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2 years ago

A light-year is _____.

Physics

1 answer:

Talja [164]2 years ago

8 0

Answer:

The time required for light to travel in one year.

Explanation:

Hope this helps!

If not, I am sorry.

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Astronomers discover a bright X-ray source in the Milky Way. They see no counterpart in the optical bands, but it has another st

VladimirAG [237]

Answer:

a. a black hole

Explanation:

X-ray emission from the central degrees of the Milky Way Bright X-ray emission traces the coherent edge brightened shell-like feature, dubbed the northern chimney, located north of Sgr A* and characterized by a diameter of about 160 pc. On the opposite side, the southern chimney appears as a bright linear feature. Bright X-ray emission is observed at high latitude

8 0

3 years ago

A 25 kg child is riding on a swing. If the child travels 8.9 m/s at the bottom of their swing, how high into the air is the chil

Setler [38]

Answer:

h = 4.04 m

Explanation:

Given that,

Mass of a child, m = 25 kg

The speed of the child at the bottom of the swing is 8.9 m/s

We need to find the height in the air is the child is able to swing. Let the height is h. Using the conservation of energy such that,

$mgh=\dfrac{1}{2}mv^2\\\\h=\dfrac{v^2}{2g}$

Put all the values,

$h=\dfrac{(8.9)^2}{2\times 9.8}\\\\h=4.04\ m$

So, the child is able to go at a height of 4.04 m.

7 0

3 years ago

The input work done on a machine is 9.63 × 103 joules, and the output work is 3.0 × 103 joules. What is the percentage efficienc

notsponge [240]

Input work = 9.63×10³ J.
Output work = 3.0×10³ J

By definition,
Efficiency = (Output work)/(Input work)
= (3.0×10³)/(9.63×10³)
= 0.31 = 31%

Answer: 31%

5 0

3 years ago

Read 2 more answers

magine an astronaut on an extrasolar planet, standing on a sheer cliff 50.0 m high. She is so happy to be on a different planet,

Mama L [17]

Answer:

$\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'} } )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g} } )$

Explanation:

Given:

height above which the rock is thrown up, $\Delta h=50\ m$

initial velocity of projection, $u=20\ m.s^{-1}$

let the gravity on the other planet be g'

The time taken by the rock to reach the top height on the exoplanet:

$v=u+g'.t'$

where:

$v=$ final velocity at the top height = 0 $m.s^{-1}$

$0=20-g'.t'$ (-ve sign to indicate that acceleration acts opposite to the velocity)

$t'=\frac{20}{g'}\ s$

The time taken by the rock to reach the top height on the earth:

$v=u+g.t$

$0=20-g.t$

$t=\frac{20}{g} \ s$

Height reached by the rock above the point of throwing on the exoplanet:

$v^2=u^2+2g'.h'$

where:

$v=$ final velocity at the top height = 0 $m.s^{-1}$

$0^2=20^2-2\times g'.h'$

$h'=\frac{200}{g'}\ m$

Height reached by the rock above the point of throwing on the earth:

$v^2=u^2+2g.h$

$0^2=20^2-2g.h$

$h=\frac{200}{g}\ m$

The time taken by the rock to fall from the highest point to the ground on the exoplanet:

$(50+h')=u.t_f'+\frac{1}{2} g'.t_f'^2$ (during falling it falls below the cliff)

here:

$u=$ initial velocity= 0 $m.s^{-1}$

$\frac{200}{g'}+50 =0+\frac{1}{2} g'.t_f'^2$

$t_f'^2=\frac{400}{g'^2}+\frac{100}{g'}$

$t_f'=\sqrt{\frac{400}{g'^2}+\frac{100}{g'} }$

Similarly on earth:

$t_f=\sqrt{\frac{400}{g^2}+\frac{100}{g} }$

Now the required time difference:

$\Delta t=(t'+t_f')-(t+t_f)$

$\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'} } )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g} } )$

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3 years ago

How are the movements of the Moon in the sky and Earth's rotation about its own axis related? A) The Moon appears to move from E

JulsSmile [24]

The moon orbits quite fast: it moves about 0.5 degrees per hour in the sky. In 24 hours it moves 13 degrees. The moon's observed motion eastward results from its physical motion of the moon along its orbit around the Earth. so I think its A .

7 0

3 years ago

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