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solmaris [256]
3 years ago
6

Counter argument for why engineers dont make a positive change in the world.​

Engineering
1 answer:
egoroff_w [7]3 years ago
3 0

Answer:

They don't make a positive change in the world because they have made mistakes that aren't able to be made fixed and there are a lot of engineers who haven't study enough and know the important basis of coming to engineer.

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A 24-tooth gear has AGMA standard full-depth involute teeth with diametral pitch of 12. Calculate the pitch diameter, circular p
torisob [31]

Answer:

Explanation:

Given:

Tooth Number, N = 24  

Diametral pitch pd = 12

pitch diameter, d = N/pd = 24/12 = 2in

circular pitch, pc = π/pd  = 3.142/12 = 0.2618in

Addendum, a  = 1/pd = 1/12 =0.08333in

Dedendum, b = 1.25/pd = 0.10417in

Tooth thickness, t = 0.5pc = 0,5 * 0.2618  = 0.1309in

Clearance, c = 0.25/pd = 0.25/12 = 0.02083in

5 0
3 years ago
Read 2 more answers
A batch of 1000 is split into 10 smaller batches of equal size 100. The processing time of each unit is 2
Vika [28.1K]

The lead time of the actual batch will be in

  • 2950 in minutes

<h3>What is Processing Time?</h3>

This refers to the amount of time which is taken for a processor to run a procedure and return a result.

We can see that a batch of 1000 is split so that they each have 10 smaller batches which has an equal size of 100 each, then if the processing time is 2 mins per machine and the set up time is 30 mins.

Hence, when this batch is processed over a serial line of 5 machines, then the lead time of the actual batch would be 2950 in minutes

Read more about processing time here:

brainly.com/question/18444145

4 0
1 year ago
A mass weighing 22 lb stretches a spring 4.5 in. The mass is also attached to a damper with Y coefficient . Determine the value
Dominik [7]

Answer:

Cc= 12.7 lb.sec/ft

Explanation:

Given that

m = 22 lb

g= 32 ft/s²

m = \dfrac{22}{32}=0.6875\ s^2/ft

x= 4.5 in

1 in = 0.083 ft

x= 0.375 ft

Spring constant ,K

K=\dfrac{m}{x}=\dfrac{22}{0.375}

K= 58.66  lb/ft

The damper coefficient for critically damped system

C_c=2\sqrt{mK}

C_c=2\sqrt{0.6875\times 58.66}

Cc= 12.7 lb.sec/ft

5 0
3 years ago
Are designed to make it easier for employees to get health and safety Information about
iren [92.7K]

Answer:

what the options

Explanation:

4 0
3 years ago
Consider the expansion of a gas at a constant temperature in a water-cooled piston-cylinder system. The constant temperature is
Leona [35]

Answer:

Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})

Explanation:

According to the first thermodynamic law, the energy must be conserved so:

dQ = dU - dW

Where Q is the heat transmitted to the system, U is the internal energy and W is the work done by the system.

This equation can be solved by integration between an initial and a final state:

(1) \int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU - \int\limits^1_2 {} \, dW

As per work definition:

dW = F*dr

For pressure the force F equials the pressure multiplied by the area of the piston, and considering dx as the displacement:

dW = PA*dx

Here A*dx equals the differential volume of the piston, and considering that any increment in volume is a work done by the system, the sign is negative, so:

dW = - P*dV

So the third integral in equation (1) is:

\int\limits^1_2 {- P} \, dV

Considering the gas as ideal, the pressure can be calculated as P = \frac{n*R*T}{V}, so:

\int\limits^1_2 {- P} \, dV = \int\limits^1_2 {- \frac{n*R*T}{V}} \, dV

In this particular case as the systems is closed and the temperature constant, n, R and T are constants:

\int\limits^1_2 {- \frac{n*R*T}{V}} \, dV = -nRT \int\limits^1_2 {\frac{1}{V}} \, dV

Replacion this and solving equation (1) between state 1 and 2:

\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU + nRT \int\limits^1_2 {\frac{1}{V}} \, dV

Q_{2} - Q_{1} = U_{2} - U_{1} + nRT(ln V_{2} - ln V_{1})

Q_{2} - Q_{1} = U_{2} - U_{1} + nRT ln \frac{V_{2}}{V_{1}}

The internal energy depends only on the temperature of the gas, so there is no internal energy change U_{2} - U_{1} = 0, so the heat exchanged to the system equals the work done by the system:

Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})

4 0
3 years ago
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