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valentinak56 [21]
3 years ago
13

If the mass of a material is 111 grams and the volume of the material is 23 cm3, what would the density of the material be? g/cm

3
Physics
1 answer:
Butoxors [25]3 years ago
4 0
Density can be kg/m^3 or g/cm3
In g/cm3 density =mass /volume =111g/23cm3
=4.826g/cm3.

In kg/m3,density=mass/volume. converting mass in grams to kg, 1000g=1kg,111g=0.111kg.
cm3 to m3, 1cm3=10^-6m3, 23cm3=0.000023m3
density=0.111kg/0.000023m3 or 2.3*10^-5=4,826.1kg/m3.

the other is a long process.
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8 0
2 years ago
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
The machine in the figure is ideal and an effort force of
DENIUS [597]

Answer:

1.5 m

Explanation:

Let the distance from the box to the pivot be c.

Let the distance from the pivot to the effort be y.

From the question given above, the following data were obtained:

Effort force (Fₑ) = 7 N

Force of resistance (Fᵣ) = 14 N

Distance from the box to the pivot (c) = 0.75 m

Distance from the pivot to the effort (y) =?

Clockwise moment = Fₑ × y

Anticlock wise moment = Fᵣ × c

Clockwise moment = Anticlock wise moment

Fₑ × y = Fᵣ × c

7 × y = 14 × 0.75

7 × y = 10.5

Divide both side by 7

y = 10.5 / 7

y = 1.5 m

Therefore, the distance from the pivot to the effort is 1.5 m

5 0
2 years ago
The force of attraction between two oppositely charged pith is 5mx 10 to the -6th power newtons. If the charge on the two is 6.7
My name is Ann [436]

Answer:

0.28 m

Explanation:

The following data were obtained from the question:

Force (F) = 5×10¯⁶ N

Charge 1 (q₁) = 6.7×10¯⁹ C

Charge 2 (q₂) = 6.7×10¯⁹ C

Electrical constant (K) = 9×10⁹ Nm²C¯²

Distance apart (r) =?

Thus, the distance between the two charges can be obtained as follow:

F = Kq₁q₂/r²

5×10¯⁶ = 9×10⁹ × 6.7×10¯⁹ × 6.7×10¯⁹/r²

5×10¯⁶ = 4.0401×10¯⁷ / r²

Cross multiply

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Divide both side by 5×10¯⁶

r² = 4.0401×10¯⁷ / 5×10¯⁶

Take the square root of both side

r = √(4.0401×10¯⁷ / 5×10¯⁶)

r = 0.28 m

Therefore, the distance between the two charges is 0.28 m

4 0
3 years ago
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