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Ilia_Sergeevich [38]
2 years ago
14

public interface Frac { /** @return the denominator of this fraction */ int getDenom(); /** @return the numerator of this fracti

on */ int getNum(); } Which of the interfaces, if correctly implemented by a Fraction class, would be sufficient functionality for a user of the Fraction class to determine if two fractions are equivalent

Engineering
1 answer:
True [87]2 years ago
7 0

Answer:

The Full details of the answer is attached.

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It is desired to produce and aligned carbon fiber-epoxy matrix composite having a longitudinal tensile strength of 630 MPa. Calc
ratelena [41]

Answer:

The answer is below

Explanation:

Given that:

Diameter (D) = 0.03 mm = 0.00003 m, length (L) = 2.4 mm = 0.0024 m, longitudinal tensile strength (\sigma_{cd})=630\ MPa = 630*10^6\ Pa, Fracture strength

(\sigma_f)=5100\ MPa=5100*10^6\ Pa,fiber-matrix\ stres(\sigma_m)=17.5\ MPa=17.5*10^6\ Pa,matrix\ strength=\tau_c=17\ MPa=17 *10^6\ Pa

a) The critical length (L_c) is given by:

L_c=\sigma_f*(\frac{D}{2*\tau_c} )=5100*10^6*\frac{0.00003}{2*17*10^6}=0.0045\ m=4.5\ mm

The critical length (4.5 mm) is greater than the given length, hence th composite can be produced.

b) The volume fraction (Vf) is gotten from the formula:

\sigma_{cd}=\frac{L*\tau_c}{D}*V_f+\sigma_m(1-V_f)\\\\V_f=\frac{\sigma_{cd}-\sigma_{m}}{\frac{L*\tau_c}{D}-\sigma_{m}}  \\\\Substituting:\\\\V_f=\frac{630*10^6-17.5*10^6}{\frac{0.0024*17*10^6}{0.00003} -17.5*10^6} \\\\V_f=0.456

6 0
2 years ago
Will mark brainliest!
sukhopar [10]

Ethics in production is a subset of business ethic that is meant to ensure that the production function or activities are not damaging to the consumer or the society. Like other ethics there is a certain code of conduct or standards to be followed, however ensuring that the ethics are complied with is often difficult.

<h2>i HOPE IT'S HELP </h2>
6 0
2 years ago
Read 2 more answers
Technician A says that mechanical shifting controls can wear out over time. Technician B says that vacuum control rubber diaphra
diamong [38]

Based on the information, both technician A and technician B are correct.

<h3>How to depict the information?</h3>

From the information given, Technician A says that mechanical shifting controls can wear out over time.

Technician B says that vacuum control rubber diaphragms can deteriorate over time.

In this case, both technicians are correct as the information depicted is true.

Learn more about technicians on:

brainly.com/question/1548867

#SPJ12

8 0
1 year ago
A Pitot-static probe is used to measure the speed of an aircraft flying at 3000 m. If the differential pressure reading is 3200
coldgirl [10]

Answer:

Speed of aircraft ; (V_1) = 83.9 m/s

Explanation:

The height at which aircraft is flying = 3000 m

The differential pressure = 3200 N/m²

From the table i attached, the density of air at 3000 m altitude is; ρ = 0.909 kg/m3

Now, we will solve this question under the assumption that the air flow is steady, incompressible and irrotational with negligible frictional and wind effects.

Thus, let's apply the Bernoulli equation :

P1/ρg + (V_1)²/2g + z1 = P2/ρg + (V_2)²/2g + z2

Now, neglecting head difference due to high altitude i.e ( z1=z2 ) and V2 =0 at stagnation point.

We'll obtain ;

P1/ρg + (V_1)²/2g = P2/ρg

Let's make V_1 the subject;

(V_1)² = 2(P1 - P2)/ρ

(V_1) = √(2(P1 - P2)/ρ)

P1 - P2 is the differential pressure and has a value of 3200 N/m² from the question

Thus,

(V_1) = √(2 x 3200)/0.909)

(V_1) = 83.9 m/s

4 0
3 years ago
Design a ductile iron pumping main carrying a discharge of 0.35 m3/s over a distance of 4 km. The elevation of the pumping stati
snow_tiger [21]

Answer:

D=0.41m

Explanation:

From the question we are told that:

Discharge rate V_r=0.35 m3/s

Distance d=4km

Elevation of the pumping station h_p= 140 m

Elevation of the Exit point h_e= 150 m

Generally the Steady Flow Energy Equation SFEE is mathematically given by

h_p=h_e+h

With

P_1-P_2

And

V_1=V-2

Therefore

h=140-150

h=10

Generally h is give as

h=\frac{0.5LV^2}{2gD}

h=\frac{8Q^2fL}{\pi^2 gD^5}

Therefore

10=\frac{8Q^2fL}{\pi^2 gD^5}

D=^5\frac{8*(0.35)^2*0.003*4000}{3.142^2*9.81*10}

D=0.41m

8 0
3 years ago
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