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3 years ago

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At a depth of 10.9 km, the Challenger Deep in the Marianas Trench of the Pacific Ocean is the deepest site in any ocean. Yet, in

1960, Donald Walsh and Jacques Piccard reached the Challenger Deep in the bathyscaph Trieste. Assuming that seawater has a uniform density of 1024 kg/m3, approximate the hydrostatic pressure (in atmospheres) that the Trieste had to withstand. (Even a slight defect in the Trieste structure would have been disastrous.)

2 answers:

gavmur [86]3 years ago

8 0

Answer:7434.47 atmosphere

Explanation:

hydrostatic pressure (in atmospheres) = atmospheric pressure + pressure at depth

specific gravity = density of substance / density of water = 1024/1000= 1.024

specific weight = 1.024 * 9.79 kN/m^3 = 10.02496 kN/m^3

= 14.7 psi + 10.02496 kN/m^3* 10.9 km * 1000m = 109286.764 psi (1km = 1000m)

to atmosphere = 109286.764 psi/14.7 psi * 1 atmosphere = 7434.47 atmosphere

KATRIN_1 [288]3 years ago

7 0

Answer:

$P = 1081.304\,atm$

Explanation:

The hydrostic pressure is equal to:

$P = P_{atm} + P_{man}$

$P = 1\,atm + \left(\frac{1\,atm}{101,325\,Pa} \right)\cdot \left(1,024\,\frac{kg}{m^{3}} \right)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (10,900\,m)$

$P = 1081.304\,atm$

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(a) The energy of the photon is 1.632 x $10^{-8}$ J.

(b) The wavelength of the photon is 1.2 x $10^{-17}$ m.

(c) The frequency of the photon is 2.47 x $10^{25}$ Hz.

Explanation:

Let;

$E_{1}$ = -13.60 ev

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(a) Energy of the emitted photon can be determined as;

$E_{2}$ - $E_{1}$ = -3.40 - (-13.60)

= -3.40 + 13.60

= 10.20 eV

= 10.20(1.6 x $10^{-9}$ )

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The energy of the emitted photon is 10.20 eV (or 1.632 x $10^{-8}$ Joules).

(b) The wavelength, λ, can be determined as;

E = (hc)/ λ

where: E is the energy of the photon, h is the Planck's constant (6.6 x $10^{-34}$ Js), c is the speed of light (3 x $10^{8}$ m/s) and λ is the wavelength.

10.20(1.6 x $10^{-9}$ ) = (6.6 x $10^{-34}$ * 3 x $10^{8}$ )/ λ

λ = $\frac{1.98*10^{-25} }{1.632*10^{-8} }$

= 1.213 x $10^{-17}$

Wavelength of the photon is 1.2 x $10^{-17}$ m.

(c) The frequency can be determined by;

E = hf

where f is the frequency of the photon.

1.632 x $10^{-8}$ = 6.6 x $10^{-34}$ x f

f = $\frac{1.632*10^{-8} }{6.6*10^{-34} }$

= 2.47 x $10^{25}$ Hz

Frequency of the emitted photon is 2.47 x $10^{25}$ Hz.

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