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Akimi4 [234]
2 years ago
5

Fffffrrrrreeeeee points you can have them!

Physics
2 answers:
stiv31 [10]2 years ago
3 0

Answer:

Thank you so much!

Explanation:

mafiozo [28]2 years ago
3 0

Answer:

joe

Explanation:

mama

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Polar molecules have ionic bonds.<br> True<br> False
Nadusha1986 [10]
Polar molecules do have ionic bonds
6 0
2 years ago
Suppose we hang a heavy ball with a mass 13 kg (so the weight is ) from a steel wire 3.9 m long that is 3.1 mm in diameter (radi
Katena32 [7]

Answer:

1.635×10^-3m

Explanation:

Young modulus is the ratio of the tensile stress of a material to its tensile strain.

Young modulus = Tensile stress/tensile strain

Tensile stress = Force/Area

Given force = 130N

Area = Πr² = Π×(1.55×10^-3)²

Area = 4.87×10^-6m²

Tensile stress = 130/4.87×10^-6 = 8.39×10^7N/m²

Tensile strain = extension/original length

Tensile strain = e/3.9

Substituting in the young modulus formula given young modulus to be 2×10¹¹N/m²

2×10¹¹N/m² = 8.39×10^7/{e/3.9)}

2×10¹¹ = (8.39×10^7×3.9)/e

2×10¹¹e = 3.27×10^8

e = 3.27×10^8/2×10¹¹

e = 1.635×10^-3m

The stretch of the steel wire will be

1.635×10^-3m

7 0
3 years ago
The three main types of nuclear radiation are alpha, beta, and gamma
valkas [14]

Answer:

Gamma beta alpha

Explanation:

6 0
2 years ago
Read 2 more answers
Anyone willing to help?
grigory [225]

Answer:

I think it is "A" because the wind systems are created by uneven heating of Earth's surface.It also may help form large global wind patterns.

Explanation:

5 0
2 years ago
Read 2 more answers
Some hypothetical alloy is composed of 12.5 wt% of metal A and 87.5 wt% of metal B. If the densities of metals A and B are 4.27
densk [106]

Answer:

The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

Explanation:

Given that,

Weight of metal A = 12.5%

Weight of metal B = 87.5%

Length of unit cell = 0.395 nm

Density of A = 4.27 g/cm³

Density of B= 6.35 g/cm³

Weight of A = 61.4 g/mol

Weight of B = 125.7 g/mol

We need to calculate the density of the alloy

Using formula of density

\rho=n\times\dfrac{m}{V_{c}\times N_{A}}

n=\dfrac{\rho\timesV_{c}\times N}{m}....(I)

Where, n = number of atoms per unit cells

m = Mass of the alloy

V=Volume of the unit cell

N = Avogadro number

We calculate the density of alloy

\rho=\dfrac{1}{\dfrac{12.5}{4.27}+\dfrac{87.5}{6.35}}\times100

\rho=5.98

We calculate the mass of the alloy

m=\dfrac{1}{\dfrac{12.5}{61.4}+\dfrac{87.5}{125.7}}\times100

m=111.15

Put the value into the equation (I)

n=\dfrac{5.9855\times(0.395\times10^{-9}\times10^{2})^3\times6.023\times10^{23}}{111.15}

n=1.99\approx 2\ atoms/cell

Hence, The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

5 0
3 years ago
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