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velikii [3]
2 years ago
5

What is the gravity of sombrero galaxy compared to earth?

Physics
2 answers:
Aleksandr [31]2 years ago
7 0
the Sombrero galaxy, is one of the largest mosaics ever assembled from Hubble observations. The hallmark of the nearly edge-on galaxy is a brilliant, white, bulbous core encircled by thick dust lanes comprising the spiral structure of the galaxy. This dust lane is the site of star formation in the galaxy. The center of M104 is thought to be home to a massive black hole. I hope this helped
postnew [5]2 years ago
4 0
The Sombrero galaxy, is one of the largest mosaics ever assembled from Hubble observations. The hallmark of the nearly edge-on galaxy is a brilliant, white, bulbous core encircled by thick dust lanes comprising the spiral structure of the galaxy. This dust lane is the site of star formation in the galaxy. The center of M104 is thought to be home to a massive black hole.

Hubble easily resolves the Sombrero galaxy’s rich system of globular clusters, estimated to be nearly 2,000 in number — 10 times more than the number of globular clusters in our Milky Way galaxy.
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Which approach would be the most interested in studying Phineas Gage
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3 years ago
A piano string having a mass per unit length of 5.00 g/m is under a tension of 1350 N. Determine the speed of transverse waves i
padilas [110]

Answer:

The speed of transverse waves in this string is 519.61 m/s.

Explanation:

Given that,

Mass per unit length = 5.00 g/m

Tension = 1350 N

We need to calculate the speed of transverse waves in this string

Using formula of speed of the transverse waves

v=\sqrt{\dfrac{T}{\mu}}

Where, \mu = mass per unit length

T = tension

Put the value into the formula

v = \sqrt{\dfrac{1350}{5.00\times10^{-3}}}

v =519.61\ m/s

Hence, The speed of transverse waves in this string is 519.61 m/s.

6 0
3 years ago
A football is kicked from the ground with a velocity of 38m/s at an angle of 40 degrees and eventually lands at the same height.
Anastasy [175]

Initially, the velocity vector is \langle 38cos(40^{\circ}),38sin(40^{\circ}) \rangle=\langle 29.110, 24.426 \rangle. At the same height, the x-value of the vector will be the same, and the y-value will be opposite (assuming no air resistance). Assuming perfect reflection off the ground, the velocity vector is the same. After 0.2 seconds at 9.8 seconds, the y-value has decreased by 4.9(0.2)^2, so the velocity is \langle 29.110, 24.426-0.196 \rangle = \langle 29.110, 24.23 \rangle.

Converting back to direction and magnitude, we get \langle r,\theta \rangle=\langle \sqrt{29.11^2+24.23^2},tan^{-1}(\frac{29.11}{24.23}) \rangle = \langle 37.87,50.2^{\circ}\rangle

4 0
3 years ago
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