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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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Veseljchak [2.6K]

Incomplete question as the unit of volume is not written correctly.So the complete question is here:

A straightforward method of finding the density of an object is to measure its mass and then measure its volume by submerging it in a graduated cylinder. What is the density of a 240-g rock that displaces 89.0 cm³?

Answer:

d_{Density}=2.7g/cm^{3}

Explanation:

Given data

Mass m=240g

Volume V=89.0 cm³

To find

Density d

Solution

If rock displaces 89.0 cm³ of water means volume of rock is also 89cm³

So

d_{Density}=\frac{mass}{volume}\\d_{Density}=\frac{240g}{89.0cm^{3} } \\d_{Density}=2.7g/cm^{3}

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3 years ago
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Korolek [52]

Answer:

Second Option

Explanation:

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Hope this helps.

8 0
3 years ago
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Se coloca una piedra de 600 g en una Honda de 50 cm y se la hace girar a una velocidad de 4 m/s. Dibuja la fuerza que ejerce la
Dima020 [189]

Answer:

Fc = 19.2 N

Explanation:

In this case, the force of the Honda over the rock, is a centripetal force. Then, you have:

F_c=m\frac{v^2}{r}

m: mass of the rock = 600g = 0.6 kg

v: tangential velocity of the Honda = 4m/s

r: radius of the Honda = 50cm = 0.5m

You replace the values of m, r and v in the equation for Fc:

F_c=(0.6kg)\frac{(4m/s)^2}{0.5m}\\\\F_c=19.2N

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A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien
n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

\phi=NBA

Put the value into the formula

\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2

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Using formula of induced emf

\epsilon=\dfrac{d\phi}{dt}

Put the value into the formula

\epsilon=\dfrac{7.85\times10^{-7}}{dt}

Put the value of emf from ohm's law

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Idt=dq

dq=1.57\times10^{-6}\ C

We need to calculate the voltage across the capacitor

Using formula of charge

dq=C dV

dV=\dfrac{dq}{C}

Put the value into the formula

dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}

dV=1.57\ V

Hence, The voltage across the capacitor is 1.57 V.

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