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astraxan [27]
3 years ago
9

si comparamos la gota de agua cayendo en una gotera, y el sonido del agua hirviendo,ambos con la misma intensidad ¿cuál sonido t

iene más energía? justifique su respuesta
Physics
1 answer:
mrs_skeptik [129]3 years ago
8 0

Answer:

ambas

Explanation:

Tendrían el mismo sonido ya que la intensidad es la misma.

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Water flows over a section of Niagara Falls at rate of 1.1×10^6 kg/s and falls 49.4 m. How much power is generated by the fallin
DerKrebs [107]

Answer:

<em>The power generated is =  5.33×10⁸ Watt. </em>

Explanation:

Power: Power can be defined as the time rate of doing work. The S.I unit of power is <em>Watt(W).</em>

<em>Mathematically,</em>

<em>Power (P) = Work done/time or Energy/time</em>

P = mgh/t............................... Equation 1

P = δgh............................. Equation 2

Where δ = fall rate, g = acceleration due to gravity, h = height.

<em>Given: </em>δ = 1.1×10⁶ kg/s, h = 49.4 m g = 9.81 m/s²

Substituting these values into equation 2

P = 1.1×10⁶×49.4×9.81

P = 533.08×10⁶

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5 0
3 years ago
The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, what should be the focal length an
yuradex [85]

Answer:

The focal length of the appropriate corrective lens is 35.71 cm.

The power of the appropriate corrective lens is 0.028 D.

Explanation:

The expression for the lens formula is as follows;

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}

Here, f is the focal length, u is the object distance and v is the image distance.

It is given in the problem that the given lens is corrective lens. Then, it will form an upright and virtual image at the near point of person's eye. The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, the corrective lens is used.

Put v= -71.4 cm and u= 24.0 cm in the above expression.

\frac{1}{f}=\frac{1}{24}+\frac{1}{-71.4}

\frac{1}{f}=0.028

f= 35.71 cm

Therefore, the focal length of the corrective lens is 35.71 cm.

The expression for the power of the lens is as follows;

p=\frac{1}{f}

Here, p is the power of the lens.

Put f= 35.71 cm.

p=\frac{1}{35.71}

p=0.028 D

Therefore, the power of the corrective lens is 0.028 D.

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