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yarga [219]
4 years ago
11

Which of these best describes how an appropriate star chart is selected to locate objects in the sky?

Physics
2 answers:
lisov135 [29]4 years ago
7 0

Answer:

match the day and time of the year and location

Explanation:

The stars that we see on the sky change with the station of the year, they change their position depending of the day of the year and the location, mainly the latitude since on that depends what "side" of the earth the sky is facing towards, at certain points the north hemisphere sees a different sky than the south, but the east and west as long as they are in the same latitude do see the same sky.

Sergio [31]4 years ago
6 0
The layout of the stars in the sky is determined by the date, time of night, and your location (mainly latitude). So to pick the best star chart, you should go with the one that's closest to the present date and your location, then make allowance for what time it is. Everything in the sky moves about a degree every 4 minutes.
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The hottest would be the O type and the coolest is M
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3 years ago
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What is kepler's law??​
Elan Coil [88]
<h2>QUESTION:- </h2>

➜what is kepler's law??

\huge\red{\boxed{\huge\mathbb{\red A \pink{N}\purple{S} \blue{W}\orange{ER}}}}

Kepler gave the three laws or theorems of motion of the orbitals bodies

{\huge {\bold{ \red{ \star}}}}{ \blue{ \bold{FIRST \: \: \: LAW}}}

This law state that the celestial bodies revolves around the stars in elliptical orbit and star as a single focus.

Example :- Earth revolves around the Sun as assuming it as single focus

This also shows that earth revolves around the sun in elliptical orbit.

{\huge {\bold{ \blue{ \star}}}}{ \green{ \bold{SECOND \: \: \: LAW}}}

Area covered by the planet is equal in equal duration of time irrespective of the position of the planet.

It also states that Angular momentum is constant

As Angular momentum is constant it means areal velocity is also constant.

\frac{ \triangle \: A}{ \triangle \: T} = \frac{L}{2m}△T△A=2mL

where:-

A is the area.

T is the time.

L is the angular momentum.

M is the mass of the body.

{\huge {\bold{ \green{ \star}}}}{ \purple{ \bold{THIRD \: \: \: LAW}}}

square of the time of the revolution is directly proportional to the cube of the distance between the planet and star in Astronomical unit.

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where:-

T = time of revolution

a is the distance between the planet and star.

\purple\star \: {Thanks \: And \: Brainlist} \blue \star \\ {\orange{ \star}}{if \: U \: Like d \: My \: Ans} {\green{ \star }}

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3 years ago
A motorcycle cover a distance of 1.8 km in 5 minute. calculate its average velocity?​
OlgaM077 [116]

<em>Answer:</em>

<h2><em>6</em><em> </em><em>m/</em><em>s</em></h2>

<em>solution \\ distance \: travelled = 1.8 \: km \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  =  1.8 \times 1000m \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  = 1800 \: m \\ time \: taken = 5 \: minute \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  = 5 \times 60 \: seconds \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  = 300 \: seconds \\ average \: velocity =  \frac{distance \: travelled}{time \: taken}  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  =  \frac{1800}{ 300}  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  = 6 \:  {ms}^{ - 1}</em>

<em>hope </em><em>this </em><em>helps.</em><em>.</em>

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Anuta_ua [19.1K]

Answer:

T_0 = 338.916 Degree\ celcius

Explanation:

Given data:

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initial temperature = 25 degree celcius

Assumption

Thermal properties remain constant, transfer of heat by radiation is negligible.

from the information given in the question we have

T_S ≈T_∞ , it implies we have h → ∞

from table 4.2 Biot number → ∞ the value of

\lambda_1 = 1.5708 and A_1= 1.2732

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t = \frac{\alpha t}{l^2} = \frac{97.1\times 10^{-6} \times 15}{0.05^2} = 0.5826

Temperature at center after 15 second of heating

\theta _{0 wall} = \frac{T_0 - T_{\infity}}{T_i -T_{\infity}} = A_i e^{\lambda_1^2 t}

T_0 = T_i -T_{\infity} \times A_i e^{\lambda_1^2 t}

T_0 = (25 - 475) 1.2732 e^{-1.5708^2 \times 0.5826} +  475  = 356 degree celcius

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