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Assoli18 [71]
3 years ago
15

What is the mean free path for the molecules in an ideal gas when the pressure is 100 kPa and the temperature is 300 K given tha

t the collision cross-section for the molecules of that gas is 2.0 � 10-20 m2? The Boltzmann constant is 1.38 � 10-23J/K, Avogadro's number is 6.02 � 1023 molecules/mole, and the ideal gas constant is R = 8.314 J/mol � K = 0.0821 L � atm/mol � K.

Physics
1 answer:
larisa86 [58]3 years ago
7 0

Explanation:

Below is an attachment containing the solution.

You might be interested in
An atom with seven valence electrons would most likely lose an electron to become stable true or fasle
mel-nik [20]

Answer:

false

Explanation:

since it has seven valence electrons it means it's a non metal and non metals always gain electrons.

5 0
2 years ago
A wheel of mass 4kg is pulled up a plane inclined at 30° to the horizontal by a force of 45N applied to the axle and parallel to
Len [333]

Answer:

v = 10 m/s

Explanation:

Let's assume the wheel does not slip as it accelerates.

Energy theory is more straightforward than kinematics in my opinion.

Work done on the wheel

W = Fd = 45(12) = 540 J

Some is converted to potential energy

PE = mgh = 4(9.8)12sin30 = 235.2 J

As there is no friction mentioned, the remainder is kinetic energy

KE = 540 - 235.2 = 304.8 J

KE = ½mv² + ½Iω²

ω = v/R

KE = ½mv² + ½I(v/R)² = ½(m + I/R²)v²

v = √(2KE / (m + I/R²))

v = √(2(304.8) / (4 + 0.5/0.5²)) = √101.6

v = 10.07968...

5 0
3 years ago
The dot or scalar product of two (3d) vec- tors ⃗a = ⟨a1,a2,a3⟩ and ⃗b = ⟨b1,b2,b3⟩ is defined as
Neko [114]

Yes, yes, we know all of that.  It certainly took you long enough to
get around to asking your question.

If
     a = (14, 10.5, 0)
and
     b = (4.62, 9.45, 0) ,

then, to begin with, neither vector has a z-component, and they
 both lie in the x-y plane.

Their dot-product  a · b = (14 x 4.62) + (10.5 x 9.45) =

                                             (64.68)   +   (99.225)  =  163.905 (scalar)          


I feel I earned your generous 5 points just reading your treatise and
finding your question (in the last line).  I shall cherish every one of them.                     

7 0
3 years ago
U = 3, 9 , v = 4, 2 (a) find the projection of u onto v. (b) find the vector component of u orthogonal to v.
ExtremeBDS [4]

Answer:

(a) At U = 3, 9 , v = 4, 2, the projection of u onto v is w1=<2,8>

(b)At U = 3, 9 , v = 4, 2,  the vector component of u orthogonal to v is w2 = <4,-1>

Explanation:

A

The projection of u onto v is given by:

w1= projvu = (u⋅v||v||2)v

Given that  u= <6,7> and v=<1,4>, we can find the projection of u onto v as shown below:

w1= projvu = (u⋅v||v||2v=(<6,7>⋅<1,4><1,4>⋅<1,)

=(6⋅1+7⋅41⋅1+4⋅4)<1,4>

=3417<1,4>

=<2,8>

Part B

The vector component of u orthogonal to v is given by:

Using the given vectors and the projection found in part (a), we can find the vector component of u orthogonal to v as shown below:

w2=u−projvu

=<6,7≻<2,8>

=<(6−2),(7−8)>

=<4,−1>

To learn more about vector component, click brainly.com/question/17016695

#SPJ4

5 0
2 years ago
The metal gold crystallizes in a face centered cubic unit cell with one atom per lattice point. When X-rays with λ = 1.436 Å are
Umnica [9.8K]

Answer:

 r =  1.45 Å

Explanation:

given,

λ = 1.436 Å

θ = 20.62°

d = a

n = 2

metal gold crystallizes in a face centered cubic unit cell

Radius of the gold atom = ?

using Bragg's Law

 n λ = 2 d sin θ

 2 x 1.436 Å = 2 a sin 20.62°

 a = 4.077 Å

We know relation of radius for face centered cubic unit cell

 a = \dfrac{4r}{\sqrt{2}}

 4.077= \dfrac{4\times r}{\sqrt{2}}

 r =  1.45 Å

the radius of a(n) gold atom. is equal to 1.45 Å

7 0
3 years ago
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