Ask question

Ask question

All categories

3 years ago

8

Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs by the propagation of an el

liptically shaped surface crack of length 0.28 mm and that has a tip radius of curvature of 0.002 mm when a stress of 1430 MPa is applied.

1 answer:

Allisa [31]3 years ago

3 0

Answer:

theoretical fracture strength = 16919.98 MPa

Explanation:

given data

Length (L) = 0.28 mm = 0.28 × 10⁻³ m

radius of curvature (r) = 0.002 mm = 0.002 × 10⁻³ m

Stress (s₀) = 1430 MPa = 1430 × 10⁶ Pa

solution

we get here theoretical fracture strength s that is express as

theoretical fracture strength = $s_{0} \times \sqrt{\frac{L}{r} }$ .............................1

put here value and we get

theoretical fracture strength = $1430 \times 10^6\times \sqrt{\frac{0.28\times 10^{-3}}{0.002\times 10^{-3}} }$

theoretical fracture strength = $16919.98 \times 10^6$

theoretical fracture strength = 16919.98 MPa

You might be interested in

The boost converter of Fig. 6-8 has parameter Vs 20 V, D 0.6, R 12.5 , L 10 H, C 40 F, and the switching frequency is 200 kHz. (

mr Goodwill [35]

Answer:

a) the output voltage is 50 V

b)

- the average inductor current is 10 A

- the maximum inductor current is 13 A

- the maximum inductor current is 7 A

c) the output voltage ripple is 0.006 or 0.6%V₀

d) the average current in the diode under ideal components is 4 A

Explanation:

Given the data in the question;

a) the output voltage

V₀ = V $_s$ /( 1 - D )

given that; V $_s$ = 20 V, D = 0.6

we substitute

V₀ = 20 / ( 1 - 0.6 )

V₀ = 20 / 0.4

V₀ = 50 V

Therefore, the output voltage is 50 V

b)

- the average inductor current

$I_L$ = V $_s$ / ( 1 - D )²R

given that R = 12.5 Ω, V $_s$ = 20 V, D = 0.6

we substitute

$I_L$ = 20 / (( 1 - 0.6 )² × 12.5)

$I_L$ = 20 / (( 0.4)² × 12.5)

$I_L$ = 20 / ( 0.16 × 12.5 )

$I_L$ = 20 / 2

$I_L$ = 10 A

Therefore, the average inductor current is 10 A

- the maximum inductor current

$I_{Lmax$ = [V $_s$ / ( 1 - D )²R] + [ V

given that, R = 12.5 Ω, V $_s$ = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz

we substitute

$I_{Lmax$ = [20 / (( 1 - 0.6 )² × 12.5)] + [ (20 × 0.6 × 5) / (2 × 10) ]

$I_{Lmax$ = [20 / 2 ] + [ 60 / 20 ]

$I_{Lmax$ = 10 + 3

$I_{Lmax$ = 13 A

Therefore, the maximum inductor current is 13 A

- The minimum inductor current

$I_{Lmax$ = [V $_s$ / ( 1 - D )²R] - [ V

given that, R = 12.5 Ω, V $_s$ = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz

we substitute

$I_{Lmin$ = [20 / (( 1 - 0.6 )² × 12.5)] - [ (20 × 0.6 × 5) / (2 × 10) ]

$I_{Lmin$ = [20 / 2 ] -[ 60 / 20 ]

$I_{Lmin$ = 10 - 3

$I_{Lmin$ = 7 A

Therefore, the maximum inductor current is 7 A

c) the output voltage ripple

ΔV₀/V₀ = D/RCf

given that; R = 12.5 Ω, C = 40 μF = 40 × 10⁻⁶ F, D = 0.6, f = 200 Khz = 2 × 10⁵ Hz

we substitute

ΔV₀/V₀ = 0.6 / (12.5 × (40 × 10⁻⁶) × (2 × 10⁵) )

