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3 years ago

8

A Permanent Electrical Safety Device (PESD) becomes a real safety device only after it is installed as part of a Lockout/Tagout

procedure. Until this is done it is nothing but another electrical component.A. TrueB. False

1 answer:

Nostrana [21]3 years ago

7 0

Answer:

True

Explanation:

Permanent electrical safety devices (PESD) acts as a layer of protection between the electrical worker and the hazardous voltage.

Permanent electrical safety devices (PESDs) are deployed to reduce the risks in isolating electrical energy.

Electrical safety can be improved if a worker determines a zero electrical energy state irrespective of any voltage exposure to themselves.

The given statement is true

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Find the resultant of two forces 130 N and 110 N respectively, acting at an angle whose tangent

monitta

Answer:

$F_r = 200N$

Explanation:

Given

Let the two forces be

$F_1 = 130N$

$F_2 = 110N$

and

$\tan(\theta) = \frac{12}{5}$

Required

Determine the resultant force

Resultant force (Fr) is calculated using:

$F_r^2 = F_1^2 + F_2^2 + 2F_1F_2\cos(\theta)$

This means that we need to first calculate $\cos(\theta)$

Given that:

$\tan(\theta) = \frac{12}{5}$

In trigonometry:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

By comparing the above formula to $\tan(\theta) = \frac{12}{5}$

$Opposite = 12$

$Adjacent = 5$

The hypotenuse is calculated as thus:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 12^2 + 5^2$

$Hypotenuse^2 = 144 + 25$

$Hypotenuse^2 = 169$

$Hypotenuse = \sqrt{169$

$Hypotenuse = 13$

$\cos(\theta)$ is then calculated using:

$\cos(\theta)= \frac{Adjacent}{Hypotenuse}$

$\cos(\theta)= \frac{5}{13}$

Substitute values for $F_1$ , $F_2$ and $cos(\theta)$ in

$F_r^2 = F_1^2 + F_2^2 + 2F_1F_2\cos(\theta)$

$F_r^2 = 130^2 + 110^2 + 2*130*110*\frac{5}{13}$

$F_r^2 = 16900 + 12100 + 11000$

$F_r^2 = 40000$

Take square roots of both sides

$F_r = \sqrt{40000$

$F_r = 200N$

<em>Hence, the resultant force is 200N</em>

4 0

3 years ago

An engineer who invested very well plans to retire now because she has $2,000,000 in her ORP account. How long will she be able

Ede4ka [16]

Answer:

20 years

Explanation:

You do 2,000,000 ÷ 100,000 but you can simplify that to 20 ÷ 1 = 20

During those 20 years, the profits you earn will be 80,000 since when you do 2,000,000 x 0.04 but you can simplify that to 20,000 x 4 getting 80,000 and that quite doesn't reach 100,000 dollars

8 0

3 years ago

The solid steel shaft DF has a diameter of 25 mm and is supported by smooth bearings at D and E. It is coupled to a motor at F,

Black_prince [1.1K]

Answer:

attached below

Explanation:

7 0

3 years ago

Two advantages of deforming steel at room temperature rather than at elevated temperatures are: (select 2 answers from the optio

Aleks [24]

Answer:

A AND C

Explanation:

Two advantages of deforming steel at room temperature rather than at elevated temperatures are: (select 2 answers from the options below)

A. better dimensional accuracy to allow form complex parts with complex geometries.

C. a smoother surface finish to allow elimination of finish machining and grinding operations.

7 0

3 years ago

Water at 15°C is to be discharged from a reservoir at a rate of 18 L/s using two horizontal cast iron pipes connected in series

love history [14]

Answer:

The required pumping head is 1344.55 m and the pumping power is 236.96 kW

Explanation:

The energy equation is equal to:

$\frac{P_{1} }{\gamma } +\frac{V_{1}^{2} }{2g} +z_{1} =\frac{P_{2} }{\gamma } +\frac{V_{2}^{2} }{2g} +z_{2}+h_{i} -h_{pump} , if V_{1} =0,z_{2} =0\\h_{pump} =\frac{V_{2}^{2}}{2} +h_{i}-z_{1}$

For the pipe 1, the flow velocity is:

$V_{1} =\frac{Q}{\frac{\pi D^{2} }{4} }$

Q = 18 L/s = 0.018 m³/s

D = 6 cm = 0.06 m

$V_{1} =\frac{0.018}{\frac{\pi *0.06^{2} }{4} } =6.366m/s$

The Reynold´s number is:

$Re=\frac{\rho *V*D}{u} =\frac{999.1*6.366*0.06}{1.138x10^{-3} } =335339.4$

$\frac{\epsilon }{D} =\frac{0.00026}{0.06} =0.0043$

Using the graph of Moody, I will select the f value at 0.0043 and 335339.4, as 0.02941

The head of pipe 1 is:

$h_{1} =\frac{V_{1}^{2} }{2g} (k_{L}+\frac{fL}{D} )=\frac{6.366^{2} }{2*9.8} *(0.5+\frac{0.0294*20}{0.06} )=21.3m$

For the pipe 2, the flow velocity is:

$V_{2} =\frac{0.018}{\frac{\pi *0.03^{2} }{4} } =25.46m/s$

The Reynold´s number is:

$Re=\frac{\rho *V*D}{u} =\frac{999.1*25.46*0.03}{1.138x10^{-3} } =670573.4$

$\frac{\epsilon }{D} =\frac{0.00026}{0.03} =0.0087$

The head of pipe 1 is:

$h_{2} =\frac{V_{2}^{2} }{2g} (k_{L}+\frac{fL}{D} )=\frac{25.46^{2} }{2*9.8} *(0.5+\frac{0.033*36}{0.03} )=1326.18m$

The total head is:

hi = 1326.18 + 21.3 = 1347.48 m

The required pump head is:

$h_{pump} =\frac{25.46^{2} }{2*9.8} +1347.48-36=1344.55m$

The required pumping power is:

$P=Q\rho *g*h_{pump} =0.018*999.1*9.8*1344.55=236965.16W=236.96kW$

8 0

3 years ago