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3 years ago

A que profundidad esta nadando una persona dentro de una alberca si la presión absoluta sobre ésta es de 156kPa?

Physics

1 answer:

lara31 [8.8K]3 years ago

5 0

Answer:

La persona está nadando en la alberca a una profundida de 5.575 metros.

Explanation:

La presión absoluta ( $P_{tot}$ ) experimentada por la persona es la suma de la presión atmosférica ( $P_{atm}$ ) y la presión hidrostática de la columna de agua de la alberca ( $P_{h}$ ), medidas en kilopascales. Es decir,

$P_{tot} = P_{atm}+P_{h}$ (1)

$P_{tot} = P_{atm} + \frac{\rho\cdot g \cdot z}{1000}$ (2)

Donde:

$\rho$ - Densidad del fluido de la alberca, medida en kilogramos por metro cúbico.

$g$ - Aceleración gravitacional, medida en metros por segundo al cuadrado.

$z$ - Profundidad de la persona en la alberca, medida en metros.

Si sabemos que $P_{atm} = 101.325\,kPa$ , $\rho = 1000\,\frac{kg}{m^{3}}$ , $g = 9.807\,\frac{m}{s^{2}}$ y $P_{tot} = 156\,kPa$ , entonces la profundidad de la persona en la alberca es:

$156 = 101.325 +\frac{(1000)\cdot (9.807)\cdot z}{1000}$

$54.675 = 9.807\cdot z$

$z = 5.575\,m$

La persona está nadando en la alberca a una profundida de 5.575 metros.

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<h3>What is total electric potential?</h3>

The amount of labor required to convey a unit of electric charge from a reference point to a given place in an electric field is known as the electric potential (also known as the electric field potential, potential drop, or the electrostatic potential).
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To learn more about total electric potential with the given link

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2 years ago

The noble gas neon is used for filling neon signs. Like other noble elements, it has a full octet (complete outer energy level)

DanielleElmas [232]

Answer:

C. unlikely to combine with other elements.

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3 years ago

What is the answer? Explain best and I will mark brainliest!

iVinArrow [24]

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3 years ago

Read 2 more answers

The half-life of Iodine-131 is 8.0252 days. If 14.2 grams of I-131 is released in Japan and takes 31.8 days to travel across the

MakcuM [25]

Answer:

Explanation:

Half-life problems are modeled as exponential equations. The half-life formula is $P=P_o\left (\dfrac{1}{2} \right)^{\frac{t}{k}}$ where $P_o$ is the initial amount, $k$ is the length of the half-life, $t$ is the amount of time that has elapsed since the initial measurement was taken, and $P$ is the amount that remains at time $t$ .

$P=14.2\left (\dfrac{1}{2} \right)^{\frac{t}{8.0252}}$

<u>Deriving the half-life formula</u>

If one forgets the half-life formula, one can derive an equivalent equation by recalling the basic an exponential equation, $y=a b^{t}$ , where $t$ is still the amount of time, and $y$ is the amount remaining at time $t$ . The constants a and b can be solved for as follows:

Knowing that amount initially is 14.2g, we let this be time zero:

$y=a b^{t}$

$(14.2)=ab^{(0)}$

$14.2=a *1$

$14.2=a$

So, $a=14.2$ , which represents out initial amount of the substance, and our equation becomes: $y=14.2 b^{t}$

Knowing that the "half-life" is 8.0252 days (note that the unit here is "days", so times for all future uses of this equation must be in "days"), we know that the amount remaining after that time will be one-half of what we started with:

$\left(\frac{1}{2} *14.2 \right)=14.2 b^{(8.0252)}$

$\dfrac{7.1}{14.2}=\dfrac{14.2 b^{8.0252}}{14.2}$

$0.5=b^{8.0252}$

$\sqrt[8.0252]{\frac{1}{2}}=\sqrt[8.0252]{b^{8.0252}}$

$\sqrt[8.0252]{\frac{1}{2}}=b$

Recalling exponent properties, one could find that $\left ( \frac{1}{2} \right )^{\frac{1}{8.0252}}=b$ , which will give the equation identical to the half-life formula. However, recalling this trivia about exponent properties is not necessary to solve this problem. One can just evaluate the radical in a calculator:

$b=0.9172535661...$

Using this decimal approximation has advantages (don't have to remember the half-life formula & don't have to remember as many exponent properties), but one minor disadvantage (need to keep more decimal places to reduce rounding error).

So, our general equation derived from the basic exponential function is:

$y=14.2* (0.9172535661)^t$ or $y=14.2*(0.5)^{\frac{t}{8.0252}}$ where y represents the amount remaining at time t.