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4 years ago

As part of PSM standards, hazard communication standards require employers who use hazardous chemicals to _____.

1 answer:

8 0

As part of PSM standards, hazards communication standards require employers of different companies who use hazardous, unsafe and life threatening chemicals "to check, evaluate, classify and label each of those chemicals". These standards are developed and established by OSHA (Occupational Safety and Health Organization).

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You have an insurance policy with a $300 premium and a $500 deductible. how much should you expect to pay the insurance company

frosja888 [35]

$300 is the correct answer. The premium is the amount that you pay to an insurance company every month for them to provide you with insurance. The deductible, on the other hand, is the amount that the person paying for the insurance has to pay when they make a claim if they want the insurance company to pay out. Having a deductible makes premiums lower as it reduces the chance of people making smaller claims, and therefore reduces the insurance company's spending.

7 0

3 years ago

Read 2 more answers

Bethany needs to borrow $8,000. She can borrow money at 6.9% simple interest for 3 yr or she can borrow at 6.5% with interest co

Evgesh-ka [11]

The loan options which would result in less total interest is borrowing money at 6.9% simple interest.

Given the following data:

Principal, P = $8,000.

Interest rate, R = 6.9%

Time, T = 3 years

Interest rate 2 = 6.5%

To determine which of the loan options would result in less total interest:

For simple interest:

Mathematically, simple interest is given by the formula:

$S.I = \frac{PRT}{100}\\\\S.I = \frac{8000 \times 6.9 \times 3}{100}\\\\S.I = 80 \times 6.9 \times 3$

S.I = $1,656.

For compound interest:

Mathematically, an interest that is compounded continuously given by the formula:

$A = Pe^{rt}\\\\A = 8000 \times e^{0.065 \times 3}\\\\A = 8000 \times e^{0.195}\\\\A = 8000 \times 1.2153$

A = $9,722.49

$Interest = A -P\\\\Interest = 9722.49-8000$

Interest = $1,722.49

Read more on simple interest here: brainly.com/question/16992474

4 0

3 years ago

Would you prefer to buy an existing business or start from scratch? Why?

vladimir2022 [97]

Answer:

I would start from scratch Because It helps u feel accomplished about your work

8 0

3 years ago

Ardel Co. budgeted to sell 218,000 units of Zbox in September. Production of one unit of Zbox requires two pounds of aluminum an

FromTheMoon [43]

Answer:

207,000

Explanation:

Data provided

Sold units = 218,000

Ending inventory = 13,000

Opening units = 24,000

The computation of units during September is shown below:-

Number of units manufactured during the year = Sold units + Ending inventory - Opening units

= 218,000 + 13,000 - 24,000

= 231,000 - 24,000

= 207,000

Therefore for computing the number of units manufactured during the year we simply applied the above formula.

6 0

4 years ago

A cylindrical part of diameter d is loaded by an axial force p. this causes a stress of pya, where a 5 pd2y4. if the load is kno

raketka [301]

Here is the correct question.

A cylindrical part of diameter d is loaded by an axial force p. this causes a stress of P/A, where A= πd²/4. if the load is known with an uncertainty of ±10 percent, the diameter is known within ±5 percent (tolerances), and the stress that causes failure (strength) is known within ±15 percent, determine the minimum design factor that will guarantee that the part will not fail.

Answer:

the minimum design factor that will guarantee that the part will not fail. = 1.434

Explanation:

Looking at the uncertainty; loss of strength must be raised to $\dfrac{1}{0.85}$ due to the stress that causes the failure (strength) is known within ±15% uncertainty.

Looking at the uncertainty; the maximum allowable load must be reduced to $\dfrac{1}{1.1}$ because the load is known with an uncertainty of ±10.

Looking at the uncertainty; the diameter must be raised to $\dfrac{1}{0.95}$ because the diameter is known within an uncertainty of ±5.

The decrease in the maximum allowable stress can be estimated as:

$\sigma' = \dfrac{P'}{A'}$

where,

$\sigma =$ stress

P = load

A = cross-sectional area of the cylinder

∴

$\sigma' = \dfrac{P'}{\dfrac{\pi}{4}(d')^2}$

replacing P' with $\dfrac{1}{1.1}P$ and d' with $\dfrac{1}{0.95}d$ , we have:

$\sigma' = \dfrac{(\dfrac{1}{1.1})\times p }{\dfrac{\pi}{4}(\dfrac{1}{0.95 } d)^2 }$

$\sigma' =\dfrac{P}{\dfrac{\pi}{4}d^2} (\dfrac{\dfrac{1}{1.1} }{(\dfrac{1}{0.95})^2 }) }$

$\sigma' =\sigma \times (\dfrac{\dfrac{1}{1.1} }{(\dfrac{1}{0.95})^2 }) }$

$\sigma' =\sigma \times 0.82045$

$\dfrac{\sigma' }{\sigma } =0.82045$

Thus, the uncertainty in diameter and the load of the allowable stress needs to decrease to 0.82045

Now, the minimum design factor that will ascertain that the part will not fail can be computed as:

$n_d = \dfrac{loss \ of \ function \ parameter }{maximum \ allowable \ parameter}$

where;

the design factor = $n_d$

$n_d =\dfrac{\dfrac{1}{0.85} }{0.82045}$

the design factor $n_d$ = 1.434.

Thus, the minimum design factor that will guarantee that the part will not fail. = 1.434

7 0

3 years ago