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strojnjashka [21]
3 years ago
13

How can you drop two eggs the feweHow can you drop two eggs the fewest amount of times, without them breaking? ...st amount of t

imes, without them breaking? ...
Engineering
2 answers:
Lesechka [4]3 years ago
8 0

Answer:

How can you drop two eggs the fewest amount of times, without them breaking it

Explanation:

This is done on any floor, the highest or the lowest. Simply drop the egg from one inch above your foot and it will not break.

AveGali [126]3 years ago
3 0

Answer:

answer

Explanation:

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What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 3.5 × 10
babunello [35]

Answer:

The magnitude of the maximum stress that exists at the tip of the internal crack is 3900.183MPa (565703.38 psi).

Explanation:

The magnitude of the maximum stress that exists at the tip of an internal crack can be estimated using fracture mechanics. If a material that have internal crack is loaded with tensile stress, the crack tends to widen due to the tensile force created by the stress, and this leads to a concentration of stress near the tip of the crack, thereby expanding the crack length and finally leading to failure of the material.

Fracture mechanics involves the study of propagation of cracks in a material. It is used to predict the conditions at which failure of a material will likely occur.

A crack may appear in the surface of the material (surface crack) or at the interior of the material (internal crack), but the same formula can be used to estimate the magnitude of the maximum stress at the tip of the crack.

The formula is given as Πm=2Πo√(a/r)

Where,

Πm is the maximum stress

Πo is the tensile stress

r is the radius of curvature

a is the crack length.

The major difference between the surface crack and the internal crack is the value for their crack length.

Crack length can be understood as the length of a crack at which the crack becomes unstable under certain applied stress.

For surface crack, crack length=a

For internal crack, crack length=2a because internal cracks forms two surfaces of a complete sphere, while the surface crack forms one surface of a sphere.

Given in the question,

Radius of curvature, r=3.5×10^(-4) mm=0.00035mm

Crack length, a=5.5×10^(-2) mm

For internal crack, a=[(crack length)/2]={[5.5×10^(-2)]/2}=0.0275mm

Tensile stress, Πo=220Mpa

Maximum stress=Πm

Using,

Πm=2Πo√(a/r)

=(2×220)√(0.0275/0.00035)

=440√78.571

=3900.183MPa (565703.38 psi)

4 0
3 years ago
Why would a Military Base be considered a Planned Community?
ira [324]
What he said is right ^^^
5 0
3 years ago
Read 2 more answers
A car is moving at 68 miles per hour. The kinetic energy of that car is 5 × 10 5 J.How much energy does the same car have when i
Blababa [14]

Answer:

The car has an energy of 1.017 × 10⁵ J when it moves at 97 miles per hour

Explanation:

Given the data in the question;

Initial velocity v₁ = 68 miles per hour = 30.398 meter per seconds

let mass of the car be m

kinetic energy of that car is 5 × 10⁵ J

so

E₁ = \frac{1}{2}mv²

we substitute

5 × 10⁵  = \frac{1}{2} × m × ( 30.398 )²

5 × 10⁵  = \frac{1}{2} × m × ( 30.398 )²

5 × 10⁵ = m × 462.019

m =  5 × 10⁵ / 462.019

m = 1082.2065 kg

Now, Also given that; v₂ = 97 miles per hour = 43.362 meter per seconds

E₂ = \frac{1}{2}mv₂²

we substitute

E₂ = \frac{1}{2} × 1082.2065 × ( 43.362 )²

E₂ = \frac{1}{2} × 1082.2065 × 1880.263

E₂ = 1.017 × 10⁵ J

Therefore, The car has an energy of 1.017 × 10⁵ J when it moves at 97 miles per hour

6 0
3 years ago
Which of the following best characterizes a scientific law?
sesenic [268]

Answer:

subjective im pretty sure

Explanation:

6 0
3 years ago
The mechanical properties of a metal may be improved by incorporating fine particles of its oxide. Given that the moduli of elas
Sedaia [141]

Answer:

a) 254.6 GPa

b) 140.86 GPa

Explanation:

a) Considering the expression of rule of mixtures for upper-bound and calculating the modulus of elasticity for upper bound;

Ec(u) = EmVm + EpVp

To calculate the volume fraction of matrix, 0.63 is substituted for Vp in the equation below,

Vm + Vp = 1

Vm = 1 - 0.63

Vm = 0.37

In the first equation,

Where

Em = 68 GPa, Ep = 380 GPa, Vm = 0.37 and Vp = 0.63,

The modulus of elasticity upper-bound is,

Ec(u) = EmVm + EpVp

Ec(u) = (68 x 0.37) + (380 x 0.63)

Ec(u) = 254.6 GPa.

b) Considering the express of rule of mixtures for lower bound;

Ec(l) = (EmEp)/(VmEp + VpEm)

Substituting values into the equation,

Ec(l) = (68 x 380)/(0.37 x 380) + (0.63 x 68)

Ec(l) = 25840/183.44

Ec(l) = 140.86 GPa

6 0
4 years ago
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