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2 years ago

What is apparent magnitude and what can affect it?

1 answer:

Montano1993 [528]2 years ago

8 0

Apparent magnitude means how bright a star APPEARS to us on Earth. It can be affected by ...

... how bright the star really is
... how far the star is from us
... how much gas and dust is between us
... how much of the star's total output is in visible light

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A 75 kg Spider Man running at 3.0 m/s jumps onto a trash can lid that has a mass of 10kg and that is already moving in the same

QveST [7]

In order to find the final velocity of the skier and the trash can lid, we may apply the principle of conservation of momentum, which states that the total momentum of a system remains constant. Mathematically, in this case:
m₁v₁ + m₂v₂ = m₃v₃
Where m₃ and v₃ are the combined mass and velocity.

75*3 + 10*2 = (75 + 10)*v₃
v₃ = 2.88 m/s

The final velocity is 2.88 m/s

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2 years ago

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Which statements correctly describe the properties of air? Select three options. Air has mass but no weight. Air pressure is cre

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Answer:

2,4,5

Air pressure is created by the weight of the atmosphere pushing on Earth’s surface.

Denser air is heavier than less dense air.

Air is less dense at higher altitudes.

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3 years ago

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Temperature and pressure of a region upstream of a shockwave are 295 K and 1.01* 109 N/m². Just downstream the shockwave, the te

seraphim [82]

Answer:

change in internal energy 3.62*10^5 J kg^{-1}

change in enthalapy 5.07*10^5 J kg^{-1}

change in entropy 382.79 J kg^{-1} K^{-1}

Explanation:

adiabatic constant $\gamma =1.4$

specific heat is given as $=\frac{\gamma R}{\gamma -1}$

gas constant =287 J⋅kg−1⋅K−1

$Cp = \frac{1.4*287}{1.4-1} = 1004.5 Jkg^{-1} k^{-1}$

specific heat at constant volume

$Cv = \frac{R}{\gamma -1} = \frac{287}{1.4-1} = 717.5 Jkg^{-1} k^{-1}$

change in internal energy $= Cv(T_2 -T_1)$

$\Delta U = 717.5 (800-295) = 3.62*10^5 J kg^{-1}$

change in enthalapy $\Delta H = Cp(T_2 -T_1)$

$\Delta H = 1004.5*(800-295) = 5.07*10^5 J kg^{-1}$

change in entropy

$\Delta S =Cp ln(\frac{T_2}{T_1}) -R*ln(\frac{P_2}{P_1})$

$\Delta S =1004.5 ln(\frac{800}{295}) -287*ln(\frac{8.74*10^5}{1.01*10^5})$

$\Delta S = 382.79 J kg^{-1} K^{-1}$

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2 years ago

The spacing and distribution of individuals within a population is known as

4vir4ik [10]

Answer:

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Species dispersion patterns—or distribution patterns—refer to how the individuals in a population are distributed in space at a given time. The individual organisms that make up a population can be more or less equally spaced.

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The answer is <span>oxbow lake</span>

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2 years ago

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