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lord [1]
3 years ago
7

Using a do-while loop, program a factorial calculator

Engineering
1 answer:
Elanso [62]3 years ago
3 0

hope it help

please mark as brainliest

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In your new role at Wayne Industries, you have been given the freedom to propose and develop your own project ideas. You have an
gulaghasi [49]

Answer:

461.65 KJ/Kg

Explanation:

In this question, we are asked to calculate the values of heat transferred in the process.

Please check attachment for complete solution and step by step explanation

5 0
4 years ago
3. (5%) you would like to physically separate different materials in a scrap recycling plant. describe at least one method that
Likurg_2 [28]

One of the methods that are used to separate polymers, aluminium alloys, and steels from one another is the Gravitation Separation method.

One straightforward technique is to run the mixture through a magnet, which will keep the steel particles on the magnet and separate them from the polymer.

What is the Gravitation Separation method?

When it is practicable to separate two components using gravity, i.e., when the combination's constituent parts have different specific weights, gravity separation is a technique used in industry. The components can be in suspension or in a dry granular mixture.

Polymers, Steel and Aluminium alloys can be readily split apart. The technique depends on how the two components are combined. The approach used is gravitational density. Due to the significant difference in relative specific mass values between steel and polymers (which range from 1.0 to 1.5), it is possible to separate them using flotation in a liquid that is safe and has the right density.

Therefore, the Gravitation Separation method is used to separate polymers, aluminium alloys and steels.

To learn more about the Polymer from the given link

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8 0
2 years ago
Use a passband of 0 to 5 kHz with 5 kHz cutoff frequency and filter to attenuate all frequencies at and above 10 kHz by at least
Alinara [238K]

Answer:

See attached picture for answer.

Explanation:

See attached picture for explanation.

5 0
3 years ago
The hydrofoil boat has an A-36 steel propeller shaft that is 100 ft long. It is connected to an in-line diesel engine that deliv
Nataliya [291]

The question is incomplete. The complete question is :

The hydrofoil boat has an A-36 steel propeller shaft that is 100 ft long. It is connected to an in-line diesel engine that delivers a maximum power of 2590 hp and causes the shaft to rotate at 1700 rpm . If the outer diameter of the shaft is 8 in. and the wall thickness is $\frac{3}{8}$  in.

A) Determine the maximum shear stress developed in the shaft.

$\tau_{max}$ = ?

B) Also, what is the "wind up," or angle of twist in the shaft at full power?

$ \phi $ = ?

Solution :

Given :

Angular speed, ω = 1700 rpm

                              $ = 1700 \frac{\text{rev}}{\text{min}}\left(\frac{2 \pi \text{ rad}}{\text{rev}}\right) \frac{1 \text{ min}}{60 \ \text{s}}$

                              $= 56.67 \pi \text{ rad/s}$

Power $= 2590 \text{ hp} \left( \frac{550 \text{ ft. lb/s}}{1 \text{ hp}}\right)$

          = 1424500 ft. lb/s

Torque, $T = \frac{P}{\omega}$

                 $=\frac{1424500}{56.67 \pi}$

                 = 8001.27 lb.ft

A). Therefore, maximum shear stress is given by :

Applying the torsion formula

$\tau_{max} = \frac{T_c}{J}$

        $=\frac{8001.27 \times 12 \times 4}{\frac{\pi}{2}\left(4^2 - 3.625^4 \right)}$

      = 2.93 ksi

B). Angle of twist :

     $\phi = \frac{TL}{JG}$

         $=\frac{8001.27 \times 12 \times 100 \times 12}{\frac{\pi}{2}\left(4^4 - 3.625^4\right) \times 11 \times 10^3}$

         = 0.08002 rad

         = 4.58°

6 0
3 years ago
A coil of wire 8.6 cm in diameter has 15 turns and carries a current of 2.7 A. The coil is placed in a magnetic field of 0.56 T.
SVEN [57.7K]

Answer:

Explanation:

it is given that diameter = 8.6 cm

radius =\frac{8.6}{2}=4.3\ cm=4.3\times 10^{-2}\ m

current =2.7 ampere

number of turns = 15

area =\pi r^2=3.14\times \left ( 4.3\times 10^{-2} \right )^{2}=0.005806 m^{2}

magnetic field =0.56 T

maximum torque= BINASINΘ  for maximum torque sinΘ=1

so maximum torque==0.56×2.7×0.005806×15=0.13174 Nm

4 0
3 years ago
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