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HACTEHA [7]
3 years ago
15

A student claims that titanium is the strongest material to fracture. Based on this interaction, which evidence would best suppo

rt the student's claim?
a. Titanium takes the longest time to reach fracture.
b. Titanium has the highest fracture stress value.
c. Titanium deforms the least before fracture.
d. Titanium increases its fracture strength as the load increases.
Engineering
1 answer:
Nataliya [291]3 years ago
8 0

Answer:

b. Titanium has the highest fracture stress value.

Explanation:

Titanium is the strongest material to fracture, according to the student. To support his claim, the evidence should be .. the titanium has the highest stress value for fracture.  

If the materials is having high value of fracture stress it means that it is strong material and it is not easy to break that type of materials.

Therefore the answer will be B.

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The speed of an aircraft is given to be 260 m/s in air. If the speed of sound at that location is 330 m/s, the flight of the air
navik [9.2K]

Answer:

Ma=\frac{260}{330} \\Ma=0.7878

So, Ma < 1  Flow is Subsonic

Explanation:

Mach Number:

Mach Number is the ratio of speed of the object to the speed of the sound. It is used to categorize the speed of the object on the basis of mach number as sonic, supersonic and hyper sonic. (It is a unit less quantity)

Mach < 1       Subsonic

Mach > 1       Supersonic

Ma= Speed of the object/Speed of the sound

Ma=\frac{260}{330} \\Ma=0.7878

So, Ma < 1 Flow is Subsonic

8 0
3 years ago
Two well-known NP-complete problems are 3-SAT and TSP, the traveling salesman problem. The 2-SAT problem is a SAT variant in whi
Hitman42 [59]

3-SAT ≤p TSP

If P ¹ NP, then no NP-complete problem can be solved in polynomial time.

both the statements are true.

<u>Explanation:</u>

  • 3-SAT ≤p TSP due to any  complete problem of NP to other problem by exits of reductions.
  • If P ¹ NP, then 3-SAT ≤p 2-SAT are the polynomial time algorithm are not for 3-SAT. In P, 2-SAT is found, 3- SAT polynomial time algorithm implies the exit of reductions. 3 SAT does not have polynomial time algorithm when P≠NP.
  • If P ¹ NP, then no NP-complete problem can be solved in polynomial time. because for the NP complete problem individually gets the polynomial time algorithm for the others. It may be in P for all the problems, the implication of latter is P≠NP.
7 0
3 years ago
The unit weight of a soil is 14.9kN/m3. The moisture content of the soil is17% when the degree of saturation is 60%. Determine:
Serggg [28]

Answer:

a) 2622.903 N/m^3

b) 1.38233

c)4.878811765

Explanation:

Find the void ratio using the formula:

y = \frac{G_{s}*y_{w} + w*G_{s}*y_{w} }{1+e} ....... Eq1

Here;

G_{s} is specific gravity of soil solids

y_{w} is unit weight of water = 998 kg/m^3

w is the moisture content = 0.17

e is the void ratio

y is the unit weight of soil = 14.9KN/m^3

Saturation Ratio Formula:

w*G_{s} = S*e  ..... Eq2

S is saturation rate

Substitute Eq 2 into Eq 1

y = \frac{(\frac{S*e}{w}) * y_{w} + S*e*y_{w}  }{1+e}

14900 = \frac{3522.352941*e + 598.8*e }{1+e} = \frac{4121.152941*e}{1+e}\\\\ e= 1.38233

Specific gravity of soil solids

G_{s} = \frac{S*e}{w} = \frac{0.6*1.38233}{0.17} = 4.878811765

Saturated Unit Weight

y_{s} = \frac{(G_{s} + e)*y_{w}  }{1+e} \\=\frac{(4.878811765 + 1.38233)*998  }{1+1.38233}\\\\= 2622.902571 N/m^3

7 0
3 years ago
What should always be done before beginning any diagnosis?
vladimir2022 [97]

Answer:

c

Explanation:

if someone is wrong that they can help with

4 0
3 years ago
The output side of an ideal transformer has 35 turns, and supplies 2.0 A to a 24-W device. Ifthe input is a standard wall outlet
Crank

Answer:

The current drawn from the outlet is 0.2 A

The number of turns on the input side is 350 turns

Explanation:

Given;

number of turns of the secondary coil, Ns = 35 turns

the output current, I_s = 2 A

power supplied, P_s = 24 W

the standard wall outlet in most homes = 120 V = input voltage

For an ideal transformer; output power = input power

the current drawn from the outlet is calculated;

I_pV_p = P_s\\\\I_p = \frac{P_s}{V_p} = \frac{24}{120} = 0.2 \ A

The number of turns on the input side is calculated as;

\frac{N_p}{N_s} = \frac{I_s}{I_p}  \\\\N_p = \frac{N_sI_s}{I_p} \\\\N_p = \frac{35 \times 2}{0.2} \\\\N_p = 350 \ turns

4 0
3 years ago
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