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2 years ago

8

Determina la presion hidrostatica en el fondo de una alberca olimpica de 8metros de profundidad

1 answer:

Dimas [21]2 years ago

6 0

Plz write it in English

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At what rate must a cylindrical spaceship rotate if occupants are to experience simulated gravity of 0.50 gg? Assume the spacesh

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Answer:

The time needed is $T = 16.8 s$

Explanation:

From the question we are told that

The magnitude of the stimulated acceleration due gravity is $a = 0.5 g$

The diameter of the spaceship is $d = 35m$

Generally the force acting on the spaceship is

$F = ma$

Given that the spaceship is rotating it implies that the force experienced by the occupant is a centripetal force so

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Thus

$ma = \frac{mv^2}{r}$

=> $\frac{v^2}{r} = a$

Generally the speed of this spaceship is mathematically represented as

$v = \frac{2 \pi}{T}$

=> $v^2 = [\frac{2\pi}{T}] ^2$

=> $\frac{\frac{4\pi^2 r^2}{T^2} }{r} = 0.5g$

=> $\frac{4 \pi^2 r }{T^2} = 0.5 g$

=> $T = \sqrt{ \frac{4\pi^2 r}{0.5g}}$

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A mountain-climber friend with a mass of 74 kg ponders the idea of attaching a helium-filled balloon to himself to effectively r

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Answer:

$V=16.65 m^3$

Explanation:

The volume of the balloon can be find compared the force in each cases so:

reduce 25% from 74kg

$R=\frac{25}{100}*74kg=18.5kg$

So the net force uproad on the balloon is

$F_b=18.5kg*g$

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$P_A=1.29 kg/m^3$

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$F_b=m_a*g-m_h*g$

$m=P/V$

$18.5kg*g=(1.29kg/m^3-0.179kg/m^3*)V*g$

$V=\frac{18.5kg}{(1.29-0.179)kg/m^3}$

$V=16.65 m^3$

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