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3 years ago

The light of a star that can be seen from Earth is called what? (Apparent or Absolute Magnitude)

1 answer:

3 0

A glance at the night sky above Earth shows that some stars are much brighter than others. Apparent brightness is how we see the star's brightness from Earth. The absolute brightness is what a star would look like if there was a standard distance that we were looking at all of the stars from.

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Why do astronomers use the word on to describe angles on the sky rather than angles in the sky?

Sever21 [200]

The use of the word on instead of the word in when referring to the angular distance between celestial objects comes about because all of the objects appear to be on the celestial sphere and at an indeterminable distance. While we know that objects are at different distances in the sky, their distance from Earth is irrelevant in determining the angular distance between the two objects as viewed from Earth.

8 0

3 years ago

What is 0.0025 m^2 in cm^2? 1m = 100cm.

PSYCHO15rus [73]

Answer:

25 cm²

Explanation:

Meters and centimeters are both the units for measuring length. The SI unit of measuring length is meters.

Area is the quantity which measures the cross-section occupied by the object.

Thus,

Given that = Area = 0.0025 m²

To convert into cm²

1 m = 100 cm

So, 1 m² = 10000 cm²

So,

Area = 0.0025 × 10000 cm² = 25 cm²

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2 years ago

Sapting Leng You have a string with a mass of 13.7 q. You stretch the string with a force of 8.39 N, giving it a length of 1.87

marshall27 [118]

Answer:

a) $\lambda=0.935\ \textup{m}$

b) $f=36.19\approx 36\ \textup{Hz}$

Explanation:

Given:

String vibrates transversely fourth dynamic, thus n = 4

mass of the string, m = 13.7 g = 13.7 × 10⁻¹³ kg

Tension in the string, T = 8.39 N

Length of the string, L = 1.87 m

a) we know

$L= n\frac{\lambda}{2}$

where,

$\lambda$ = wavelength

on substituting the values, we get

$1.87= 4\times \frac{\lambda}{2}$

$\lambda=0.935\ \textup{m}$

b) Speed of the wave (v) in the string is given as:

$v =f\lambda$

also,

$v=\sqrt\frac{T}{(\frac{m}{L})}$

equating both the formula for 'v' we get,

$f\lambda=\sqrt\frac{T}{(\frac{m}{L})}$

on substituting the values, we get

$f\times 0.935=\sqrt\frac{8.39}{(\frac{13.7\times 10^{3}}{1.87})}$

$f=\frac{33.84}{0.935}$

$f=36.19\approx 36\ \textup{Hz}$

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2 years ago

No matter how well we get along with our parents as adolescents, there comes a time when we need to break away and assert our ow

KonstantinChe [14]

Answer: I never said anything to them but if i did it would be "I need a break from you, i love you but sometimes I have to find the path on my own." i would say this if i wasn't scared to hurt them.

Explanation:

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3 years ago

Which statement best explains why radon and krypton do not bond easily with other elements?

mote1985 [20]

Answer.B

Explanation:

6 0

3 years ago

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