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3 years ago

Suppose an astronaut were to visit a planet where the force of gravity is half that of Earth. His mass on that planet would be:

1 answer:

8 0

"Mass" is what YOU ARE. It doesn't matter whether you're on Earth, on the moon, on an asteroid, on a meteoroid, on the sun, swimming in chicken soup on Saturn, or in a space capsule falling from one solar system body to another one. Your mass DOESN'T change.

Whatever mass the astronaut had after he ate that steak dinner just before he was launched, THAT's the mass he has while he's on the way to another planet, and THAT's the same mass he'll have when he gets there. It makes no difference what the gravity is like on the other planet. His mass will not change.<em> (choice-B)</em>

Only his WEIGHT changes when he's in different places.

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larisa [96]

I do a yoga mat that could be different to your answer

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3 years ago

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In 1995 a research group led by Eric Cornell and Carl Wiemann at the University of Colorado successfully cooled Rubidium atoms t

saveliy_v [14]

Answer:

0.00493 m/s

Explanation:

T = Temperature of the isotope = 85 nK

R = Gas constant = 8.341 J/mol K

M = Molar mass of isotope = 86.91 g/mol

Root Mean Square speed is given by

$v_r=\sqrt{\dfrac{3RT}{M}}\\\Rightarrow v_r=\sqrt{\dfrac{3\times 8.314\times 85\times 10^{-9}}{86.91\times 10^{-3}}}\\\Rightarrow v_r=0.00493\ m/s$

The Root Mean Square speed is 0.00493 m/s

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3 years ago

A camera lens with focal length f = 50 mm and maximum aperture f>2

Brut [27]

Answer:

The minimum distance between two points on the object that are barely resolved is 0.26 mm

The corresponding distance between the image points = 0.0015 m

Explanation:

Given

focal length f = 50 mm and maximum aperture f>2

s = 9.0 m

aperture = 25 mm = 25 *10^-3 m

Sin a = 1.22 *wavelength /D

Substituting the given values, we get –

Sin a = 1.22 *600 *10^-9 m /25 *10^-3 m

Sin a = 2.93 * 10 ^-5 rad

Now

Y/9.0 m = 2.93 * 10 ^-5

Y = 2.64 *10^-4 m = 0.26 mm

Y’/50 *10^-3 = 2.93 * 10 ^-5

Y’ = 0.0015 m

8 0

3 years ago

A student falls off a cliff into the lake 54.0 m below. What is the final velocity of the student?

Vladimir79 [104]

Answer:

$v_{y}$ = -32.53 m / s

this velocity is directed downwards

Explanation:

This is a free fall exercise, let's use the expression

$v_{y}^{2}$ = v_{oy}^{2} + 2 g (y -yo)

where we are assuming that there is friction with the air, as the body falls its initial velocity is zero

v_{oy} = √ 2g (y - y₀)

let's calculate

v_{y} = √ (2 9.8 (0-54.0))

$v_{y}$ = -32.53 m / s

this velocity is directed downwards

6 0

4 years ago

What is cohesion force

olchik [2.2K]

Answer:

The force of attraction between the molecules of the same substance

Explanation:

3 0

3 years ago