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yaroslaw [1]
3 years ago
8

Technician A says that proper footwear may include both leather and steel-toed shoes. Technician B says that leather-soled shoes

provide slip resistance. Who is correct
Engineering
1 answer:
Amanda [17]3 years ago
5 0

Given:

We have given two statements.

Statement 1: Proper footwear may include both leather and steel-toed shoes.

Statement 2:  Leather-soled shoes provide slip resistance.

Find:

Which statement is true.

Solution:

A slip-resistant outsole is smoother and more slip-resistant than other outsole formulations when exposed to water and oil. A smoother outsole in rubber ensures a slip-resistant shoe can handle a slippery floor more effectively.

Slip resistant shoes have an interlocked tread pattern that does not close the water in, enabling the slip resistant sole to touch the floor to provide better slip resistance.

Leather-soled shoes don't provide slop resistance.

Therefore, both the Technicians are wrong.

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Evan notices a small fire in his workplace. Since the fire is small and the atmosphere is not smoky he decides to fight the fire
Norma-Jean [14]

Answer:

not calling the firemean

Explanation:

7 0
3 years ago
The pressure gage on a 2.5-m^3 oxygen tank reads 500 kPa. Determine the amount of oxygen in the tank if the temperature is 28°C
s2008m [1.1K]

Answer:

19063.6051 g

Explanation:

Pressure = Atmospheric pressure + Gauge Pressure

Atmospheric pressure = 97 kPa

Gauge pressure = 500 kPa

Total pressure = 500 + 97 kPa = 597 kPa

Also, P (kPa) = 1/101.325  P(atm)

Pressure = 5.89193 atm

Volume = 2.5 m³ = 2500 L ( As m³ = 1000 L)

Temperature = 28 °C

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

T(K) = T( °C) + 273.15  

So,  

T₁ = (28.2 + 273.15) K = 301.15 K  

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

5.89193 atm × 2500 L = n × 0.0821 L.atm/K.mol × 301.15 K  

⇒n = 595.76 moles

Molar mass of oxygen gas = 31.9988 g/mol

Mass = Moles * Molar mass = 595.76 * 31.9988 g = 19063.6051 g

7 0
3 years ago
Brainliest need help
insens350 [35]

Answer:

answer c

Explanation:

4 0
2 years ago
Design a plate and frame heat exchanger for the following problem:
qwelly [4]

Answer:

See explaination and attachment.

Explanation:

Iteration method is a repetitive method applied until the desired result is achieved.

Let the given equation be f(x) = 0 and the value of x to be determined. By using the Iteration method you can find the roots of the equation. To find the root of the equation first we have to write equation like below

x = pi(x)

Let x=x0 be an initial approximation of the required root α then the first approximation x1 is given by x1 = pi(x0).

Similarly for second, thrid and so on. approximation

x2 = pi(x1)

x3 = pi(x2)

x4 = pi(x3)

xn = pi(xn-1).

please go to attachment for the step by step solution.

8 0
3 years ago
Air is saturated with water vapor at 35.0 oC and a total pressure of 1.50 atmospheres. If the molar flow rate of the dry air in
Marizza181 [45]

Answer:

11.541 mol/min

Explanation:

temperature = 35°C

Total pressure = 1.5 * 1.013 * 10^5 = 151.95 kPa

note : partial pressure of water in mixture = saturation pressure of water at T = 35°c )

from steam table it is = 5.6291 Kpa

calculate the mole fraction of H_{2}o ( YH_{2}o )

= 5.6291 / 151.95

= 0.03704

calculate the mole fraction of air ( Yair )

 = 1 - mole fraction of water

= 1 - 0.03704 = 0.9629

Now to determine the molar flow rate of water vapor in the stream

lets assume N = Total molar flow rate

NH_{2}o = molar flow rate of water

Nair = molar  flow rate of air = 300 moles /min

note : Yair * n = Nair

therefore n = 300 / 0.9629 = 311.541  moles /min

Molar flowrate of water

=  n -  Nair

= 311.541 - 300 = 11.541 mol/min

4 0
3 years ago
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