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3 years ago

Select the properties and typical applications for the high carbon steels.

Engineering

1 answer:

yanalaym [24]3 years ago

3 0

Answer:

<u>Option-(A)</u>

Explanation:

<u>Typical applications for the high carbon steels includes the following;</u>

It is heat treatable, relatively large combinations of mechanical characteristics. Typical applications: railway wheels and tracks, gears, crankshafts, and machine parts.

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what is an example of an innovative solution to an engineering problem? Explain briefly why you chose this answer.

Leviafan [203]

Answer:

robotic technology

Explanation:

Innovation is nothing but the use of various things such as ideas, products, people to build up a solution for the benefit of the human. It can be any product or any solution which is new and can solve people's problems.

Innovation solution makes use of technology to provide and dispatch new solutions or services which is a combination of both technology and ideas.

One such example of an innovative solution we can see is the use of "Robots" in medical science or in any military operations or rescue operation.

Sometimes it is difficult for humans to do everything or go to everywhere. Thus scientist and engineers have developed many advance robots or machines using new ideas and technology to find solutions to these problems.

Using innovations and technologies, one can find solutions to many problems which is difficult for the peoples. Robots can be used in any surveillance operation or in places of radioactive surrounding where there is a danger of humans to get exposed to such threats. They are also used in medical sciences to operate and support the patient.

3 0

3 years ago

For laminar flow over a hot flat plate, the local heat transfer coefficient decreases with distance because (select all that are

kotegsom [21]

Answer:

B. The thickness of the heated region near the plate is increasing.

Explanation:

First we know that, a boundary layer is the layer of fluid in the immediate vicinity of a bounding surface where the effects of viscosity are significant. The fluid is often slower due to the effects of viscosity. Advection i.e the transfer of heat by the flow of liquid becomes less since the flow is slower, thereby the local heat transfer coefficient decreases.

From law of conduction, we observe that heat transfer rate will decrease based on a smaller rate of temperature, the thickness therefore increases while the local heat transfer coefficient decreases with distance.

3 0

3 years ago

Make two lists of applications of matrices, one for those that require jagged matrices and one for those that require rectangula

Agata [3.3K]

Answer:

Explanation:

You can utilize barbed clusters to store inadequate grids. On the off chance that there are a great many lines yet each line has just 4 or 5 associations with different segments, at that point as opposed to utilizing a 1000x1000 cluster you can utilize a 1000 line rough exhibit while you simply store the components that the present section has association with another segment. Other utilization can be done on account of query tables. Query tables will be tables which have different qualities concerning a solitary key where the quantity of qualities isn't fixed. Aside from this, barbed clusters have an exceptionally set number of utilization cases. Multidimensional exhibits then again have plenty of utilizations. It is utilized to store a great deal of information reliably on the grounds that the greater part of the information is put away is steady concerning which section compares to what information. Aside from that it very well may be utilized to make thick diagrams or sparse(not effective), plotting information. Another utilization case would be used as an impermanent stockpiling for the figurings that need to tail them and utilize the past information like in powerful programming.

3 0

3 years ago

The current entering the positive terminal of a device is i(t)= 6e^-2t mA and the voltage across the device is v(t)= 10di/dtV.

liberstina [14]

Answer:

a) 2,945 mC

b) P(t) = -720*e^(-4t) uW

c) -180 uJ

Explanation:

Given:

i (t) = 6*e^(-2*t)

v (t) = 10*di / dt

Find:

( a) Find the charge delivered to the device between t=0 and t=2 s.

( b) Calculate the power absorbed.

( c) Determine the energy absorbed in 3 s.

Solution:

- The amount of charge Q delivered can be determined by:

dQ = i(t) . dt

$Q = \int\limits^2_0 {i(t)} \, dt = \int\limits^2_0 {6*e^(-2t)} \, dt = 6*\int\limits^2_0 {e^(-2t)} \, dt$

- Integrate and evaluate the on the interval:

$= 6 * (-0.5)*e^-2t = - 3*( 1 / e^4 - 1) = 2.945 C$

- The power can be calculated by using v(t) and i(t) as follows:

v(t) = 10* di / dt = 10*d(6*e^(-2*t)) /dt

v(t) = 10*(-12*e^(-2*t)) = -120*e^-2*t mV

P(t) = v(t)*i(t) = (-120*e^-2*t) * 6*e^(-2*t)

P(t) = -720*e^(-4t) uW

- The amount of energy W absorbed can be evaluated using P(t) as follows:

$W = \int\limits^3_0 {P(t)} \, dt = \int\limits^2_0 {-720*e^(-4t)} \, dt = -720*\int\limits^2_0 {e^(-4t)} \, dt$

- Integrate and evaluate the on the interval:

$W = -180*e^-4t = - 180*( 1 / e^12 - 1) = -180uJ$

6 0

3 years ago

You are riding in an elevator that is going up at 10 ft/s. You are holding your cell phone 5 ft above the floor when it suddenly

Luba_88 [7]

Answer:

It falls at the same speed in both cases.

Explanation:

If I were standing still the phone would be in free fall after slipping out of my hand.

I set a frame of reference with origin on the ground and the positive Y axis pointing up.

It would slip at t0 = 0, from a position Y0 = 5 ft, with a speed of Vy0 = 0.

It would be subject to an gravitational acceleration of -32.2 ft/s^2.

Since acceleration is constant:

Y(t) = Y0 + Vy0 * t + 1/2 * 4 * t^2

When it hits the floor at t1 it will be at Y(t1) = 0

0 = 5 + 0 * t1 - 16.1 * t1^2

16.1 * t1^2 = 5

t1^2 = 5 / 16.1

$t1 = \sqrt{0.31} = 0.55 s$

If the elevator is standing still it would take 0.55 s to hit the ground.

Now, if the elevator is moving up at 10 ft/s.

The frame of reference will have its origin at the place the floor of the elevator is at t = 0, and stay there as the elevetor moves. The floor of trhe elevator will have a position of Ye = 10 * t

Vy0 = 10 ft/s because it will be moving initially at the same speed as the elevator.

And it will hit the floor of the elevator not at 0, but at

Ye = 10 * t2

So:

10 * t2 = 5 + 10 * t2 - 16.1 * t2^2

0 = 5 - 16.1 * t2^2

16.1 * t1^2 = 5

t1^2 = 5 / 16.1

$t1 = \sqrt{0.31} = 0.55 s$

It falls at the same speed in both cases.

4 0

3 years ago