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Gekata [30.6K]
3 years ago
10

Because of interstellar dust, astronomers can see at most about 5 kpc into the disk of the galaxy at visual wavelengths. What pe

rcentage of the galactic disk does that include? (Hint: Consider the area of the entire disk versus the area visible from Earth. The diameter of the Milky Way Galaxy is approximately 25 kpc.)
Physics
1 answer:
NNADVOKAT [17]3 years ago
5 0

Answer:

96%

Explanation

Let A the total area of the galaxy, is modeled as a disc:

A = πR^2 = π (25 kpc)^2

And let a be the area that astronomers are able to see:

a = πr^2 = π(5 kpc)^2

The percentage that can be seen is equal to 100 times the ratio of the areas, of the galaxy and the "visible" part:

P = 100 a/A = (5/25)^2 = 100/25 = 4%

Therefore, the percentage of the galaxy not included, i.e. not seen is:

(100-4)% = 96%

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Answer:

r = 20 m

Explanation:

The formula for the angular momentum of a rotating body is given as:

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r = radius of merry-go-round

Therefore,

10000 kg.m²/s = mr(2 m/s)

m r = (10000 kg.m²/s)/(2 m/s)

m r = 5000 kg.m   ------------- equation 1

Now, the moment of inertia of a solid uniform disc about its axis through its center is given as:

I = (1/2) m r²

where,

I = moment of inertia = 50000 kg.m²

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50000 kg.m² = (1/2)(m r)(r)

using equation 1, we get:

50000 kg.m² = (1/2)(5000 kg.m)(r)

(50000 kg.m²)/(2500 kg.m) = r

<u>r = 20 m</u>

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3 years ago
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A piece of wood 400N in weight and 50cm X 30cm X 20cm in size lies on 50cm X 20cm face. Calculate the pressure exerted by it.
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Answer:

P = 4000 [Pa]

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1000[cm^{2}]*\frac{1^{2}m^{2}  }{100^{2}m^{2}  }=0.1[m^{2} ]

And the pressure is:

P=\frac{F}{A} \\P=400/0.1\\P=4000[N/m^{2} ]or 4000[Pa]

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The kinetic energy of the phone right before it hits the ground is 9J.

<h3>Kinetic energy of the phone</h3>

The kinetic energy of the phone right before it hits the ground is calculated as follows;

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Thus, the kinetic energy of the phone right before it hits the ground is 9J.

Learn more about kinetic energy here: brainly.com/question/25959744

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