Engineering College

Read through the paragraph on personal protective equipment on page 2 and fill in the blanks using the list of words below. Each word is used only once.


Sewers
Ear defenders
Replaces
Drop
Hazards
Dust mask
Hands
Employer
Safety boots
85db
Protect
Falling objects
Tread
Helmet
Gauntlets
Responsibility
Noise
PPE

I have just started a new job, and it is the responsibility of my ____________________ to supply the correct PPE for me to wear. PPE is designed to ___________________ me against workplace _____________________ . When we go on site I need good strong ______________ _____________________ to protect my feet from things like dampness, in case I ____________ something on my foot and also in case I _____________________ on an up-turned nail. My boss always asks me to use safety gloves to protect my ______________________. In fact, once I had to wear arm-length ____________________ when we were working on some _______________ |On site I also have to wear a safety ____________________ all the time. This is a good thing because it protects my head from any ___________________ ________________________. Working with other trades, such the carpenters, can be hazardous because they make lots of ______________________ with their power tools. If the noise is above ____________________ we have to use ______________ ______________________, otherwise my hearing could be damaged. Because of all the sawdust they create when they are cutting wood, they regularly have to wear a _____________ ___________________, which stops them breathing in the dust.

I have to look after all the ______________________ my boss gives me, as it is my ______________________. If something gets broken, he _______________________ it for me as he does not want to get in to trouble with the HSE.

Answers

Answer 1

Answer:

BESUE THE 35

Explanation:

Related Questions

You are playing guitar on a stool that is 22" tall. How tall is the stool if it is expressed as a combination of feet and inches?

Answers

Answer:

1 foot 10 inches

Explanation:

1 foot = 12 inches + 10 inches = 22 inches

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What test should be performed on abrasive wheels

Answers

Answer:

before wheel is put on it should be looked at for damage and a sound or ring test should be done to check for cracks, to test the wheel it should be tapped with a non metallic instrument (I looked it up)

The  test that should be performed on abrasive wheels is the ring test.

What is the purpose of the ring test on the  abrasive wheels?

The ring test can be regarded as one of the mechanical test that is used to know whether the wheel is cracked or damaged.

To carry out this test , the wheel will be arranged to be in the 45 degrees each side and it is then aligned to be at a specific diameter, this can be done by the expert in this field to know the state of that wheel.

Learn more about  ring test  on:

https://brainly.com/question/4621112

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To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105 mm^-2 . Suppose that all the dislocations in 1000 mm^3 (1 cm^3) were somehow removed and linked end to end.

Required:
a. How far (in miles) would this chain extend?
b. Now suppose that the density is increased to 1010 mm^-2 by cold working. What would be the chain length of dislocations in 1000 mm^3 of material?

Answers

Answer:

[tex]62.14\ \text{miles}[/tex]

[tex]6213727.37\ \text{miles}[/tex]

Explanation:

The distance of the chain would be the product of the dislocation density and the volume of the metal.

Dislocation density = [tex]10^5\ \text{mm}^{-2}[/tex]

Volume of the metal = [tex]1000\ \text{mm}^3[/tex]

[tex]10^5\times 1000=10^8\ \text{mm}\\ =10^5\ \text{m}[/tex]

[tex]1\ \text{mile}=1609.34\ \text{m}[/tex]

[tex]\dfrac{10^5}{1609.34}=62.14\ \text{miles}[/tex]

The chain would extend [tex]62.14\ \text{miles}[/tex]

Dislocation density = [tex]10^{10}\ \text{mm}^{-2}[/tex]

Volume of the metal = [tex]1000\ \text{mm}^3[/tex]

[tex]10^{10}\times 1000=10^{13}\ \text{mm}\\ =10^{10}\ \text{m}[/tex]

[tex]\dfrac{10^{10}}{1609.34}=6213727.37\ \text{miles}[/tex]

The chain would extend [tex]6213727.37\ \text{miles}[/tex]

the pressure rise, across a pump can be expressed as where D is the impeller diameter, p, is the fluid density, w is the rotational speed, adn q is the flowrate. determine a suitable set of dimensionless parameters

Answers

Answer:

hello your question is incomplete below is the complete question

The pressure rise Δp across a pump can be expressed as Δp = f(D, p, w, Q) where D is the impeller diameter, p is the fluid density, w is the rotational speed, and Q is the flowrate. determine a suitable set of dimensionless parameters

answer : Δp / D^2pw^2 = Ф (Q / D^3w )

Explanation:

k ( number of variables ) = 5

r ( number of reference dimensions ) = 3

applying the pi theorem

hence the number of pi terms = k - r = 5 - 3 = 2

Quadrilateral ABCD is a rectangle.
If m ZADB = 7k + 60 and mZCDB = -5k + 40, find mZCBD.

Answers

Hope this helps...........

A spherical Gaussian surface of radius R is situated in space along with both conducting and insulating charged objects. The net electric flux through the Gaussian surface is:______

Answers

Answer:

Ф = [tex]\frac{Q}{e_{0} } [ \frac{\frac{4\pi }{3 }(R)^3 }{\frac{4}{3}\pi (R)^3 } ][/tex]

Explanation:

Radius of Gaussian surface = R

Charge in the Sphere ( Gaussian surface ) = Q

lets take the radius of the sphere to be equal to radius of the Gaussian surface i.e. R

To determine the net electric flux through the Gaussian surface

we have to apply Gauci law

Ф = 4[tex]\pi r^2 E[/tex]

Ф = [tex]\frac{Q_{enc} |}{e_{0} }[/tex]

    = [tex]\frac{Q}{e_{0} } [ \frac{\frac{4\pi }{3 }(R)^3 }{\frac{4}{3}\pi (R)^3 } ][/tex]

A differential amplifier is to have a voltage gain of 100. What will be the feedback resistance required if the input resistances are both 1 kΩ?

Answers

Answer:

required feedback resistance ( R2 ) = 100 k Ω

Explanation:

Given data :

Voltage gain = 100

input resistance ( R1 ) = 1 k ohms

calculate feedback resistance required

voltage gain of differential amplifier

[tex]\frac{Vout}{V2 - V1 } = \frac{R2}{R1}[/tex]

= Voltage gain =  R2/R1

= 100 = R2/1

hence required feedback resistance ( R2 ) = 100 k Ω

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