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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
Dkdkdkdkdkzkdkcldkdskkds
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ou purchase a rectangular piece of metal that has dimen- sions 5.0 * 15.0 * 30.0 mm and mass 0.0158 kg. The seller tells you tha
Natalija [7]

Answer: 7022.2kg/m³, yes, I was cheated

Explanation:

Density of an object is defined as the ratio of the mass of the object to its volume. Mathematically;

Density = Mass/Volume

Note that the unit of both mass and volume must be standard unit.

Given mass = 0.0158kg

Dimension of the metal = 5mm×15mm×30mm

Note that 1mm = 0.001m

The volume of the metal will be

0.005×0.015×0.03

= 0.00000225m³

Density = 0.0158/0.00000225

Average density of the metal = 7022.2kg/m³

Since the standard density of Gold is 19,320kg/m³ and is higher than the density prescribed for me, it shows the I was cheated.

4 0
4 years ago
Will mark Brainliest!
DENIUS [597]
I think it’s A protons .. hope this helps
8 0
3 years ago
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Bicyclists in the Tour de France do enormous amounts of work during a race. For example, the average power per kilogram generate
tigry1 [53]

Explanation:

It is given that,

Average power per unit mass generated by Lance, \dfrac{P}{m}=6.5\ W/kg

P=6.5\times 75=487.5\ W

(a) Distance to cover race, d = 160\ km =160\times 10^3\ m

Average speed of the person, v = 11 m/s

If t is the time taken to cover the race.

t=\dfrac{d}{v}

t=\dfrac{160\times 10^3\ m}{11\ m/s}

t = 14545.46 s

Let W is the work done. The relation between the work done and the power is given by :

P=\dfrac{W}{t}

W=P\times t

W=487.5\times 14545.46

W = 7090911.75 J

(b) Since, 1\ J=2.389\times 10^{-4}\ calories

So, in 7090911.75 J, W=7090911.75 \times 2.389\times 10^{-4}

W = 1694.01 J

Hence, this is the required solution.

6 0
3 years ago
An experimenter finds that standing waves on a string fixed at both ends occur at 24 Hz and 32 Hz , but at no frequencies in bet
Vera_Pavlovna [14]

Answer:

8 Hz

Explanation:

Given that

Standing wave at one end is 24 Hz

Standing wave at the other end is 32 Hz.

Then the frequency of the standing wave mode of a string having a length, l, is usually given as

f(m) = m(v/2L), where in this case, m could be 1. 2. 3. 4 etc

Also, another formula is given as

f(m) = m.f(1), where f(1) is the fundamental frequency..

Thus, we could say that

f(m+1) - f(m) = (m + 1).f(1) - m.f(1) = f(1)

And as such,

f(1) = 32 - 24

f(1) = 8 Hz

Then, the fundamental frequency needed is 8 Hz

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3 years ago
According to Newton's 3rd law of motion, an action creates _____.
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C.an equal and opposite reaction
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