Explain what the engineering team should advise in the following scenario.

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

Allisa [31]

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

3 0

3 years ago

Nancy ate a 500 Cal lunch. Neglecting efficiency issues (i.e., assuming 100% conversion of energy to work), to what height could

VladimirAG [237]

Answer:

$4265.04\ \text{m}$

$2.38\times 10^{10}\ \text{W}$

Explanation:

PE = Energy of food = 500 cal = $500\times4184=2.092\times10^6\ \text{J}$

m = Mass of object = 50 kg

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

Potential energy of food is given by

$PE=mgh\\\Rightarrow h=\dfrac{PE}{mg}\\\Rightarrow h=\dfrac{2.092\times 10^6}{50\times 9.81}\\\Rightarrow h=4265.04\ \text{m}$

Nancy could raise the weight to a maximum height of $4265.04\ \text{m}$ .

Mass of $H_2$ used per year = $25\times 10^{9}\ \text{kg/year}$

Energy of $H_2$ = $\dfrac{30\times10^9}{1000}=30\times 10^6\ \text{J/kg}$

Power

$P=25\times 10^{9}\ \text{kg/year}\times 30\times 10^6\ \text{J/kg}\\\Rightarrow P=7.5\times 10^{17}\ \text{J/year}\\\Rightarrow P=\dfrac{7.5\times 10^{17}}{365.25\times 24\times 60\times 60}\\\Rightarrow P=2.38\times 10^{10}\ \text{W}$

The power requirement is $2.38\times 10^{10}\ \text{W}$ .

6 0

3 years ago

Under the right conditions, it is possible, due to surface tension, to have metal objects float on water. Consider placing a sho

stiv31 [10]

Answer:

D = 0.060732 in

Explanation:

given data

sp. wt. = 500 lb/ft³

diameter = 0.036 in

solution

we get here maximum diameter of rod that is express as

D = $\sqrt{\frac{8 \sigma }{\pi y}}$ ......................1

here $\sigma$ surface tension of water at 60⁰f = 5.03 × $10^{-3}$ lb/ft and y = 500 lb/ft³

so put here value and we will get

D = $\sqrt{\frac{8 \times 5.03 \times 10^{-3} }{\pi \times 500}}$

D = 0.005061 ft

D = 0.060732 in

4 0

3 years ago

QUICK ASAP!!!

nignag [31]

Answer:

Im pretty sure the answer is B

Explanation:

Paul will need to persuade his client of the benefits of the different filter.

6 0

2 years ago

Admission to an aquarium is $14 per person. There is also an IMAX theatre in the building, which charges $8 per ticket for a 3D

adoni [48]

Answer:

x=14a+8b+16.5c

Explanation:

x=total cost

a=number of people in just aquarium

b=number of people in just IMAX

c=number of people in both

c=(14+8)×0.75=22×0.75=16.5

8 0

3 years ago