Mining is an example of this type of business

Engineers create a new metal that is stronger than steel but much lighter. This material is also significantly cheaper than what

zysi [14]

The best step for the engineers to make next is option D. Begin to design an airplane using this metal.

<h3>What is the metallic is plane parts?</h3>

Aluminum and its alloys are nevertheless very famous uncooked substances for the production of business planes, because of their excessive electricity at exceedingly low density. Currently, excessive-electricity alloy 7075, which includes copper, magnesium and zinc, is the only used predominantly withinside the plane industry.

The solution is D, due to the fact even as it's far crucial to marketplace the fabric and ensure humans are inquisitive about buying, they first want to attempt to layout aircraft the usage of this fabric. There isn't anyt any use promoting an aircraft constituted of this material_ if a aircraft can not be built.

Read more about the aircraft:

brainly.com/question/5055463

#SPJ1

7 0

2 years ago

before adjusting drive-belt tension, technician a checks for proper pulley alignment. technician b looks up the specified belt t

Vsevolod [243]

Answer:

Technician b is correct

Explanation:

Before adjusting drive-belt tension, it is very important to check the vehicle workshop manual for specified belt tension, so that you can match your reading against the specification in the vehicle's service manual. If the tension reading you have matches the suggested reading in the vehicle's service manual and the belt is not damaged then you do not need to proceed any further. But if the reading does not match, then you can adjust the belt tension.

Therefore, technician b is correct.

5 0

3 years ago

A small lake with volume of 160,000 m^3 receives agricultural drainage waters that contain 150 mg / L total dissolved solids (TD

Stels [109]

Answer:

Explanation:

Given that : -

The desirable limit is 500 mg / l , but

allowable upto 2000 mg / l.

The take volume is V = 160.000 m3

V = 160 , 000 x 103 l

The crainage gives 150 mg / l and lake has initialy 100 mg / l

Code of tpr frpm drawn = 150 x 60, 000 x 1000

Ci = 9000 kg / gr

Cl = 100 x 160,000 x 1000

Cl = 16, 000 kg

Since allowable limit = 2000 mg / l

Cn = ( 2000 x 160, 00 x 1000 )

= 320, 000 kg

so, each year the rate increases, by 9000 kg / yr

Read level = ( 320, 000 - 16,000 )

Li = 304, 000 kg

Tr=<u>304,000</u>

900

=33.77

5 0

3 years ago

Read 2 more answers

If gain of the critically damped system is increased, the system will behave as a) Under damped b) Over damped c) Critically dam

Ganezh [65]

Answer:

a) Under damped

Explanation:

Given that system is critically damped .And we have to find out the condition when gain is increased.

As we know that damping ratio given as follows

$\zeta =\dfrac{C}{C_c}$

Where C is the damping coefficient and Cc is the critical damping coefficient.

$C_c=2\sqrt{mK}$

So from above we can say that

$\zeta =\dfrac{C}{2\sqrt{mK}}$

$\zeta \alpha \dfrac{1}{\sqrt K}$

From above relationship we can say when gain (K) is increases then system will become under damped system.

7 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

So $V = \frac{0.000694}{800} =8.68\times10^{-7} m3/sec$

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

3 years ago