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ryzh
3 months ago
8

Which of the following types of protective equipment protects workers who are passing by from stray sparks or metal while anothe

r worker is welding?
Engineering
1 answer:
choli [298]3 months ago
7 0

Answer:

Flame-resistant clothing and aprons

Explanation:

Workers involved with welding are generally mandated to wear flame-resistant clothing and aprons to shield them from various hazards, including heat, flames, burns, and potential radiation. In the context of welding, this gear protects individuals from flying sparks that can ignite and cause fires. Hence, such clothing helps to prevent accidents in these situations.

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The top 3 most popular male names of 2017 are Oliver, Declan, and Henry according to babynames. Write a program that modifies th
Viktor [391]

Answer:

The Python code is provided below with suitable comments

Explanation:

# convert list to set

male_names = set(['Oliver','Declan','Henry'])

# get name for removal and addition from user

remove_name = input("Enter remove name: ")

add_name = input("Enter add name: ")

# remove specified name from set

male_names.remove(remove_name)

# add new name to set

male_names.add(add_name)

# sort the resulting set

a = sorted(male_names)

# output the sorted set

print(a)

7 0
3 months ago
An open vat in a food processing plant contains 500 L of water at 20°C and atmospheric pressure. If the water is heated to 80°C,
Mrrafil [318]

Answer:

Volume change percentage is 2.60%

Water level increase is 4.138 mm

Explanation:

Provided data

Water volume V = 500 L

Initial temperature T1 = 20°C

Final temperature T2 = 80°C

Diameter of the vat = 2 m

Objective

We aim to determine percentage change in volume and the rise in water level.

Solution

We will apply the bulk modulus equation, which relates the change in pressure to the change in volume.

It can similarly relate to density changes.

Thus,

E = -\frac{dp}{dV/V}................1

And -\frac{dV}{V} = \frac{d\rho}{\rho}............2

Here, ρ denotes density. The density at 20°C = 998 kg/m³.

The density at 80°C = 972 kg/m³.

Plugging in these values into equation 2 gives

-\frac{dV}{V} = \frac{d\rho}{\rho}

-\frac{dV}{500*10^{-3} } = \frac{972-998}{998}

dV = 0.0130 m³

Therefore, the percentage change in volume will be

dV % = -\frac{dV}{V}  × 100

dV % = -\frac{0.0130}{500*10^{-3} }  × 100

dV % = 2.60 %

Hence, the percentage change in volume is 2.60%

Initial volume v1 = \frac{\pi }{4} *d^2*l(i)................3

Final volume v2 = \frac{\pi }{4} *d^2*l(f)................4

From equations 3 and 4, subtract v1 from v2.

v2 - v1 =  \frac{\pi }{4} *d^2*(l(f)-l(i))

dV = \frac{\pi }{4} *d^2*dl

Substituting all values yields

0.0130 = \frac{\pi }{4} *2^2*dl

Thus, dl = 0.004138 m.

Consequently, the water level rises by 4.138 mm.

8 0
2 months ago
The molecular weight of a 10g rubber band
Viktor [391]
The response to this query is 1 * 10 g/mole = 10.
8 0
2 months ago
The Machine Shop has received an order to turn three alloy steel cylinders. Starting diameter = 250 mm and length = 625 mm. Feed
alex41 [359]

Response:

The cutting speed is calculated at 365.71 m/min

Clarification:

Given parameters include

diameter D = 250 mm

length L = 625 mm

Feed f = 0.30 mm/rev

cut depth = 2.5 mm

n = 0.25

C = 700

To find

the cutting speed that ensures the tool life coincides with the cutting time for the three parts

The formula for cutting time is given as

Tc =

....................1

where D refers to diameter, L refers to length and f refers to feed while V represents speed \frac{\pi DL}{1000*f*V}

Thus, we derive

Tc = \frac{\pi (250)*625}{1000*0.3*V}

Tc = \frac{1636.25}{V}

Given the tool life is expressed as

T = 3 × Tc............................2

where T denotes tool life and Tc is the cutting duration

Calculating tool life by substituting values into equation 2 yields

T = 3 × \frac{1636.25}{V}

According to the Taylor tool formula, cutting speed is expressed as

VT^{0.25} = 700

 × V × 8.37 = 700

This yields V = 365.71

Thus, the cutting speed calculates to 365.71 m/min

5 0
2 months ago
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