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2 years ago

Chemical manufacturers must present which information on the product's label?

Engineering

1 answer:

Wewaii [24]2 years ago

6 0

Answer:

hazardous chemicals leaving the workplace is labeled, tagged or marked with the following information: product identifier; signal word; hazard statement

Explanation:

this is so you know what chemicals are in it

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What is the smallest variable type I can use to represent the number 27?

oksano4ka [1.4K]

Answer:3

Explanation:

Cuz

3 0

2 years ago

Consider the following list. list = {24, 20, 10, 75, 70, 18, 60, 35} Suppose that list is sorted using the insertion sort algori

Greeley [361]

Answer:

Option B

10,20,24,75,70,18,60,35

Explanation:

The first, second and third iteration of the loop will be as follows

insertion sort iteration 1: 20,24,10,75,70,18,60,35

insertion sort iteration 2:10,20,24,75,70,18,60,35

insertion sort iteration 3: 10,20,24,75,70,18,60,35

8 0

2 years ago

The ignition temperature of 87-octane gasoline vapor is about 430 ∘C and, assuming that the working gas is diatomic and enters t

Flura [38]

Answer:

I have solved the problem below. I hope it will let you clear the concept.

For any inquiries ask me in the comments.

Explanation:

6 0

2 years ago

A stainless steel (AISI 304) tube used to transport a chilled pharmaceutical has an inner diameter of 36 mm and a wall thickness

Bingel [31]

Answer:

a) heat gain per unit tube length = $\frac{23-6}{1.35} = 12.6W/m$

b) heat gain per unit tube length = $\frac{23-6}{2.20} = 7.7W/m$

Explanation:

Assumptions:

Constant properties
Steady state conditions
Negligible effect of radiation
Negligible constant resistance between tube and insulation
one dimensional radial conduction

a) What is the heat gain per unit tube length

$R_{conv,i}'=\frac{1}{2\pi r_1h_i}$

$d_1=36mm$ Therefore $r_1=\frac{d_1}{2} =36/2=18mm=18*10^{-3}$

$r_2=2mm=2*10^{-3}m$

$k_{st}=14.2W/m.k$

$h_o=6W/m^2$

$h_i=400W/m^2$

$R_{conv,i}'=\frac{1}{2\pi * 1.8*10^{-3}*400}= 0.221m.K/W$

$R_{cond,st}'=\frac{ln(r_2/r_1)}{2\pi k_{st}} =\frac{ln(20/18)}{2\pi *14.2} =1.18*10^{-3}m.K/W$

$R_{conv,o}'=\frac{1}{2\pi r_2h_0}=\frac{1}{2\pi *2*10^{-3}*6}=1.33m.K/W$

$R_{tot}'=R_{conv,i}'+R_{cond,st}'+R_{conv,o}'=0.221+(1.18*10^{-3})+1.33=1.35m.K/W$

heat gain per unit tube length = $\frac{23-6}{1.35} = 12.6W/m$

b) What is the heat gain per unit length if a 10-mm-thick layer of calcium silicate insulation (k_ins = 0.050 W/m.K) is applied to the tube

$r_3=r_1+r_2+10mm=30mm=0.03m$

$R_{conv,i}'$ and $R_{cond,st}'$ are the same, but $R_{conv,o}'$ changes.

Therefore:

$R_{conv,o}'=\frac{1}{2\pi r_3h_0} = \frac{1}{2\pi *0.03*6}=0.88m.K/W$

$R_{conv,ins}'=\frac{ln(r_3/r_)}{2\pi k_{ins}} =\frac{ln(30/20)}{2\pi *0.05} =1.29m.K/W$

The total resistance $R_{tot}'=R_{conv,i}'+R_{cond,st}'+R_{conv,ins}'+R_{conv,o}'=0.221+(1.18*10^{-3})+1.29+0.88=2.20m.K/W$

heat gain per unit tube length = $\frac{23-6}{2.20} = 7.7W/m$

8 0

2 years ago

Read 2 more answers

The velocity components u and v in a two-dimensional flow field are given by: u = 4yt ft./s, v = 4xt ft./s, where t is time. Wha

zmey [24]

Answer:

vec(a) = 16 i + 16 j

mag(a) = 22.63 ft/s^2

Explanation:

Given,

- The two components of velocity are given for fluid flow:

u = 4*y ft/s

v = 4*x ft/s

Find:

What is the time rate of change of the velocity vector V (i.e., the acceleration vector) for a fluid particle at x = 1 ft. and y = 1 ft. at time t = 1 second?

Solution:

- The rate of change of velocity is given to be acceleration. We will take derivative of each components of velocity with respect to time t:

a_x = du / dt

a_x = 4*dy/dt

a_y = dv/dt

a_y = 4*dx/dt

- The expressions dx/dt is the velocity component u and dy/dt is the velocity component v:

a_x = 4*(4*y) = 16y

a_y = 4*(4*x) = 16x

- The acceleration vector can be expressed by:

vec(a) = 16y i + 16x j

- Evaluate vector (a) at x = 1 and y = 1:

vec(a) = 16*1 i + 16*1 j = 16 i + 16 j

- The magnitude of acceleration is given by:

mag(a) = sqrt ( a^2_x + a^2_y )

mag(a) = sqrt ( 16^2 + 16^2 )

mag(a) = 22.63 ft/s^2

7 0

2 years ago