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gayaneshka [121]

3 years ago

8

If the 78.0 kg astronaut were in a spacecraft 6R from the center of the earth, what would the astronaut's weight be on earth? 76

4.4 N760 N764 N in the spacecraft? 21.24 N21 N21.3 N

1 answer:

den301095 [7]3 years ago

6 0

(a) 764.4 N

The weight of the astronaut on Earth is given by:

$F=mg$

where

m is the astronaut's mass

g is the acceleration due to gravity

Here we have

m = 78.0 kg

g = 9.8 m/s^2 at the Earth's surface

So the weight of the astronaut is

$F=(78.0)(9.8)=764.4 N$

(b) 21.1 N

The spacecraft is located at a distance of

$r=6R$

from the center of Earth.

The acceleration due to gravity at a generic distance r from the Earth's center is

$g=\frac{GM}{r^2}$

where G is the gravitational constant and M is the Earth's mass.

We know that at a distance of r = R (at the Earth's surface) the value of g is 9.8 m/s^2, so we can write:

$GM=9.8R^2$ (1)

the acceleration due to gravity at r=6R instead will be

$g'=\frac{GM}{(6R)^2}$

And substituting (1) into this formula,

$g'=\frac{9.8R^2}{36R^2}=0.27 m/s^2$

So the weight of the astronaut at the spacecratf location is

$F'=mg'=(78.0 kg)(0.27 m/s^2)=21.1 N$

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3 years ago

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Black_prince [1.1K]

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3 years ago

A horizontal pipe contains water at a pressure of 110 kPa flowing with a speed of 1.4 m/s. When the pipe narrows to one half its

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a

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b

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The speed of the water is $v_1 = 1.4 \ m/s$

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The new area of the pipe is $A_2 = \pi * \frac{[\frac{d}{2} ]^2}{4} = \pi * \frac{\frac{d^2}{4} }{4} = \pi \frac{d^2}{16}$

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Here $v_2$ is the new velocity

So

$\pi * \frac{d^2}{4} * 1.4 = \pi * \frac{d^2}{16} * v_2$

=> $\frac{d^2}{4} * 1.4 = \frac{d^2}{16} * v_2$

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=> $1.4 = 0.25 * v_2$

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Here $\rho$ is the density of water with value $\rho = 1000 \ kg /m^3$

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=> $P_2 = 110 *10^{3} + \frac{1}{2} * 1000 * [ 1.4 ^2 - 5.6 ^2 ]$

=> $P_2 = 80600 \ Pa$

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8 0

3 years ago