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Tom [10]
3 years ago
10

A three-point flexure test is conducted on a cylindrical specimen of aluminum oxide. The specimen radius is 5.0 mm and the dista

nce between the two supportsis 15.0 mm. If the specimen breaks when the applied force is 7500 N and the center deflection is 0.01 mm, what are the elastic modulus andthe flexural strength of the material?

Engineering
1 answer:
kondaur [170]3 years ago
7 0

Answer:

Detailed solution is given in the attached diagram

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A sports car has a drag coefficient of 0.29 and a frontal area of 20 ft2, and is travelling at a speed of 120 mi/hour. How much
Andrej [43]

Answer:

Power required to overcome aerodynamic drag is 50.971 KW

Explanation:

For explanation see the picture attached

4 0
2 years ago
Engineering Questions
valkas [14]
5 is the correct one to choose for this
6 0
2 years ago
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A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28
Delicious77 [7]

Answer:

5984.67N

Explanation:

A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28 psi and a pressure drop of 2 psi occurs through the contraction if the upstream velocity is 4.0 ft/sec. What is the magnitude of the resultant force (lbs) needed to hold the pipe in place?

from continuity equation

v1A1=v2A2

equation of continuity

v1=4ft /s=1.21m/s

d1=14 inch=.35m

d2=14-2=0.304m

A1=pi*d^2/4

0.096m^2

a2=0.0706m^2

from continuity once again

1.21*0.096=v2(0.07)

v2=1.65

force on the pipe

(p1A1- p2A2) + m(v2 – v1)

from bernoulli

p1 + ρv1^2/2 = p2 + ρv2^2/2

difference in pressure or pressure drop

p1-p2=2psi

13.789N/m^2=rho(1.65^2-1.21^2)/2

rho=21.91kg/m^3

since the pipe is cylindrical

pressure is egh

13.789=21.91*9.81*h

length of the pipe is

0.064m

AH=volume of the pipe(area *h)

the mass =rho*A*H

0.064*0.07*21.91

m=0.098kg

(193053*0.096- 179263.6* 0.07) + 0.098(1.65 – 1.21)

force =5984.67N

4 0
3 years ago
Investigating how slime molds reproduce is an example of applied research.<br> True<br> False
azamat

Answer:

false

Explanation:

3 0
3 years ago
How would you describe what would happen to methane if the primary bonds were to break?
erastova [34]

Answer:

All the bonds in methane (CH4CH4) are equivalent, and all have the same dissociation energy.

The product of the dissociation is methyl radical (CH3CH3). All the bonds in methyl radical are equivalent, and all have the same dissociation energy.

The product of that dissociation is methylene (CH2CH2). All the bonds in methylene are equivalent, and all have the same dissociation energy.

The product of that dissociation is methyne (CHCH) .

The C-H bonds in methane do not have the same dissociation energy as C-H bonds in methyl radical, which in turn do not have the same dissociation energy as the C-H bonds in methylene, which are again different from the C-H bond in methyne.

If (by some miracle) you were able to get all four bonds in methane to dissociate absolutely simultaneously, they would all show the same dissociation energy… but that energy, per bond broken, would be different than the energy required to break just one C-H bond in methane, because the products are different.

(In this case, it’s CH4→C+4HCH4→C+4H versus CH4→CH3+HCH4→CH3+H.)

To alter hydrocarbons you add enough energy to break a C-H bond. Why does only one bond break? What concentrates the energy on one C-H bond?

the weakest CH bond is the one that breaks. in plain alkanes it has to do with the molecular orbital interactions between neighboring carbon atoms. look at propane for example. the middle carbon has two C-C bonds, and each of those C-C bonds is strengthened by slight electron delocalization from the C-H bonds overlapping with the antibonding orbitals of the adjacent carbons.

since the C-H bonds on the middle carbon donate electron density to both of its neighbors, those two are weakest.

one of them will break preferentially.

which one actually breaks depends on the reaction conditions (kinetics). frankly it's whichever one ramdomly approaches a nucleophile first. when the nucleophile pulls of one of the H's, the other C-H bonds start to share (delocalize) the negative charge across the whole molecule. so while the middle C feels the majority of the negative charge character, the other two C's take on a fair amount as well...

by the way, alkanes don't really like to break and form anions like that.

a better example would be something like isopropyl iodide, where the C-I bond breaks and the I carries away the electron pair, forming a carbocation (also not particularly stable, but more so than the carbanion).

7 0
2 years ago
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