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OLEGan [10]
3 years ago
14

Astronomers have recently observed stars orbiting at very high speeds around an unknown object near the center of our galaxy. Fo

r stars orbiting at distances of about 1014 m from the object, the orbital velocities are about 106 m/s. Assume the orbits are circular, and estimate the mass of the object, in units of the mass of the sun (MSun = 2x1030 kg). If the object was a tightly packed cluster of normal stars, it should be a very bright source of light. Since no visible light is detected coming from it, it is instead believed to be a supermassive black hole.
Physics
1 answer:
nataly862011 [7]3 years ago
6 0

Answer:

<em>The mass of the object is 745000 units of the sun</em>

Explanation:

We know that the centripetal force with which the stars orbit the object is represented as

F_{c} = \frac{mv^{2} }{r}

and this centripetal force is also proportional to

F_{c} = \frac{kMm}{r^{2} }

where

m is the mass of the stars

M is the mass of the object

v is the velocity of the stars = 10^6 m/s

r is the distance between the stars and the object = 10^14 m

k is the gravitational constant = 6.67 × 10^-11 m^3 kg^-1 s^-2

We can equate the two centripetal force equations to give

\frac{mv^{2} }{r} = \frac{kMm}{r^{2} }

which reduces to

v^{2} = \frac{kM}{r}

and then finally

M = \frac{rv^{2} }{k}

substituting values, we have

M = \frac{10^{14}*(10^{6})^{2}  }{6.67*10^{-11} } = 1.49 x 10^36 kg

If the mass of the sun is 2 x 10^30 kg

then, the mass of the the object in units of the mass of the sun is

==> (1.49 x 10^36)/(2 x 10^30) = <em>745000 units of sun</em>

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Answer:

Explanation:

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MA_775_DIABLO [31]

Answer:

a)   D_ total = 18.54 m,   b)        v = 6.55 m / s

Explanation:

In this exercise we must find the displacement of the player.

a) Let's start with the initial displacement, d = 8 m at a 45º angle, use trigonometry to find the components

           sin 45 = y₁ / d

           cos 45 = x₁ / d

           y₁ = d sin 45

           x₁ = d sin 45

           y₁ = 8 sin 45 = 5,657 m

           x₁ = 8 cos 45 = 5,657 m

The second offset is d₂ = 12m at 90 of the 50 yard

            y₂ = 12 m

            x₂ = 0

total displacement

          y_total = y₁ + y₂

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          x_total = x₁ + x₂

          x_total = 5,657 + 0

          x_total = 5,657 m

          D_total =   17.657 i^+ 5.657 j^  m

          D_total = Ra (17.657 2 + 5.657 2)

          D_ total = 18.54 m

b) the average speed is requested, which is the offset carried out in the time used

           v = Δx /Δt

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         v = 6.55 m / s

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