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Alex Ar [27]
3 years ago
12

I need help please so fast

Physics
1 answer:
Simora [160]3 years ago
7 0

Answer:

DUumb

Explanation:

Duumb

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On an asteroid, the density of dust particles at a height of 3 cm is ~30% of its value just above the surface of the asteroid. A
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From the law of atmosphere

N_v(y) = n_0*e^{-\frac{mgy}{Kb*T}}

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n_0 = constant and is number density where the height y = 0cm

n_V = Number density at height y=3cm

Kb = Boltzmann constant = 1.38*10^{-23}J/K

T=20K

m = 10^{-19}kg

Re-arranging the equation to have the value of the gravity,

\frac{N_v(y)}{n_0} = e^{-\frac{mghy}{KbT}}

ln(\frac{N_v(y)}{n_0}) = -\frac{mgy}{KbT}

Since it is 30% of value above surface, therefore N_v = 0.3n_0

ln(\frac{0.3n_0}{n_0}) = -\frac{mgy}{KbT}

g = -\frac{KbT ln(0.3)}{my}

g = -\frac{(1.38*10^{-23}J/K)(20K)(Ln(0.3))}{10^{-19}(3*10^{-2})}

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g = 1.104*10^{-1}m/s^2

g = 0.1m/s^2

Therefore the correct answer is C.

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Which of the following are neutral particles?<br> Protons<br> Neutrons<br> Electrons<br> Atoms
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A brave child decides to grab onto an already spinning merry‑go‑round. The child is initially at rest and has a mass of 25.5 kg.
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Answer:

72.75 kg m^2

Explanation:

initial angular velocity, ω = 35 rpm

final angular velocity, ω' =  19 rpm

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Let the moment of inertia of the merry go round is I.

Use the concept of conservation of angular momentum

I ω = I' ω'

where I' be the moment of inertia of merry go round and child

I x 35 = ( I + md^2) ω'

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