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user100 [1]
3 years ago
11

Infrared observations of the orbits of stars close to the galactic center indicate a small object at the center with a mass of a

bout how many solar masses?
Physics
1 answer:
MariettaO [177]3 years ago
5 0

Answer:

(4.31±0.38) million Solar masses.

Explanation:

The galactic center is the center of the milky way around which the galaxy rotates. It is most likely the location of a supermassive black hole which has a mass of (4.31±0.38) million Solar masses. The location is called Sagittarius A*.

As there is interstellar dust in our line of sight from the Earth infrared observations need to be taken.

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You throw a glob of putty straight up toward the ceiling, which is 3.60 m above the point where the putty leaves your hand. The
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Answer:

Explanation:

Given

Height of ceiling is h=3.6\ m

Initial speed of Putty u=9.5\ m/s

Speed of Putty just before it strike the ceiling is given by

v^2-u^2=2as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

v^2-9.5^2=2\times (-9.8)\times 3.6

v^2=19.69

v=4.43\ m/s

time taken by putty to reach the ceiling

v=u+at

4.43=9.5-9.8\times t

t=\frac{5.07}{9.8}

t=0.517\ s

8 0
3 years ago
Which letter on the diagram below represents the trough of a wave?
Nikitich [7]

A would be the wavelength, C would be a crest, D would be the amplitude, leaving B which is the trough.

8 0
3 years ago
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Which of the following type of personal debts is usually considered wise and justified? A. credit card purchases of luxuries tha
NeTakaya

Answer:

C. Money to pay off other debts and interest

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3 years ago
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Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el
rewona [7]

Answer:

a) x = 0.200 m

b)E = 3.84*10^{-4} N/C

Explanation:

q_1 = 21.0\mu C

q_1 = 47.0\mu C

DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m

by relation for electric field we have following relation

E = \frac{kq}{x}^2

according to question E = 0

FROM FIGURE

x is the distance from left point charge where electric field is zero

\frac{k21}{x}^2 = \frac{k47}{0.5-x}^2

solving for x we get

\frac{0.5}{x} = 1+ \sqrt{\frac{47}{21}}

x = 0.200 m

b)electric field at half way mean x =0.25

E =\frac{k*21*10^{-6}}{0.25^2} -\frac{k*47*10^{-6}}{0.25^2}

E = 3.84*10^{-4} N/C

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3 years ago
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Please answer will give brainliest
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3b. No

4a. 50N


I hope this helps


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3 years ago
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