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Blababa [14]
3 years ago
10

PLEASE HELPPPPPPP!!!!,

Engineering
2 answers:
s2008m [1.1K]3 years ago
8 0
Hydraulic engineering
raketka [301]3 years ago
4 0


software engineers

hardware engineers

metallurgic engineers

biomechanical engineers
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Damien wants to study the effect on materials let 60 after it is submerged in water for varying durations what type of graphical
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Answer:

please help you are not the intended recipient

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3 years ago
The steel bracket is used to connect the ends of two cables. if the allowable normal stress for the steel is sallow = 30 ksi, de
garri49 [273]

The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb

<h3>The static equilibrium is given as:</h3>

F = P (Normal force)

Formula for moment at section

M = P(4 + 1.5/2)

= 4.75p

Solve for the cross sectional area

Area = \frac{\pi d^{2} }{4}

d = 1.5

\frac{\pi *1.5^{2} }{4}

= 1.767 inches²

<h3>Solve for inertia</h3>

\frac{\pi *0.75^4}{4}

= 0.2485inches⁴

Solve for the tensile force from here

\frac{F}{A} +\frac{Mc}{I}

30x10³ = \frac{P}{1.767} +\frac{4.75p*0.75}{0.2485} \\\\

30000 = 14.902 p

divide through by 14.902

2013.15 = P

The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb

Read more on tensile force here: brainly.com/question/25748369

4 0
2 years ago
For this problem, calculate the following by hand and show the procedure for how you obtained the results. Subsequently, solve p
Gemiola [76]

Answer:wat

Explanation:

4 0
3 years ago
Advantages, disadvantages and applications of dsb-sc​
allochka39001 [22]

Answer:

<h3>advantages: </h3>

<em>lower power consumption, modulation system is simple</em>

<h3>disadvantages<em>:</em></h3>

<em>complex detection</em>

<h3><em>applications:</em></h3>

analog TV systems: to transmit color information

<h3><em /></h3>

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Explanation:

3 0
3 years ago
Assume the average fuel flow rate at the peak torque speed (1500 rpm) is 15kg/hr for a sixcylinder four-stroke diesel engine und
sveticcg [70]

Answer:

Q = 8.845 DEGREE

Explanation:

given data:

combine Mass for 6 cylinder (M) =15 Kg/hr

mass of  each cylinder (m) = 15/6 = 2.5 Kg/hr = 0.000694 Kg/ sec

Engine speed (N)= 1500rpm

Diameter of one nozzle hole ( d) = 200 micrometer = 0.0002 m

Discharge Coefficient (Cd) = 0.75

Pressure difference = 100 MPa

Density of fuel = 800 kg/m^3

velocity of fuel is v  = cd\sqrt{\frac{2*P}{p}}

v = 0.75 \sqrt{\frac{2\times 100\times 10^6}{800}} = 375 m/sec

injected fuel volume  (V) =Area of given  Orifices × Fuel velocity × time of single injection × no of injection/sec

we know that p = m/ V

SoV = \frac{0.000694}{800} =8.68\times10^{-7} m3/sec

putting these value in volume equation and solve for Discharge 8.68\times 10^{-7} = (\frac{(3.14}{4})\times 6\times( .0002\times .0002) \times  375 \times  \frac{(Q}{360}) \times \frac{30}{750} \times \frac{(750}{60)}

Q = 8.845 DEGREE

4 0
2 years ago
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