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Fantom [35]
3 years ago
7

Cng containers need to be inspected

Engineering
1 answer:
nevsk [136]3 years ago
3 0
Texas: The owner of the vehicle must prove that the container meets the inspection criteria of FMVSS 304. (Every 36 months or 36,000 miles, whichever occurs first.) 5. Utah: Inspection is required every 36 months or 36,000 miles, whichever occurs first.
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Which of the following answer options are your employer's responsibility?
tino4ka555 [31]

Answer:

Develop a written hazard communication program

Implement a hazard communication program

Maintain a written hazard communication program

Explanation:

To find - Which of the following answer options are your employer's responsibility?  Select all that apply.

Develop a written hazard communication program

Implement a hazard communication program

Maintain a written hazard communication program

Solution -

The correct options are -

Develop a written hazard communication program

Implement a hazard communication program

Maintain a written hazard communication program

All are the Responsibilities of an employer

Reason -

The most important duty of the employer is to stay alert and implement a correctly and efficiently written communication program related to hazards of the substances in the workplace.

He also has to maintain the program so that employees do not get affected.

3 0
3 years ago
What are the top 4 solar inventions, how they are used, and how they are better than the original way of powering them
OlgaM077 [116]
Yes I will answer soon
7 0
3 years ago
Water vapor at 100 psi, 500 F and a velocity of 100 ft./sec enters a nozzle operating at steady sate and expands adiabatically t
almond37 [142]

Answer:

a)exit velocity of the steam, V2 = 2016.8 ft/s

b) the amount of entropy produced is 0.006 Btu/Ibm.R

Explanation:

Given:

P1 = 100 psi

V1 = 100 ft./sec

T1 = 500f

P2 = 40 psi

n = 95% = 0.95

a) for nozzle:

Let's apply steady gas equation.

h_1 + \frac{(v_1) ^2}{2} = h_2 + \frac{(v_2)^2}{2}

h1 and h2 = inlet and exit enthalpy respectively.

At T1 = 500f and P1 = 100 psi,

h1 = 1278.8 Btu/Ibm

s1 = 1.708 Btu/Ibm.R

At P2 = 40psi and s1 = 1.708 Btu/Ibm.R

1193.5 Btu/Ibm

Let's find the actual h2 using the formula :

n = \frac{h_1 - h_2*}{h_1 - h_2}

n = \frac{1278.8 - h_2*}{1278.8 - 1193.5}

solving for h2, we have

h_2 = 1197.77 Btu/Ibm

Take Btu/Ibm = 25037 ft²/s²

Using the first equation, exit velocity of the steam =

(1278.8 * 25037) + \frac{(100)^2}{2}= (1197.77*25037)+ \frac{(V_2)^2}{2}

Solving for V2, we have

V2 = 2016.8 ft/s

b) The amount of entropy produced in BTU/ lbm R will be calculated using :

Δs = s2 - s1

Where s1 = 1.708 Btu/Ibm.R

At h2 = 1197.77 Btu/Ibm and P2 =40 psi,

S2 = 1.714 Btu/Ibm.R

Therefore, amount of entropy produced will be:

Δs = 1.714Btu/Ibm.R - 1.708Btu/Ibm.R

= 0.006 Btu/Ibm.R

3 0
3 years ago
The path taken to move items from their origin to their destination is known as which of the following?
mrs_skeptik [129]

Answer:d

Explanation:

7 0
3 years ago
The bar DA is rigid and is originally held in the horizontal position when the weight W is supported from C. The wires are made
Rom4ik [11]

Answer:

W = 112 lb

Explanation:

Given:

- δb = 0.025 in

- E = 29000 ksi      (A-36)

- Area A_de = 0.002 in^2

Find:

Compute Weight W attached at C

Solution:

- Use proportion to determine δd:

                              δd/5 = δb/3

                              δd = (5/3) * 0.025

                              δd = 0.0417 in

- Compute εde i.e strain in DE:

                               εde = δd / Lde

                               εde = 0.0417 / 3*12

                               εde = 0.00116

- Compute stress in DE, σde:

                               σde = E*εde

                               σde = 29000*0.00116

                               σde = 33.56 ksi

- Compute the Force F_de:

                               F_de = σde *A_de

                               F_de = 33.56*0.002

                               F_de = 0.0672 kips

- Equilibrium conditions apply:

                               (M)_a = 0

                               3*W - 5*F_de = 0

                               W = (5/3)*F_de

                              W = (5/3)* 0.0672 = 112 lb

4 0
3 years ago
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