Which one of the following activities is not an example of incident coordination

You are preparing to work with Chemical A. You open the appropriate storage cabinet, and notice Chemical B, as well as Chemical

PtichkaEL [24]

The correct answer is; Stability and reactivity.

Further Explanation:

The stability and reactivity section of the SDS sheets is where to check for the possibility of hazardous reactions for the chemicals. This also lists the chemical stability of each chemical that people may be using. This can be found in section 10 of the OSHA Quick Card.

The SDS sheets has 16 sections for employees to use. Since 2015, the sections can be found in uniform format for easier and faster ways to find the section needed. The 16 sections for the SDS sheets are:

Identification
Hazard(s) identification
Composition/information on ingredients
First-aid measures
First-aid measures
First-aid measures
Handling and storage
Exposure controls/personal protection
Physical and chemical properties
Stability and reactivity
Toxicological information
Ecological information
Disposal considerations
Transport information
Regulatory information
Other information

Learn more about SDS sheets at brainly.com/question/9753408

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5 0

3 years ago

If Ella decided to become a children’s doctor, what career cluster would she belong in?

vodomira [7]

D . Health Services

8 0

3 years ago

The period of a pendulum T is assumed to depend only on the mass m, the length of the pendulum `, the acceleration due to gravit

zzz [600]

Answer:

The expression is shown in the explanation below:

Explanation:

Thinking process:

Let the time period of a simple pendulum be given by the expression:

$T = \pi \sqrt{\frac{l}{g} }$

Let the fundamental units be mass= M, time = t, length = L

Then the equation will be in the form

$T = M^{a}l^{b}g^{c}$

$T = KM^{a}l^{b}g^{c}$

where k is the constant of proportionality.

Now putting the dimensional formula:

$T = KM^{a}L^{b} [LT^{-} ^{2}]^{c}$

$M^{0}L^{0}T^{1} = KM^{a}L^{b+c}$

Equating the powers gives:

a = 0

b + c = 0

2c = 1, c = -1/2

b = 1/2

so;

a = 0 , b = 1/2 , c = -1/2

Therefore:

$T = KM^{0}l^{\frac{1}{2} } g^{\frac{1}{2} }$

T = $2\pi \sqrt{\frac{l}{g} }$

where k = $2\pi$

8 0

3 years ago

You are traveling upstream on a river at dusk. You see a buoy with the number 5 and a flashing green light . What should you do?

nalin [4]

Answer:

please give brainliest my brother just got the corona virus

Explanation:

this is my brothers account he wants to get 5 brainliest

8 0

3 years ago

A car is traveling at sea level at 78 mi/h on a 4% upgrade before the driver sees a fallen tree in the roadway 150 feet away. Th

Dmitrij [34]

Answer: V = 47.7 mi/hr

Explanation:

first we calculate elements of aero-dynamic resistance

Ka = p/2 * CD * A.f

p is the density of air(0.002378 slugs/ft^3) for zero altitude, CD is the drag coefficient(0.35) and A.f is the front region of the vehicle

so we substitute

Ka = 0.002378/2 * 0.35 * 18

Ka = 0.00749

Now we calculate the final speed of the vehicle (V2) using the relation;

S = (YbW/2gKa)In[ (UW + KaV1^2 + FriW ± Wsinθg) / (UW + KaV2^2 + FriW ± Wsinθg)

so

WE SUBSTITUTE

150 = (1.04 * 2700 / 2 * 32.2 * 0.0075) In [(0.8 * 2700 + 0.0075 *(78mil/hr * 5280ft/1min * 1hr/3600s)^2 + 0.017 * 2700 ± 2700 * 0.04) / (0.8 * 2700 + 0.0075 * V2^2 + 0.017 * 2700 ± 2700 * 0.04)]

150 = (2808/0.483) In [(2160 + 98.16 + 153.9) / ( 2160 + 0.0075V2^2 + 153.9)]

150 = 5813.66 In [ (2160 + 98.16 + 153.9) / ( 2160 + 0.0075V2^2 + 153.9)]

divide both sides by 5813.66

0.0258 = In [ (2412.06) / ( 0.0075V2^2 + 2313.9)]

take the e^ of both side

e^0.0258 = (2412.06) / ( 0.0075V2^2 + 2313.9)

1.0261 = (2412.06) / ( 0.0075V2^2 + 2313.9)]

(0.0075V2^2 + 2313.9) = 2412.06 / 1.0261

(0.0075V2^2 + 2313.9) = 2350.7

0.0075V2^2 = 2350.7 - 2313.9

0.0075V2^2 = 36.8

V2^2 = 36.8 / 0.0075

V2^2 = 4906.6666

V2 = √4906.6666

V2 = 70.0476 ft/s

converting to miles per hour

V2 = 70.0476 ft/s * 1 mil / 5280 ft * 3600s / 1hr

V = 47.7 mi/hr

8 0

3 years ago