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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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R S ( M ) = 2 G M c 2 , where G is the gravitational constant and c is the speed of light. It is okay if you do not follow the d
padilas [110]

The provided question's answer is "Schwarzschild radius".

The conversion factor between mass and energy is the speed of light squared.

GM/r stands for gravitational potential energy, also known as energy per unit mass.

GM/rc² then has "mass per unit mass" units. In other words, as mass/mass splits out in a dimensional analysis, "dimensionless per unit."

The derivation yields a formula for time or space coordinate ratios requiring sqrt(1 - 2GM/rc²). This number becomes 0 when r=2GM/c2, or the formula becomes infinite if in the denominator. However, there is no justification for using c² as a conversion factor there. Consider the initial expression sqrt(1 - 2GM/rc²).

Assume that m is used as the test particle's mass instead of 1. Then you have sqrt(m - 2GMm/rc² and mass units. This expression denotes that the rest energy of the test mass m you introduced into the gravitational field is "gone" at that radius.

The 2 would be absent if the gravitational field were Newtonian. However, at the event horizon, Einstein gravity is slightly stronger than Newton gravity, resulting in the factor 2 in qualitative terms.

So, the given equation is of Schwarzschild radius.

Learn more about Schwarzschild radius here:

brainly.com/question/12647190

#SPJ10

3 0
2 years ago
Show that the electric potential along the axis of a uniformly charged disk of radius R and charge density sigma is given by by
OlgaM077 [116]

Explanation:

Area of ring \ 2{\pi} a d a

Charge of on ring d q=-(\ 2{\pi} a d a)

Charge on disk

Q=-\left(\pi R^{2}\right)

\begin{aligned}d v &=\frac{k d q}{\sqrt{x^{2}+a^{2}}} \\&=2 \pi-k \frac{a d a}{\sqrt{x^{2}+a^{2}}} \\v(1) &=2 \pi c k \int_{0}^{R} \frac{a d a}{\sqrt{x^{2}+a^{2}}} \cdot_{2 \varepsilon_{0}}^{2} R \\&=2 \pi \sigma k[\sqrt{x^{2}+a^{2}}]_{0}^{2} \\&=\frac{2 \pi \sigma}{4 \pi \varepsilon_{0}}[\sqrt{z^{2}+R^{2}}-(21)] \\&=\frac{\sigma}{2}(\sqrt{2^{2}+R^{2}}-2)\end{aligned}

Note: Refer the image attached

8 0
3 years ago
Does fire have ways to protect itself from changes in the environment
Alexandra [31]

Answer:

I think no because the critical properties distinguishing life is adaptation to changing environment and self replication of the information encoding the life process. Fire does not change its process to adapt to its environment, e.g. moving toward more fuel or storing and conserving fuel when it is in short supply.

Hope this help you!:)

6 0
3 years ago
When measuring the speed of a snail, a good unit to use is millimeters per minute. A snail takes 12 hours to travel 5 meters. Wh
Helen [10]
You first need to convert the hours into minutes, then use the speed equals distance over time formula. 
5 0
3 years ago
A girl pushes her little brother on his sled with a force of 200. N for 850. m. How much work is this iſ the fored
Eddi Din [679]

Explanation:

Given that,

A girl pushes her little brother on his sled with a force of 200. N for 850. m

We need to find the work done by the force of friction acting on the sled. The force of friction is 200 N

Net force = 200 N - 200 N = 0

Work done = 0

If force of friction is 300 N.

Net force = 300 N - 200 N = 100 N

The work done by a force is given by :

W = Fd

Put values,

W = 100 × 850

W = 85000 J

Hence, this is the required solution.

5 0
3 years ago
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