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abruzzese [7]
2 years ago
10

The density of molybdenum is 10.28 g/cm^3 and it crystallizes in the face centered cubic unit cell. Calculate the edge length of

the unit cell. (The atomic mass of Mo is 95.96 g/mole)
Physics
1 answer:
Effectus [21]2 years ago
4 0

The edge length of the unit cell at the given atomic mass and density of the molybdenum is 314.2 pm.

<h3>Volume of molybdenum</h3>

V = (zm/ρN)

where;

  • z is 2 for cubic unit cell
  • m is mass of the molybdenum
  • ρ is density of the molybdenum

V = (2 x 95.96) / (10.28 x 6.02 x 10²³)

V = 3.10 x 10⁻²³ cm³

<h3>Edge length of the unit cell</h3>

a³ = V

a = (V)^¹/₃

a = ( 3.10 x 10⁻²³)^¹/₃

a = 3.142 x 10⁻⁸ cm

a = 3.142 x 10⁻¹⁰ m

a = 314.2 x 10⁻¹² m

a = 314.2 pm

Thus, the edge length of the unit cell at the given atomic mass and density of the molybdenum is 314.2 pm.

Learn more about edge length here:

brainly.com/question/16673486

#SPJ1

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What is the difference between condensation and conduction
aniked [119]

Answer:

Convection is the heat transfer due to the bulk movement of molecules within fluids such as gases and liquids, including molten rock. Condensation is the process by which water vapor in the air is changed into liquid water. Condensation is crucial to the water cycle because it is responsible for the formation of clouds. These clouds may produce precipitation, which is the primary route for water to return to the Earth's surface within the water cycle.

Explanation:

Their a difference....... A huge One

8 0
3 years ago
g In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period
NeTakaya

The question is incomplete. The complete question is :

In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period of the oscillations is 1.2 microseconds (1.2*10^-6). What is Q?

Solution :

The underdamped RLC circuit

$v_{t} = ve^{-\frac{R}{2L}t} \cos \omega t$

$\omega = \sqrt{\frac{1}{LC}-\frac{R^2}{4L^2}}= \frac{2 \pi}{T}$

We know in one time period, v = 2v, at t = T, $v_t = 3.8 v$

so, $9.8 = 5 e^{-\frac{R}{2L}T} \cos \frac{2 \pi}{T}T$

   $e^{-\frac{R}{2L}T} = \frac{3.8}{5} \times 1$

   $\frac{R}{2L}T= \ln \frac{5}{3.8}$

  $\frac{R}{L}= \frac{2}{1.2 \times 10^{-6}} \ln \frac{5}{3.8}$

 $\frac{R}{L} = 457.3 \times 10^3$

Now, Q value $= \frac{1}{R}\sqrt{\frac{L}{C}}$

                     $=\sqrt{\frac{L}{R^2C}\times \frac{L}{L}}$

                     $=\sqrt{(\frac{L}{R})^2 \times \frac{1}{LC}}$

              $\frac{1}{LC}=27.43 \times 10^{12}$

∴ $Q=\sqrt{\left(\frac{1}{457.3 \times 10^3}\right)^2 \times 27.43 \times 10^{12}}$

  $Q=\sqrt{131.166}$

      = 11.45

4 0
2 years ago
a car with a mass of 1200 kilograms is moving around a circular curve at a uniform velocity of 20 meters per second. the centrip
qaws [65]
Well, first of all, a car moving around a circular curve is not moving
with uniform velocity.  The direction of motion is part of velocity, and
the direction is constantly changing on a curve.

The centripetal force that keeps an object moving in a circle is

         Force  =  (mass of the object) · (speed)² / (radius of the circle)

         F  =  m s² / r

We want to know the radius, to rearrange the formula to give us
the radius as a function of everything else.

                                          F     =  m s² / r

Multiply each side by 'r':       F· r  =  m · s²

Divide each side by 'F':            r  =  m · s² / F    

We know all the numbers on the right side,
so we can pluggum in:

                      r  =       m       ·        s²      /     F

                      r  =  (1200 kg) · (20 m/s)² / (6000 N) .

I'm pretty sure you can finish it up from here.

                                      


5 0
3 years ago
A car speeds upu as it rolls down a hill. Which is this an example of?
anygoal [31]
Positive acceleration
4 0
3 years ago
Read 2 more answers
Please help me!! I can't figure the answer out, AND haven't gotten past this (being the block test) for a couple weeks now!!
ololo11 [35]

The density of the nugget is 19 g/cm^3 and is made of gold

Explanation:

The density of an object can be calculated as

d=\frac{m}{V}

where

d is the density

m is the mass

V is the volume of the object

We have to note that density of an object actually depends on the material the object is made of (therefore, two objects made of the same material can have different mass and different volume, but they have same density).

For the nugget in this problem, we have:

mass: m = 38 g

volume: V=2 cm^3

So, its density is

d=\frac{38}{2}=19 g/cm^3

And by looking at the table, we see that this value corresponds approximately to the density of gold, so the nugget is made of gold.

Learn more about density:

brainly.com/question/5055270

brainly.com/question/8441651

#LearnwithBrainly

5 0
3 years ago
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