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Mamont248 [21]
3 years ago
8

Why do astronomers find it difficult to detect individual exoplanets?

Physics
2 answers:
Maurinko [17]3 years ago
6 0
Light from the stars, because the orbits make it difficult to see them. 
Tcecarenko [31]3 years ago
3 0
Astronomers find it difficult to detect individual exoplanets because these planet<span>s are not bright enough compared to the stars they orbit. </span>
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Increasing over time
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Which layer of the sun gives off the visible light normally seen from earth?
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Most of the energy we receive from Sun is the visible light or whit light emitted from the photosphere. The photosphere is one of the most coolest region of the Sun. The photosphere is the visible surface of the Sun that we are most familiar with.
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Velocity vs Time
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A and C

C

40 m/s

zero

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8 0
3 years ago
If the copper is drawn into wire whose diameter is 6.50 mm, how many feet of copper can be obtained from the ingot? the density
Kobotan [32]
Assume that an ingot of copper has a mass of 9.1 kg or 9100 g.

The cross-sectional area of the copper wire with diameter of 6.5 mm (or 0.65 cm) is
A = (π/4)*(0.65 cm)² = 0.3318 cm²

The density of copper is given as 8.94 g/cm³.
If the length of copper wire is L cm, then 
(0.3318 cm²)*(L cm)*(8.94 g/cm³) = 9100 g
L = 9100/(0.3318*8.94) = 3.0678 x 10³ cm

Note that
1 cm = 1/2.54 in = 1/2.54 in = 0.3937 in
        = 0.3937/12 = 0.03281 ft

Therefore
L = (3.0678 x 10³ cm)*(0.03281 ft/cm) = 100.65 ft

Answer: 100.65 ft

6 0
3 years ago
A hand pump is being used to inflate a bicycle tire that has a gauge pressure of 41.0 lb/in2. If the pump is a cylinder of lengt
damaskus [11]

Answer:

L - h = 12.3672 in

Explanation:

Given

P = 41.0 lb/in² = 41 P.S.I

L = 16.8 in

A = 3.00 in²

h = ?

In order that air flows into the tire, the pressure in the pump must be more than the tire pressure,  41.0  PSI.

We assume that air follows ideal gas equation, the temperature of the compressed air remains constant as the piston moves down. Taking one atmospheric pressure to be  14.6959  P.S.I , we can use the ideal gas equation

P*V = n*R*T

As number of moles of air do not change during its compression in the pump, n*R*T of the gas equation is constant. Therefore we have

P₁*V₁ = P₂*V₂    ⇒    V₂ = P₁*V₁ / P₂

where  

1  and  2  are initial and final states respectively,

V₁ = A*L = (3.00 in²)*(16.8 in)   ⇒   V₁ = 50.4 in³

P₁ = 14.6959  P.S.I

P₂ = P₁ + P = (14.6959 lb/in²) + (41.0 lb/in²) = 55.6959 lb/in²

Inserting various values we get

V₂ = (14.6959  P.S.I)*(50.4 in³) / (55.6959 lb/in²)

⇒  V₂ = 13.2985 in³

Length of pump, measured from bottom, this volume corresponds to is

h = V₂ / A  = (13.2985 in³) / (3.00 in²)

⇒  h = 4.4328 in

Piston must be pushed down by more than

L - h = 16.8 in - 4.4328 in = 12.3672 in

4 0
3 years ago
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