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3 years ago

When would working with machinery be a common type of caught-in and caught-between hazard?

Engineering

1 answer:

tigry1 [53]3 years ago

6 0

Answer:

A working with machinery be a common type of caught-in and caught-between hazard is described below in complete detail.

Explanation:

“Caught in-between” accidents kill mechanics in a variety of techniques. These incorporate cave-ins and other hazards of tunneling activity; body parts extracted into unconscious machinery; reaching within the swing range of cranes and other installation material; caught between machine & fixed objects.

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A heat engine that rejects waste heat to a sink at 520 R has a thermal efficiency of 35 percent and a second- law efficiency of

xeze [42]

Answer:

The source temperature is 1248 R.

Explanation:

Second law efficiency of the engine is the ratio of actual efficiency to the maximum possible efficiency that is reversible efficiency.

Given:

Temperature of the heat sink is 520 R.

Second law efficiency is 60%.

Actual thermal efficiency is 35%.

Calculation:

Step1

Reversible efficiency is calculated as follows:

$\eta_{II}=\frac{\eta_{a}}{\eta_{rev}}$

$0.6=\frac{0.35}{\eta_{rev}}$

$\eta_{rev}=0.5834$

Step2

Source temperature is calculated as follows:

$\eta_{rev}=1-\frac{T_{L}}{T}$

$\eta_{rev}=1-\frac{520}{T}$

$0.5834=1-\frac{520}{T}$

T = 1248 R.

The heat engine is shown below:

Thus, the source temperature is 1248 R.

6 0

3 years ago

What is the output of a system with the transfer function s/(s + 3)^2 and subject to a unit step input at time t = 0?

Dominik [7]

Answer:

Explanation:

output =transfer function H(s) ×input U(s)

here H(s)= $\frac{s}{(s+3)^2}$

U(s)= $\frac{1}{s}$ for unit step function

output =H(s)×U(s)

= $\frac{s}{(s+3)^2}$ × $\frac{1}{s}$

= $\frac{1}{(s+3)^2}$

taking inverse laplace of output

output=t× $e^{-3t}$

at t=0 putting the value of t=0 in output

output =0

3 0

3 years ago

(c) As Engineering and Computing students, you must be familiar, with the respective professional bodies, as well as, Rules of P

guajiro [1.7K]

The professional ethics for computer engineers are:

They will Contribute to society and to human well-being.
They will Avoid harm.
Be honest and trustworthy.
They will be fair and take action that do to discriminate others.

<h3>What are the Characteristics of Code of Ethics?</h3>

The code of ethics are known to be a kind of a universal moral values, that is one that state that what a person expect of any given employee such as been trustworthy, respectful, responsible, and others.

Note that Rules of Practice, Professional Obligations and Codes of Ethics. are known to be put in place to avoid issues that may lead to conflict.

Therefore, i believe that As Engineering and Computing students, the respective professional bodies, Rules of Practice, Professional Obligations and Codes of Ethics are good and acts as a check and balance to us.

Therefore, The professional ethics for computer engineers are:

They will Contribute to society and to human well-being.
They will Avoid harm.
Be honest and trustworthy.
They will be fair and take action that do to discriminate others.

Learn more about Engineering rules from

brainly.com/question/17169621

#SPJ1

4 0

1 year ago

If the slotted arm rotates counterclockwise with a constant angular velocity of thetadot = 2rad/s, determine the magnitudes of t

astraxan [27]

Answer:

Magnitude of velocity=10.67 m/s

Magnitude of acceleration=24.62 ft/ $s^{2}$

Explanation:

The solution of the problem is given in the attachments

3 0

3 years ago

A rocket is launched from rest with a constant upwards acceleration of 18 m/s2. Determine its velocity after 25 seconds

lisabon 2012 [21]

Answer:

The final velocity of the rocket is 450 m/s.

Explanation:

Given;

initial velocity of the rocket, u = 0

constant upward acceleration of the rocket, a = 18 m/s²

time of motion of the rocket, t = 25 s

The final velocity of the rocket is calculated with the following kinematic equation;

v = u + at

where;

v is the final velocity of the rocket after 25 s

Substitute the given values in the equation above;

v = 0 + 18 x 25

v = 450 m/s

Therefore, the final velocity of the rocket is 450 m/s.

5 0

2 years ago