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Lesechka [4]
3 years ago
7

What is Astronomy , planets and stars

Physics
2 answers:
Natali [406]3 years ago
4 0

Answer:

Astronomy is a natural science that studies celestial objects and phenomena which uses maths physics and chemistry in order to explain their origin and evolution. Planets and stars are linked with astronomy.

Planets are large natural objects that orbit, or travel around, stars.  A planet must do three things: it must orbit a star, it must be big enough to have enough gravity to force a spherical shape, and it must be big enough that its gravity cleared away any objects of a similar size near its orbit.

A star is an astronomical object consisting of a luminous spheroid of plasma held together by its own gravity. Represents the most fundamental building blocks of galaxies. Stars form within amorphous clouds of gas and dust called nebulae.

belka [17]3 years ago
3 0

Answer:

Astronomy (from Greek: ἀστρονομία) is a natural science that studies celestial objects and phenomena. It uses mathematics, physics, and chemistry in order to explain their origin and evolution. Objects of interest include planets, moons, stars, nebulae, galaxies, and comets.

A planet is an astronomical body orbiting a star or stellar remnant that is massive enough to be rounded by its own gravity, is not massive enough to cause thermonuclear fusion, and has cleared its neighbouring region of planetesimals.

A star is an astronomical object consisting of a luminous spheroid of plasma held together by its own gravity. The nearest star to Earth is the Sun.

HOPE IT HELPS :)

PLEASE MARK IT THE BRAINLIEST!

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Turning the barrel of a 50-mm-focal-length lens on a manual-focus camera moves the lens closer to or farther from the sensor to
Yuliya22 [10]

Answer:

Explanation:

To focus object at .7m , the image distance can be measured as follows

object distance u = .7m

focal length f = .05 m

image distance v = ?

from lens formula

\frac{1}{v} -\frac{1}{u} = \frac{1}{f}

\frac{1}{v} +\frac{1}{.7} = \frac{1}{.05}

\frac{1}{v} =\frac{1}{.05} -\frac{1}{.7}

v = .054 m

= 54 mm

when the object is at infinity , image is formed at focus ie at distance of

50 mm .

So lens position from sensor  where image is formed , varies from 54 mm to 50 mm .

4 0
3 years ago
The magnitude of the magnetic field B a distance r from a long straight wire is B = μ 0 I 2 π r where μ 0 is the permeability co
PSYCHO15rus [73]

Answer:

The magnetic field is lowest for largest distance and highest when distance is least.

Explanation:

The magnitude of magnetic field strength at a distance 'r' from a long straight wire carrying current 'I' is given as:

B=\frac{\mu_0 I}{2\pi r}\\Where,\mu_0\to permeability\ constant\ of\ free\ space

Now, as per question, the distance 'r' is varied while keeping the current constant in the wire.

As seen from the above formula, the magnitude of magnetic field strength for a constant current varies inversely with the distance 'r'. This means that, as the value of 'r' increases, the magnitude of magnetic field strength decreases and vice-versa.

Therefore, the magnitude of magnetic field strength is maximum when the distance 'r' is least and the magnetic field is minimum for the largest distance.

Example:

If B_1, B_2,\ and\ B_3 are the magnitudes of magnetic field strengths for distances r_1,r_2, \ and\ r_3 respectively such that r_1. Now, as per the explanation above, the order of magnitudes of magnetic field strength is:

B_1>B_2>B_3

6 0
4 years ago
The displacement of a moving object in given interval of time is zero . Would the distance travelled by the object also be zero.
kirill [66]





Hi Pupil Here's Your answer ::




➡➡➡➡➡➡➡➡➡➡➡➡➡



No, on a circle in going from one point around circular path the displacement is zero but the distance covered is equal to the circumference of the circle.

i.e.


2\pi r


⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅




Hope this helps . . . . . . .

3 0
3 years ago
This is a spreading out of a wave around corners or through holes.
lesya692 [45]

Answer:

Diffraction.

Explanation:

Diffraction is the bending and scattering of a wave as it hits an object or goes through an opening.

5 0
3 years ago
Read 2 more answers
A high school physics student claims her muscle car can achieve a constant acceleration of 10 ft/s/s. Her friend develops an acc
Nikolay [14]

Answer: Ф = 17.2657 ≈ 17°

Explanation:

we simply apply ET =0 about the ending of the rod

so In.g.L/2sinФ - In.a.L/2cosФ = 0

g.sinФ - a.cosФ = 0

g.sinФ = a.cosФ

∴ tanФ = a/g

Ф =  tan⁻¹ a / g

Ф = tan⁻¹ ( 10 / 32.17405)

Ф = tan⁻¹ 0.31080948777

Ф = 17.2657 ≈ 17°

Therefore the angle of rotation of the rod during this acceleration is 17.2657 ≈ 17°

5 0
3 years ago
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