Which kind of fracture (ductile or brittle) is associated with each of the two crack propagation mechanisms?

Compute the atomic density (the number of atoms per cm3 rather than mass density g/cm3) for a perfect crystal of silicon at room

Ann [662]

Answer:

5E22 atoms/cm³

Explanation:

We need to find the number of moles of silicon per cm³

number of moles per cm³ = density/atomic weight = 2.33 g/cm ÷ 28.09 g/mol = 0.083 mol/cm³.

Since there are 6.022 × 10²³ atoms/mol, then the number of atoms of silicon per cm³ = number of atoms per mol × number of moles per cm³

= 6.022 × 10²³ atoms/mol × 0.083 mol/cm³

= 0.4995 × 10²³ atoms/cm³

= 4.995 × 10²² atoms/cm³

≅ 5 × 10²² atoms/cm³

= 5E22 atoms/cm³

5 0

3 years ago

Water flows with an average speed of 6.5 ft/s in a rectangular channel having a width of 5 ft The depth of the water is 2 ft.

lisov135 [29]

Answer:

specific energy = 2.65 ft

y2 = 1.48 ft

Explanation:

given data

average speed v = 6.5 ft/s

width = 5 ft

depth of the water y = 2 ft

solution

we get here specific energy that is express as

specific energy = y + $\frac{v^2}{2g}$ ...............1

put here value and we get

specific energy = $2 + \frac{6.5^2}{2\times 9.8\times 3.281}$

specific energy = 2.65 ft

and

alternate depth is

y2 = $\frac{y1}{2} \times (-1+\sqrt{1+8Fr^2})$

and

here Fr² = $\frac{v1}{\sqrt{gy}} = \frac{6.5}{\sqrt{32.8\times 2}}$

Fr² = 0.8025

put here value and we get

y2 = $\frac{2}{2} \times (-1+\sqrt{1+8\times 0.8025^2})$

y2 = 1.48 ft

7 0

3 years ago

if I had 5 toes and 6 toes how many toes will I have NO HINT try your best to get close to the answer give u 105 points

Rasek [7]

Answer:

11 toes on one foot? and 5 one the other or just 11 toes?

your NO HINT threw me off

Explanation:

3 0

3 years ago

Read 2 more answers

He engine in Problem 3.8 produces 57kW of brake power at 2000 RPM. Calculate:

Serjik [45]

Answer:

Answer is below attached.

Explanation:

3 0

3 years ago

A specimen of a 4340-steel alloy with a plane strain fracture toughness of 54.8 MPa sqrt(m) (50 ksi sqrt(in.)) is exposed to a s

Lunna [17]

Answer:

critical stress $\sigma _c$ = 1382.67 MPa

Explanation:

given data

plane strain fracture toughness = 54.8 MP

length of surface creak = 0.5 mm

we take here

parameter Y = 1.0

solution

we apply critical stress formula that is

critical stress $\sigma _c$ = $\frac{K}{Y\sqrt{\pi \times a} }$ .............................1

here K is design stress plane strain fracture toughness and a is length of surface creak so put all these value in equation 1

critical stress $\sigma _c$ = $\frac{54.8 \times 10^6}{1 \sqrt{\pi \times 5 \times 106{-4}}}$

solve it we get

critical stress = 1382.67 MPa

As exposed stress 1030 MPa is less than critical stress 1382 MPa

so that fracture will not be occur here

7 0

3 years ago