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3 years ago

What is the first step in the problem-solving process, as well as in the engineering design process?

Engineering

1 answer:

DENIUS [597]3 years ago

5 0

Answer:

C. Brainstorm

Explanation:

Brainstorming refers to thinking and developing ideas and solutions to problems. It involves a free-thinking approach and putting oneself in the context of the problem you are attempting to solve. You can use the empathy maps to combine ideas from interviewing other people on the same. This stage is followed by defining the problem, concept generation, developing a solution, constructing a testing a prototype, evaluating the solution and finally presenting the solution.

Answer choice A is incorrect because it is the last stage .

Answer choice B is incorrect because it is the second stage.

Answer choice D is incorrect because it is step 5 and 6.

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Suppose that you can throw a projectile at a large enough v0 so that it can hit a target a distance R downrange. Given that you

viktelen [127]

Answer:

$\theta_1=15^o\\\theta_2=75^o$

Explanation:

Projectile Motion

In projectile motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration (assuming no friction), and the acceleration in the vertical direction is always the acceleration of gravity. The basic formulas are shown below:

$V_x=V_{ox}=V_ocos\theta$

Where $\theta$ is the angle of launch respect to the positive horizontal direction and Vo is the initial speed.

$V_y=V_{oy}-gt=V_osin\theta-gt$

The horizontal and vertical distances are, respectively:

$x=V_{o}cos\theta t$

$\displaystyle y=y_o+V_{o}sin\theta t-\frac{gt^2}{2}$

The total flight time can be found as that when y = 0, i.e. when the object comes back to ground (or launch) level. From the above equation we find

$\displaystyle t_f=\frac{2V_osin\theta}{g}$

Using this time in the horizontal distance, we find the Range or maximum horizontal distance:

$\displaystyle R=\frac{V_o^2sin2\theta}{g}$

Let's solve for $\theta$

$\displaystyle sin2\theta=\frac{R.g}{V_o^2}$

This is the general expression to determine the angles at which the projectile can be launched to hit the target. Recall the angle can have to values for fixed positive values of its sine:

$\displaystyle \theta_1=\frac{asin\left(\frac{R.g}{V_o^2}\right)}{2}$

$\displaystyle \theta_2=\frac{180^o-asin\left(\frac{R.g}{V_o^2}\right)}{2}$

Or equivalently:

$\theta_2=90^o-\theta_1$

Given Vo=37 m/s and R=70 m

$\displaystyle \theta_1=\frac{asin\left(\frac{70\times 9.8}{37^2}\right)}{2}$

$\theta_1=15^o$

And

$\theta_2=90^o-15^o=75^o$

5 0

3 years ago

explain each of the following kinds of rockets: Solid-Fuel Rocket, Liquid-Fuel Rocket, Ion Rocket and Plasma Rocket.

Rudik [331]

Answer:

ur answer friend

Explanation:

answer

Solid-Fuel Rocket - a solid-propellant rocket or solid rocket is a rocket with a rocket engine that uses solid propellants. The earliest rockets were solid-fuel rockets powered by gunpowder; they were used in warfare by the Chinese, Indians, Mongols and Persians, as early as 13th century.

Liquid-Fuel Rocket - a liquid-propellant rocket or liquid rocket utilizes a rocket engine that use liqiud propellants. An inert gas stored in a tank at a high pressure is sometimes used instead of pumps in simpler small engines to force the propellants into the combustion chamber.

Ion Rocket - an ion thruster or ion drive is a form of electric propulsion used for spacecraft propulsion. It creates thrust by accelerating ions using electricity. The Deep Space 1 spacecraft, powered by an ion thruster, changed velocity by 4.3 km/s ( 9600 mph ) while consuming less than 74 kg ( 163 lb ) of xenon.

Plasma Rocket - in this type of rocket, a combination of electric and magnetic fields are used to break down the atoms and molecules of a propellant gas into a collection of particles that have either a positive charge (ions) or a negative charge (electrons). In other words, the propellant gas becomes a plasma.

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5 0

3 years ago

In a 5V system if you were asked to take one input HIGH and another LOW what would you do (i.e. where would you connect them)?

dangina [55]

Answer:

HIGH from the supply voltage

LOW from ground

Explanation:

The answer depends on the kind of system and the purpose of the signal. But for practical reasons, in a DIGITAL system where 5V is HIGH and 0 V is LOW, 5 volts can be taken from the supply voltage (usually the same as high, BUT must be verified), and the LOW signal from ground.

If the user has a multimeter, it must be set to continuous voltage on 0 to 20 V range. Then place the probe in the ground of the circuit (must be a big copper area). Finally leave one probe in the circuit ground and place the other probe in some test points to identify 5 v.

4 0

3 years ago

Suppose that H1(s) and H2(s) are two strictly proper singleinput, single-output transfer functions with controllable statespace

kvv77 [185]

Answer:

Explanation:

See attached files.