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Darya [45]
3 years ago
5

Why are different constellations of stars seen during different seasons?

Physics
2 answers:
lesya [120]3 years ago
8 0
Because of the earth's and axis, and as well as how it rotates around the sun.
slamgirl [31]3 years ago
5 0
Actually, they're not.  There's a group of stars and constellations arranged
around the pole of the sky that's visible at any time of any dark, clear night,
all year around.  And any star or constellation in the rest of the sky is visible
for roughly 11 out of every 12 months ... at SOME time of the night. 

Constellations appear to change drastically from one season to the next,
and even from one month to the next, only if you do your stargazing around
the same time every night.

Why does the night sky change at various times of the year ?  Here's how to
think about it:

The Earth spins once a day. You spin along with the Earth, and your clock is
built to follow the sun . "Noon" is the time when the sun is directly over your
head, and "Midnight" is the time when the sun is directly beneath your feet.

Let's say that you go out and look at the stars tonight at midnight, when you're
facing directly away from the sun.

In 6 months from now, when you and the Earth are halfway around on the other
side of the sun, where are those same stars ?  Now they're straight in the
direction of the sun.  So they're directly overhead at Noon, not at Midnight.

THAT's why stars and constellations appear to be in a different part of the sky,
at the same time of night on different dates.
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8 0
2 years ago
What is the value for the kinetic energy for a n = 5 Bohr orbit electron in zeptoJoules?
Ne4ueva [31]

Answer:

The kinetic energy is 86.6 zepto joules.

Explanation:

Given that,

Number of orbit n =5

We know that,

Bohr's radius for hydrogen atom is

r = 0.53\times10^{-10}\times n^2\ m

Now, put the value of n in the formula of radius

r=0.53\times10^{-10}\times5^2

r =1.33\times10^{-9}\ m

We need to calculate the kinetic energy

Using formula of kinetic energy

E_{k}=\dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{e^2}{2\times r_{s}}

Put the value into the formula

E_{k}=\dfrac{9\times10^{9}\times(1.6\times10^{-19})^2}{2\times1.33\times10^{-9}}

E_{k}=8.66\times10^{-20}\ J

We know that,

1\ zepto\ joule=1\times10^{-21}\ J

The kinetic energy is

E_{k}=\dfrac{8.66\times10^{-20}}{1\times10^{-21}}

E_{k}=86.6\ zepto\ Joules

Hence, The kinetic energy is 86.6 zepto joules.

8 0
3 years ago
The speed of a box traveling on a horizontal friction surface changes from vi = 13 m/s to vf = 11.5 m/s in a distance of d = 8.5
KiRa [710]

Answer:

0.68 s

Explanation:

We are given that

Initial velocity of box=u=13m/s

Final velocity of box=v=11.5 m/s

Distance=d=8.5 m

We have to find the time taken by box to slow by this amount.

We know that

v^2-u^2=2as

Substitute the values

(11.5)^2-(13)^2=2a(8.5)

132.25-169=17a

-36.75=17a

a=\frac{-36.75}{17}=-2.2m/s^2

We know that

Acceleration=a=\frac{v-u}{t}

Substitute the values

-2.2=\frac{11.5-13}{t}

-2.2=\frac{-1.5}{t}

t=\frac{1.5}{2.2}=0.68 s

Hence, the time taken by box to slow by this amount=0.68 s

8 0
3 years ago
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