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drek231 [11]
3 years ago
9

The luminous star Alnilam in the Orion belt is 1,340 light-years away from Earth. Use the conversion factor 1 parsec = 3.262 lig

ht-years to find the closest conversion of this distance from light-years to parsecs (pc).

Physics
2 answers:
svlad2 [7]3 years ago
8 0

The answer is 410.8 pc.

Angelina_Jolie [31]3 years ago
6 0

Answer:

The distance to the star Alnilam in the Orion belt is 410.8 Pc

Explanation:

The angle due to the change in position of a nearby object against the background stars is known as parallax. This angle is gotten when the position of the object is measured in January and then in July according to the configuration of the Earth with respect to the Sun in those months (see the image below).

       

The parallax angle can be used to find out the distance by means of triangulation. Making a triangle between the nearby star, the Sun and the Earth(as is shown in the image below), knowing that the distance between the Earth and the Sun (150000000 Km) is defined as 1 astronomical unit:

\tan{p} = \frac{1AU}{d}

Where d is the distance to the star.

p('') = \frac{1}{d}    (1)  

Equation (1) can be rewritten in terms of d:

d(pc) = \frac{1}{p('')}  (2)

Equation (2) represents the distance in a unit known as parsec (pc), which is the distance of a star with a parallax angle of 1 arc second.

For the case of the star Alnilam in the Orion belt ( 1340 light-years):

Since 1 parsec is equivalent to 3.262 light-years, therefore:

d = 1340 ly \cdot \frac{1pc}{3.262 ly} ⇒ $410.8 Pc

Notice how both , Light years and parsecs are units for distance.          

Key terms:

Parsec: Parallax of arc second.

Light-year: Distance traveled by light in one year.

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Answer:

f= 3.0 \times 10 {}^{8} \div 7.0 \times 10 {}^{7} \\ f = 4.28hz

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7 0
3 years ago
A mass of M-kg rests on a frictionless ramp inclined at 30°. A string with a linear mass density of μ=0.025" kg/m" is attached t
I am Lyosha [343]

Answer:

44.3 m/s

Explanation:

a) Draw a free body diagram of the mass M.  There are three forces:

Weight force mg pulling down,

Normal force N pushing perpendicular to the ramp,

and tension force T pulling parallel up the ramp.

Sum of forces in the parallel direction:

∑F = ma

T − Mg sin 30° = 0

T = Mg sin 30°

T = Mg / 2

Draw a free body diagram of the hanging mass m.  There are two forces:

Weight force mg pulling down,

and tension force T pulling up.

Sum of forces in the vertical direction:

∑F = ma

T − mg = 0

T = mg

Substitute:

mg = Mg / 2

m = M / 2

M = 2m

b) Velocity of a standing wave in a string is:

v = √(T / μ)

T = mg, and m = 5 kg, so T = (5 kg) (9.8 m/s²) = 49 N.  Therefore:

v = √(49 N / 0.025 kg/m)

v = 44.3 m/s

7 0
3 years ago
You are trying to hear your friend give directions to new store in town. But from your distance (1 point) of 15 m you only hear
Marat540 [252]

Answer:

option D

Explanation:

given,

Intensity of sound = 20 dB

distance = 15 m

intensity of sound is increased to = 50 dB

distance between the sound level = ?

Using relation

L_2 = L_1 - |20(log \dfrac{r_2}{r_1})|

L₁ = 20 dB        L₂ = 50 dB         r₁ = 15 m      r₂ = ?

log (\dfrac{r_2}{r_1}) = \dfrac{L_1 -L_2}{20}

\dfrac{r_2}{r_1}= 10^{\dfrac{|L_1 -L_2|}{20}}

r_2 =r_1 10^{\dfrac{|L_1 -L_2|}{20}}

r_2 =15 \times 10^{\dfrac{|20-50|}{20}}

r_2 =15 \times 10^{-1.5}

r₂ = 0.47 m

r₂ = 47 cm

hence, the correct answer is option D

7 0
3 years ago
The distance between a speaker and a listener is 8.0 m. Determine the frequency of the sound wave if exactly 20 waves are formed
dmitriy555 [2]

The frequency of the wave is 6800 Hz

<u>Explanation:</u>

Given:

Wave number, n = 20

Speed of light, v = 340 m/s

Frequency, f = ?

we know:

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20 = \frac{f}{340 m/s} \\\\f = 20 X 340 s^-^1\\\\f = 6800 Hz

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3 years ago
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Answer:

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