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Ymorist [56]
3 years ago
9

Astronomers have observed a small, massive object at the center of our Milky Way Galaxy. A ring of material orbits this massive

object; the ring has a diameter of about 16 light-years and an orbital speed of about 130 km/s. For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Black hole calculations.
Determine the mass M of the massive object at the center of the Milky Way galaxy.Take the distance of one light year to be 9.461 x 1015m.Express your answer in kilograms.
Physics
1 answer:
Burka [1]3 years ago
4 0

Answer:

1.91773\times 10^{37}\ kg

Explanation:

v = Orbital speed = 130 km/s

d = Diameter = 16 ly

r = Radius = \dfrac{d}{2}=\dfrac{16}{2}=8\ ly

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

1\ ly=9.461\times 10^{15}\ m

As the centripetal force balances the gravitational energy we have the following relation

\dfrac{GMm}{r^2}=\dfrac{mv^2}{r}\\\Rightarrow M=\dfrac{v^2r}{G}\\\Rightarrow M=\dfrac{130000^2\times 8\times 9.461\times 10^{15}}{6.67\times 10^{-11}}\\\Rightarrow M=1.91773\times 10^{37}\ kg

Mass of the the massive object at the center of the Milky Way galaxy is 1.91773\times 10^{37}\ kg

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If a car is traveling 103 miles per hour, how many miles and how much time would it take for another car to catch up to them fro
pentagon [3]

Answer:

Required distance is 103 miles and the required time is 1 hour

Explanation:

Given;

speed of the car, v = 103 miles per hour

Speed is the given as the ratio of distance traveled to time taken for the motion.

The distance it will take another car to catch up to them from a complete stop is 103 miles and the time it will take the car is 1 hour.

Therefore, required distance is 103 miles and the required time is 1 hour.

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3 years ago
A student drew the following model:
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Answer: C Tectonic Plates

Explanation:

8 0
3 years ago
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A block of unknown mass is attached to a spring with a spring constant of 7.00 N/m 2 and undergoes simple harmonic motion with a
KatRina [158]

Answers:

a) 0.80 kg

b) 2.12 s

c) 1.093 m/s^{2}

Explanation:

We have the following data:

k=7 N/m is the spring constant

A=12.5 cm \frac{1 m}{100 cm}=0.125 m is the amplitude of oscillation

V=32 cm/s=0.32 m/s is the velocity of the block when x=\frac{A}{2}=0.0625 m

Now let's begin with the answers:

<h3>a) Mass of the block</h3>

We can solve this by the conservation of energy principle:

U_{o}+K_{o}=U_{f}+K_{f} (1)

Where:

U_{o}=k\frac{A^{2}}{2} is the initial potential energy

K_{o}=0  is the initial kinetic energy

U_{f}=k\frac{x^{2}}{2} is the final potential energy

K_{f}=\frac{1}{2} m V^{2} is the final kinetic energy

Then:

k\frac{A^{2}}{2}=k\frac{x^{2}}{2}+\frac{1}{2} m V^{2} (2)

Isolating m:

m=\frac{k(A^{2}-x^{2})}{V^{2}} (3)

m=\frac{7 N/m((0.125 m)^{2}-(0.0625 m)^{2})}{(0.32 m/s)^{2}} (4)

m=0.80 kg (5)

<h3>b) Period</h3>

The period T is given by:

T=2 \pi \sqrt{\frac{m}{k}} (6)

Substituting (5) in (6):

T=2 \pi \sqrt{\frac{0.80 kg}{7 N/m}} (7)

T=2.12 s (8)

<h3>c) Maximum acceleration</h3>

The maximum acceleration a_{max} is when the force is maximum F_{max}, as well :

F_{max}=m.a_{max}=k.x_{max} (9)

Being x_{max}=A

Hence:

m.a_{max}=kA (10)

Finding a_{max}:

a_{max}=\frac{kA}{m} (11)

a_{max}=\frac{(7 N/m)(0.125 m)}{0.80 kg} (12)

Finally:

a_{max}=1.093 m/s^{2}

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3 years ago
Suppose that the radius of the circular path is r when the speed of the rocket is v and the acceleration of the rocket has magni
SCORPION-xisa [38]

Answer:

A3

Explanation:

4 0
3 years ago
A passenger on a Ferris wheel moves in a vertical circle at a constant speed. Are the forces on her balanced?
vovikov84 [41]

C.  The force is a constant change,  because her position on the Ferris wheel will constantly change.  I believe this is the answer, but use sources to double check.  I might use different vocab. then your teachers.  

6 0
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