ΔV₀/V₀ = 0.6 / 100

ΔV₀/V₀ = 0.006 or 0.6%V₀

Therefore, the output voltage ripple is 0.006 or 0.6%V₀

d) the average current in the diode under ideal components;

under ideal components; diode current = output current

hence the diode current will be;

$I_D$ = V₀/R

as V₀ = 50 V and R = 12.5 Ω

we substitute

$I_D$ = 50 / 12.5

$I_D$ = 4 A

Therefore, the average current in the diode under ideal components is 4 A

7 0

3 years ago

Carbon resistors often come as a brown cylinder with colored bands. These colored bands can be read to determine the manufacture

alexandr1967 [171]

Answer:

a) 4.7 kΩ, +/- 5%

b) 2.0 MΩ, +/- 20%

Explanation:

a) If the resistor has the following combination of color bands:

1) Yellow = 1st digit = 4

2) Violet = 2nd digit = 7

3) Red = multiplier = 10e2

4) Gold = tolerance = +/- 5%

this means that the resistor has 4700 Ω (or 4.7 kΩ), with 5% tolerance.

b) Repeating the process for the following combination of color bands:

1) Red = 1st digit = 2

2) Black = 2nd digit = 0

3) Green = multiplier = 10e5

4) Nothing = tolerance = +/- 20%

This combination represents to a resistor of 2*10⁶ Ω (or 2.0 MΩ), with +/- 20% tolerance.

7 0

3 years ago

Witch measuring tool would be used to determine the diameter of a crankshaft journal

djverab [1.8K]

Answer =

dial bore gauge

a “dial bore gauge” measures the inside of round holes, such as the bearing journals . can mesure up to 2” and 6” diameter holes .

when ( “ ) is next to a number it means inches fwi - but hope this helped have a good day :)

3 0

3 years ago

About what thickness of aluminum is needed to stop a beam of (a) 2.5-MeV electrons, (b) 2.5-MeV protons, and (c) 10-MeV alpha pa

Nana76 [90]

The thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.

<h3>Thickness of the aluminum</h3>

The thickness of the aluminum can be determined using from distance of closest approach of the particle.

$K.E = \frac{2KZe^2}{r}$

where;

Z is the atomic number of aluminium = 13
e is charge
r is distance of closest approach = thickness of aluminium
k is Coulomb's constant = 9 x 10⁹ Nm²/C²

<h3>For 2.5 MeV electrons</h3>

$r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (1.6\times 10^{-19})^2}{2.5 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m$

<h3>For 2.5 MeV protons</h3>

Since the magnitude of charge of electron and proton is the same, at equal kinetic energy, the thickness will be same. r = 1.5 x 10⁻¹⁴ m.

<h3>For 10 MeV alpha-particles</h3>

Charge of alpah particle = 2e

$r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (2 \times 1.6\times 10^{-19})^2}{10 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m$

Thus, the thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.

Learn more about closest distance of approach here: brainly.com/question/6426420

7 0

2 years ago

What is the activation energy (Q) for a vacancy formation if 10 moles of a metal have 2.3 X 10^13 vacancies at 425°C?

Yakvenalex [24]

Answer:

$Activation\ Energy=2.5\times 10^{-19}\ J$

Explanation:

Using the expression shown below as:

$N_v=N\times e^{-\frac {Q_v}{k\times T}$

Where,

$N_v$ is the number of vacancies

N is the number of defective sites

k is Boltzmann's constant = $1.38\times 10^{-23}\ J/K$

${Q_v}$ is the activation energy

T is the temperature

Given that:

$N_v=2.3\times 10^{13}$

N = 10 moles

1 mole = $6.023\times 10^{23}$

So,

N = $10\times 6.023\times 10^{23}=6.023\times 10^{24}$

Temperature = 425°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (425 + 273.15) K = 698.15 K

T = 698.15 K

Applying the values as:

$2.3\times 10^{13}=6.023\times 10^{24}\times e^{-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}$

$ln[\frac {2.3}{6.023}\times 10^{-11}]=-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}$

$Q_v=2.5\times 10^{-19}\ J$

4 0

3 years ago