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SOVA2 [1]
2 years ago
3

A neutron star has a density of about 5.9 10 17 kg/m3. What would be the approximate mass of a 1-centimeter cube (a cube 1 cm on

all sides)? 5.9 1014 kg (1300 trillion pounds) 5.9 108 kg (1.3 billion pounds) 5.9 1017 kg (1.3 quadrillion pounds) 5.9 1011 kg (1.3 trillion pounds)
Mathematics
1 answer:
egoroff_w [7]2 years ago
6 0
Density is usually stated as gm/cc
1 kilogram = 1*10^3 grams
1 cubic meter = 1*10^6 cc

5.9 x 10^17 kg/m^3 = 5.9 x 10^20 g/m^3
5.9 x 10^20 g/m^3 = 5.9 x 10^14 g/cc

So, the mass is 5.9x10^14 grams which equals
5.9 x 10^11 kilograms (which is the last answer).



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Andrei wants to fill a glass tank with marbles, and then fill the remaining space with water. WWW represents the volume of water
andre [41]

Answer:

32 Liters

Step-by-step explanation:

In the glass tank:

W = represents the volume of water Andrei uses (in liters)

n =Number of marbles

Given that: W=32-0.05n

The volume of the glass tank is the volume of the water in the tank plus the total volume of n marbles added.

Therefore:

W+0.05n=32

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2 years ago
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Two unique letters are chosen at random from the alphabet.
ElenaW [278]
A - choosing letter A first

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2 years ago
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Walt received a package that is 2 1/3 inches long, 6 3/4 inches high, and 8 1/2 inches wide. What is the surface area of the pac
grin007 [14]
Add whole numbers: 2 + 6 + 8 = 16
Add fractions:

= 1/3 + 3/4 + 1/2

= (8 + 18 + 12) ÷ 24

= 38/24 or 1 14/24 or simplified to 1 7/12

Total surface area = 16 + 1 7/12 or 17 7/12


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2 years ago
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a total of 17100 seats are still available for the next hockey game.if 62% of the tickets are sold how many seats are in the are
BartSMP [9]
If 62% of the seats are sold; 

100-62= 38% seats still available. 

17100=38% seats available

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450 seats

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Hope I helped :) 
3 0
2 years ago
A specimen of aluminum having a rectangular cross section 10 mm 12.7 mm (0.4 in. 0.5 in.) is pulled in tension with 35,500 N (80
miskamm [114]

Answer:

\epsilon = 3.958\times 10^{-3}

Step-by-step explanation:

The Young's Module of Aluminium is E = 69\times 10^{9}\,Pa. The axial stress on the specimen is:

\sigma = \frac{F}{A_{t}}

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The strain is derived of following expression:

\epsilon = \frac{\sigma}{E}

\epsilon = \frac{2.731\times 10^{8}\,Pa}{69\times 10^{9}\,Pa}

\epsilon = 3.958\times 10^{-3}

4 0
2 years ago
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