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3 years ago

There are a number of requirements that employers must do to protect their workers from caught-in or

Engineering

1 answer:

GrogVix [38]3 years ago

6 0

Answer:

B.Provide guards on power tools and other equipment with moving parts

Explanation:

A guard/ safeguard is a machine feature that is on/around an engineering equipment which can prevent the contact of a user with hazardous parts of the machines such as point of operation and rotating parts.The guard can be on power tools to enable an alternataive ways of switching off the equipment when a danger scenario occurs.When using a tool/equipment, safeguards must be on/active.

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Which of the following is NOT part of a car's drive train?

FromTheMoon [43]

Answer:

B. rotor

Explanation:

The correct answer Is rotor because the others are part of a cars drivetrain

3 0

2 years ago

Read 2 more answers

Ammonia contained in a piston-cylinder assembly, initially saturated vapor at 0o F, undergoes an isothermal process during which

Rudik [331]

ANSWERS:

$-P_{2(a)} =15.6lbf/in^2\\-P_{2(b)} =30.146lbf/in^2\\ T_{2(a)} =0^oF\\T_{2(b)} =0^oF\\x_{2(b)} =49.87percent$

Explanation:

Given:

Piston cylinder assembly which mean that the process is constant pressure process P=C.

<u>AMMONIA </u>

state(1)

saturated vapor $x_{1} =1$

The temperature $T_{1} =0^0 F$

Isothermal process $T=C$

$-V_{2} =2V_{1}$ ( double)

$-V_{2} =.5V_{2}$ (reduced by half)

To find the final state by giving the quality in lbf/in we assume the friction is neglected and the system is in equilibrium.

state(1)

using PVT data for saturated ammonia

$-P_{1} =30.416 lbf/in^2\\-v_{1} =v_{g} =9.11ft^3/lb$

then the state exists in the supper heated region.

a) from standard data

$-v_{1(a)} =2v_{1} =18.22ft^3/lb\\-T_{1} =0^oF$

$at\\P_{x} =14lbf/in^2\\-v_{x} =20.289 ft^3/kg$

$at\\P_{y} =16 lbf/in^2\\-v_{y} =17.701ft^3/kg$

assume linear interpolation

$\frac{P_{x}-P_{2(b)} }{P_{x}- P_{y} } =\frac{v_{x}-v_{1(a)} }{v_{x}-v_{y} }$

$P_{1(b)}=P_{x} -(P_{x} -P_{y} )*\frac{v_{x}- v_{1(b)} }{v_{x}-v_{y} }\\ \\P_{1(b)} =14-(14-16)*\frac{20.289-18.22}{20.289-17.701} =15.6lbf/in^2$

$-v_{2(a)} =2v_{1} =4.555ft^3/lb\\v_{g}$

from standard data

$-v_{f} =0.02419ft^3/kg\\-v_{g} =9.11ft^3/kg\\v_{f}$

then the state exist in the wet zone

$-P_{s} =30.146lbf/in^2\\v_{2(a)} =v_{f} +x(v_{g} -v_{f} )$

$x=\frac{v_{2(a)-v_{f} } }{v_{g} -v_{f} } \\x=\frac{4.555-0.02419}{9.11-0.02419} =49.87%$

3 0

3 years ago

Which is the better measure of computer system performance—a benchmark, such as SPECINT; or a processor speed measure, such as G

Vesna [10]

Answer:

A benchmark

Explanation:

Most times a benchmark serves as the better measure when assessing a computer's performance, this is because CPU speeds can only evaluate an aspect of a computer's performance whereas a benchmark offers the advantage of measuring all the aspects of a computer's performance for a specific type of computing problem.

5 0

3 years ago

A 4KJ of energy is supplied to a machine used for lifting a mass. The force required is 800N. If the machine has an efficiency o

navik [9.2K]

The height at which the mass will be lifted is; 3 meters

<h3>How to utilize efficiency of a machine?</h3>

Formula for efficiency is;

η = useful output energy/input energy

We are given

η = 60% = 0.6

Input energy = 4 KJ = 4000 J

Thus;

0.6 = useful output energy/4000

useful output energy = 0.6 * 4000

useful output energy = 2400 J

Work done in lifting mass(useful output energy) = force * distance moved

Useful output energy = 800 * h

where h is height to lift mass

Thus;

800h = 2400

h = 2400/800

h = 3 meters

Read more about Machine Efficiency at; brainly.com/question/3617034

#SPJ1

8 0

2 years ago

Given a force of 72 lbs at a distance of 15 ft, calculate the moment produced.

Elis [28]

Answer:

1425.78 N.m

Explanation:

Moments of force is calculated as ;

Moments= Force * distance

M= F*d

The S.I unit for moment of force is Newton-meter (N.m)

Given in the question;

Force = 72 lbs

1 pound = 4.45 N

72 lbs = 4.45 * 72=320.4 N

Distance= 15 ft

1ft= 0.3048 m

15 ft = 15*0.3048 = 4.57 m

d= 4.57 m

M= F*d

M=320.4*4.57 =1425.78 N.m

5 0

3 years ago