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3 years ago

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Transforma las siguientes unidades utilizando factores de conversión (no vale poner el número solo) a. 85 atm a mmHg b. 60 Pa a

atm c. 780 mmHg a Pa d. 740 mmHg a atm

1 answer:

Nostrana [21]3 years ago

3 0

Answer:

a. 6.5 × 10⁴ mmHg

b. 5.9 × 10⁻⁴ atm

c. 1.0 × 10⁵ Pa

d. 0.97 atm

Explanation:

Transformaremos las siguientes unidades utilizando factores de conversión.

a. 85 atm a mmHg

Usaremos el factor de conversion 1 atm = 760 mmHg.

85 atm × 760 mmHg/1 atm = 6.5 × 10⁴ mmHg

b. 60 Pa a atm

Usaremos el factor de conversion 1 atm = 101325 Pa.

60 Pa × 1 atm/101325 Pa = 5.9 × 10⁻⁴ atm

c. 780 mmHg a Pa

Usaremos el factor de conversion 760 mmHg = 101325 Pa.

780 mmHg × 101325 Pa/760 mmHg = 1.0 × 10⁵ Pa

d. 740 mmHg a atm

Usaremos el factor de conversion 1 atm = 760 mmHg.

740 mmHg × 1 atm/760 mmHg = 0.97 atm

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Answer:

Approximately $0.560\; {\rm m}$ , assuming that:

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Explanation:

Notations:

Let $k$ denote the spring constant of the spring.
Let $m$ denote the mass of the clay sphere.
Let $v$ denote the initial speed of the spring.
Let $g$ denote the gravitational field strength.
Let $h$ denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let $x$ denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of $x$ , the elastic potential energy $\text{PE}_{\text{spring}}$ in this spring would be:

$\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}$ .

The initial kinetic energy $\text{KE}$ of the clay sphere was:

$\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}$ .

When the spring is at the maximum compression:

The clay sphere would be right on top of the spring.
The top of the spring would be below the original position (when the spring was uncompressed) by $x$ .
The initial position of the clay sphere, however, is above the original position of the top of the spring by $h = 3.21\; {\rm m}$ .

Thus, the initial position of the clay sphere ( $h = 3.21\; {\rm m}$ above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by $(h + x)$ .

The gravitational potential energy involved would be:

$\text{GPE} = m\, g\, (h + x)$ .

No mechanical energy would be lost under the assumptions listed above. Thus:

$\text{PE}_\text{spring} = \text{KE} + \text{GPE}$ .

$\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x)$ .

Rearrange this equation to obtain a quadratic equation about the only unknown, $x$ :

$\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0$ .

Substitute in $k = 1570\; {\rm N \cdot m^{-1}}$ , $m = 4.87\; {\rm kg}$ , $v = 5.21\; {\rm m\cdot s^{-1}}$ , $g = 9.81\; {\rm m \cdot s^{-2}}$ , and $h = 3.21\; {\rm m}$ . Let the unit of $x$ be meters.

$785\, x^{2} - 47.775\, x - 219.453 \approx 0$ (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that $x \ge 0$ :

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