25 cm/s
  • area of the hose, A = 2 cm²
  • volume of the container, V = 25 L = 25,000 cm³

The volumetric flow rate of the water is calculated as follows;

Q = Av

where;

  • Q is volumetric flow rate
  • v is the speed of the water
  • A is the area of the hose

[tex]Q = Av\\\\\frac{V}{t}= Av\\\\t = \frac{V}{Av} \\\\t = \frac{25,000}{2 \times 25} \\\\t = 500 \ s\\\\t = 8.3 \min[/tex]

Thus, the time required for a 25-L container to be filled with water is 8.3 min.

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