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Akimi4 [234]
3 years ago
13

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 13 light-years and an orbital speed of about 180 km/s .A) Determine the mass of the massive object at the center of the Milky Way galaxy.Take the distance of one light year to be 9.461×1015m. I was able to get this it is 4.26×1037kg.B) Express your answer in solar masses instead of kilograms, where one solar mass is equal to the mass of the sun, which is 1.99×1030. I got this to it is 2.14×107 solar massesC)Observations of stars, as well as theories of the structure of stars, suggest that it is impossible for a single star to have a mass of more than about 50 solar masses. Can this massive object be a single, ordinary star? I got this it is no.
Physics
1 answer:
Mkey [24]3 years ago
3 0
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Answer:

1. MOON has the greatest influence over the tides

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Explanation:

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Why do you feel warm when you are standing away from a camp fire
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3 years ago
Which of these boxes will not accelerate? A. B. C. D.
serg [7]

Answer: The correct answer is Image B.

Explanation: For an object to accelerate, there should be unbalanced forces present. An object will move in the direction of net force.

Balanced forces are defined as the forces acting on the same object which are equal in magnitude but act in opposite direction. The net forces are 0.

Unbalanced forces are defined as the forces acting on the same object which are unequal in magnitude. The net force is non-zero.

For the given images:

Image A: This box will accelerate easily because the net force is non-zero and is moving in right direction.

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Image C: This box will accelerate easily because the net force is non-zero and  is acting in between the normal and applied force.

Image D: This box will accelerate easily because the net force is non-zero and is moving in right direction.

Hence, the correct option is Image B.

7 0
3 years ago
Read 2 more answers
Need help with both questions!
xenn [34]
#14 isn't really a Physics problem.  It's more of just reading a graph.

A). When speed changes, acceleration is

       (change in speed) / (time for the change) .

To be correct about it, acceleration can be positive ... when speed
is increasing ... or it can be negative ... when speed is decreasing.
So, on this graph, there are two periods of acceleration:

From zero to 2 seconds, acceleration = (8 m/s) / (4 sec) = 2 m/s² .

From 10 to 12 seconds, acceleration = (-4 m/s) / (2 sec) = -2 m/s² .

B). From 12 to16 seconds, you can read the speed right from
the graph.  It's 4 m/s .

C).  From 2 to 10 seconds, the objects speed is a steady 8 m/s.
Covering 8 m/s every second for 8 seconds, it covers 64 meters.
Do you remember that distance is the area under the speed/time
graph?  You can see that plainly on this graph.  From 2 to 10 sec,
there are 16 blocks.  Each block is (2 m/s) high and (2 sec) wide,
so its area is (2 m/s) x (2 sec) = 4 meters.  The area of 16 blocks
is (16) x (4 meters) = 64 meters.
====================================

#15.

a).  constant velocity on a distance graph is a line that slopes up;
constant velocity on a velocity graph is a horizontal line;

b). positive constant acceleration on a distance graph is a
line that curves up;
positive constant acceleration on a velocity graph is a
straight line that slopes up;

c).  "uniformly slowing down to a stop" on a distance graph
is a line that's less and less curved as time goes on, and
eventually reaches the x-axis.
"uniformly slowing down to a stop" on a velocity graph is
a straight line that slopes down, and stops when it reaches
the x-axis.




7 0
3 years ago
Answer the question with step.​
Maru [420]

Answer:

f1/f2 =W1/W2 = 1/3

.0 f2 = 3f1

As ,

1/F= 1/f1 +1/f2

...1/40 = 1/f1 - 1/3f1

f1=> 80/3 cm

... f2 = 2f1 = 3 x 80/3 = 80 cm

7 0
3 years ago
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