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2 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]2 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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What is the purpose of the graphic language?

solmaris [256]

Answer:

enables the representation, analysis and communication of various aspects of an information system. These aspects correspond to varying and incomplete views of information systems and the processes therein.

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2 years ago

All aspects of the Kirby-Bauer test are standardized to assure reliability. What might be the consequence of pouring the plates

EleoNora [17]

Answer:

it would affect the distance the antiantibodies diffuse from the disk

Explanation:

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2 years ago

Technician A says mismatching tires of the same size on a heavy vehicle will generally not affect ABS operation. Technician B sa

marysya [2.9K]

Technician A is correct

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3 years ago

An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40MPa. It has been dete

Nataly [62]

Answer:

Yes, fracture will occur

Explanation:

Half length of internal crack will be 4mm/2=2mm=0.002m

To find the dimensionless parameter, we use critical stress crack propagation equation

$\sigma_c=\frac {K}{Y \sqrt {a\pi}}$ and making Y the subject

$Y=\frac {K}{\sigma_c \sqrt {a\pi}}$

Where Y is the dimensionless parameter, a is half length of crack, K is plane strain fracture toughness, $\sigma_c$ is critical stress required for initiating crack propagation. Substituting the figures given in question we obtain

$Y=\frac {K}{\sigma_c \sqrt {a\pi}}= \frac {40}{300\sqrt {(0.002*\pi)}}=1.682$

When the maximum internal crack length is 6mm, half the length of internal crack is 6mm/2=3mm=0.003m

$\sigma_c=\frac {K}{Y \sqrt {a\pi}}$ and making K the subject

$K=\sigma_c Y \sqrt {a\pi}$ and substituting 260 MPa for $\sigma_c$ while a is taken as 0.003m and Y is already known

$K=260*1.682*\sqrt {0.003*\pi}=42.455 Mpa$

Therefore, fracture toughness at critical stress when maximum internal crack is 6mm is 42.455 Mpa and since it’s greater than 40 Mpa, fracture occurs to the material

6 0

3 years ago

1.0•10^-10 standard form

Drupady [299]

Answer:

$1.0 * 10^{-10} = 0.0000000001$

Explanation:

Given

$1.0 * 10^{-10}$

Required

Convert to standard form

$1.0 * 10^{-10}$

From laws of indices

$a^{-x} = \frac{1}{a^x}$

So, $1.0 * 10^{-10}$ is equivalent to

$1.0 * 10^{-10} = 1.0 * \frac{1}{10^{10}}$

$1.0 * 10^{-10} = 1.0 * \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}$

$1.0 * 10^{-10} = 1.0 * \frac{1}{10000000000}$

$1.0 * 10^{-10} = 1.0 * 0.0000000001$

$1.0 * 10^{-10} = 0.0000000001$

Hence, the standard form of $1.0 * 10^{-10}$ is $0.0000000001$

3 0

3 years ago