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2 years ago

the tire restraining device or barrier shall be removed immediately from service for any of these defects except

1 answer:

5 0

Restraining devices and barriers shall be visually inspected on the rim wheel components or sudden release of contained air.

Restraining device means an apparatus such as a <em>cage, rack, assemblage of bars and other components</em> that will constrain all rim wheel components.

Restraining devices and barriers shall be visually inspected on the rim wheel components or sudden release of contained air. Any restraining device or barrier exhibiting damage such as the following defects shall be immediately removed from service.

Find out more on Restraining devices at: brainly.com/question/24647450

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In digital communication technologies, what is an internal network also known as?

garri49 [273]

Intranet is a network that is internal and use internet technologies. It makes information of any company accessible to its employees and hence facilitates collaboration. Same methods can be used to get information, use resources, and update the data as that of the internet.
Hopefully this helped.

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3 years ago

Solid Isomorphous alloys strength

Sphinxa [80]

Answer:

Explanation:

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3 years ago

What is the activation energy (Q) for a vacancy formation if 10 moles of a metal have 2.3 X 10^13 vacancies at 425°C?

Yakvenalex [24]

Answer:

$Activation\ Energy=2.5\times 10^{-19}\ J$

Explanation:

Using the expression shown below as:

$N_v=N\times e^{-\frac {Q_v}{k\times T}$

Where,

$N_v$ is the number of vacancies

N is the number of defective sites

k is Boltzmann's constant = $1.38\times 10^{-23}\ J/K$

${Q_v}$ is the activation energy

T is the temperature

Given that:

$N_v=2.3\times 10^{13}$

N = 10 moles

1 mole = $6.023\times 10^{23}$

So,

N = $10\times 6.023\times 10^{23}=6.023\times 10^{24}$

Temperature = 425°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (425 + 273.15) K = 698.15 K

T = 698.15 K

Applying the values as:

$2.3\times 10^{13}=6.023\times 10^{24}\times e^{-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}$

$ln[\frac {2.3}{6.023}\times 10^{-11}]=-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}$

$Q_v=2.5\times 10^{-19}\ J$

4 0

3 years ago

Name two common fuel gases that can be used for oxyfuel cutting

zlopas [31]

Hi

Acetylene and propane

I hope this help you!

8 0

1 year ago

A reservoir delivers water to a horizontal pipeline 39 long The first 15 m has a diameter of 50 mm, after which it suddenly beco

allsm [11]

Answer:

The difference of head in the level of reservoir is 0.23 m.

Explanation:

For pipe 1

$d_1=50 mm,f_1=0.0048$

For pipe 2

$d_2=75 mm,f_2=0.0058$

Q=2.8 l/s

$Q=2.8\times 10^{-3]$

We know that Q=AV

$Q=A_1V_1=A_2V_2$

$A_1=1.95\times 10^{-3}m^2$

$A_2=4.38\times 10^{-3} m^2$

$So V_2=0.63 m/s,V_1=1.43 m/s$

head loss (h)

$h=\dfrac{f_1L_1V_1^2}{2gd_1}+\dfrac{f_2L_2V_2^2}{2gd_2}+0.5\dfrac{V_1^2}{2g}$

Now putting the all values

$h=\dfrac{0.0048\times 15\times 1.43^2}{2\times 9.81\times 0.05}+\dfrac{0.0058\times 24\times 0.63^2}{2\times 9.81\times 0.075}+0.5\dfrac{1.43^2}{2\times 9.81}$

So h=0.23 m

So the difference of head in the level of reservoir is 0.23 m.

8 0

3 years ago