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Eduardwww [97]
2 years ago
9

If the density of a diamond is 3.5 /cm^ 3 what would be the mass of a diamond whose volume is 0.5 cm ^ 3 ?

Physics
1 answer:
Elan Coil [88]2 years ago
7 0

Answer:

1.75g

Explanation:

Given parameters:

Density  = 3.5g/cm³

Volume  = 0.5cm³  

Unknown:

Mass of the diamond  = ?

Solution:

Density is the mass per unit volume of a substance

  Density  = \frac{mass}{volume}  

So;

  Mass  = 3.5 x 0.5  = 1.75g

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If I were a geologist working in South America along one of those plate boundaries, what might I see if I were looking at sedime
faust18 [17]

Answer:

you'll see rocks that have been there for years

4 0
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This problem follows up on a discussion from lecture. A wind turbine with an efficiency of 45% for converting wind energy into e
Volgvan

Answer:

4.1 m

Explanation:

10 kW = 10000 W

20mi/h = 20*1.6 km/mi = 32 km/h = 32 * 1000 (m/km) *(1/3600) hr/s = 8.89 m/s

The power yielded by the wind turbine can be calculated using the following formula

P = \frac{1}{2} \rho v^3 A C_p

where \rho = 1.2 kg/m^3 is the air density, v = 8.89 m/s is the wind speed, A is the swept area and C_p = 0.45 is the efficiency

10000 = 0.5 * 1.2 * 8.89^3 * A * 0.45

10000 = 190A

A = 10000 / 190 = 52.7 m^2

The swept area is a circle with radius r being the blade length

\pi r^2 = A = 52.7

r^2 = 52.7 / \pi = 16.79

r = \sqrt{16.79} = 4.1 m

4 0
2 years ago
What do I feed my pet fly
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Answer:

If she ___(be) rich, she _____ (buy) a new sports car.

Group of answer choices

is/buys

be/buy

were/would buy

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2 years ago
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which of the folloing statements about ionization energy is true? A elements toward the bottom of a group periodic table general
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<span>
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6 0
3 years ago
Assuming that each nucleus is roughlyspherical and that its mass is roughly equal to A (in atomic mass units {\rm u}), what is t
lara [203]

Answer:

ρ/ρ2 = 3 / R₀       the two densities are different

Explanation:

Density is defined as

       ρ = M / V

As the nucleus is spherical

       V = 4/3 π r³

Let's replace

      ρ = A / (4/3 π R₀³)

      ρ = ¾ A / π R₀³

b)

      ρ2 = F / area

The area of ​​a sphere is

     A = 4π R₀²

     ρ2 = F / 4π R₀²

     ρ2 = F / 4π R₀²

Atomic number is the number of protons in the nucleon in not very heavy nuclei. This number is equal to the number of neutrons, but changes in heavier nuclei, there are more neutrons than protons.

Let's look for the relationship of the two densities

     ρ/ρ2 = ¾ A / π R₀³ / (F / 4π R₀²)

     ρ /ρ2 = 3 (A / F) (1 / R₀)

In this case it does not say that the nucleon number is A (F = A), the relationship is

     ρ/ρ2 = 3 / R₀

I see that the two densities are different

3 0
3 years ago
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