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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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4 years ago
Provide an example of when momentum is conserved and explain your answer you can get 10 PTS if answered with a good explaination
dezoksy [38]

Answer:

m_1=8\ kg,\ m_2=6\ kg,\ v_1=12\ m/s, v_2=4\ m/s,\ v_1'=-6\ m/s,\ v_2'=28\ m/s

Explanation:

<u>Conservation of Momentum </u>

The total momentum of a system of two particles is

p=m_1v_1+m_2v_2

Where m1,m2,v1, and v2 are the respective masses and velocities of the particles at a given time. Then, the two particles collide and change their velocities to v1' and v2'. The final momentum is now

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Let's put some numbers in the problem and say

m_1=8\ kg,\ m_2=6\ kg,\ v_1=12\ m/s, v_2=4\ m/s,\ v_1'=-6\ m/s,\ v_2'=28\ m/s

(8)(12)+(6)(4)=(8)(-6)+(6)(28)

96+24=-48+168

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It means that when the particles collide, the first mass returns at 6 m/s and the second continues in the same direction at 28 m/s

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143m/s if you just perhaps by what you know you'll figure it out
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<h3>Answer:</h3><h3>we can say that:-</h3>
  1. A reading with more no of significant figures is considered to be more precise.
  2. Kyra recorded a reading of 24.3 sec. Since all non 0 digits are considered to be significant this reading has 3 significant figures.
  3. Pari recorded a reading of 24 sec. Since all non 0 digits are considered to be significant this reading has 3 significant figures.
<h3>hence we can say that kyra's reading has more significant figures nd so it is more precise.</h3>

7 0
3 years ago
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