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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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The amount of heat required to raise the temperature is 706.05J.

To find the answer, we need to know more about the specific heat capacity.

<h3>How to find the heat required to raise the temperature?</h3>
  • Specific heat can be defined as, the amount of heat needed to increase a substance's temperature by one degree for every gram.
  • We have the expression for amount of heat required to raise the temperature from T1 to T2 as,

                                 Q=msΔT

  • We have given with the following values,

                 m=2.50g=2.50*10^{-3}kg\\s=0.523 J/gK=\frac{0.523}{10^{-3}} J/kgK=523J/kgK\\T_1=65^0C=338K\\T_2=600^0C=873K\\

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            Q=2.5*10^{-3}*523*(873-333)=706.05 J

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brainly.com/question/18798150

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2 years ago
A very long wire carries a uniform linear charge density of 7.0 nC/m. What is the electric field strength 16.0 m from the
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Electric field due to a long wire is given by

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Answer:t=0.3253 s

Explanation:

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speed of balloon is u=3\ m/s

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Suppose after t sec of  throw camera reach balloon then,

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s_1=ut+\frac{1}{2}at^2

s_1=20\times t-\frac{1}{2}gt^2

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\Rightarrow 5t^2-17t+5=0

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