To prevent injury while maintaining equipment make sure the equipment is locked out and/or tagged out (i.e., de-energized).
<h3>What does locked-out equipment mean?</h3>
Equipment is locked out when it does not contain any type of change that may affect the security of a device.
This task (locked out) is fundamental to avoid any safety problems associated with a device.
In conclusion, to prevent injury while maintaining equipment make sure the equipment is locked out and/or tagged out (i.e., de-energized).
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Answer:
I think you need motivation and you have to be creative and if you love writing then go ahead, start writing a book. If you have so many ideas write them all down.
Explanation:
Answer:
Recent research, policy literature, and federal legislation suggest that comprehensive WBL programs contain three key components: the alignment of classroom and workplace learning; application of academic, technical, and employ-ability skills in a work setting; and support from classroom or workplace mentors.
Explanation:
The distance between the campsite and the rest area is 9 miles.
The given parameters:
- <em>Initial speed of the campers, u = 4.5 mph</em>
- <em>Final speed of the campers, v = 4 mph</em>
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Let the time of motion from the campsite to rest area = t (hours)
Time for return trip = t hours + 15 mins
= (t + 0.25) hours
Let the distance between the campsite and rest area = d
d = 4.5t
d = 4(t + 0.25)
4.5t = 4(t + 0.25)
4.5t = 4t + 1
4.5t - 4t = 1
0.5t = 1
t = 2 hours
The distance between the campsite and the rest area is calculated as follows;
d = 4.5t
d = 4.5 x 2
d = 9 miles
Thus, the distance between the campsite and the rest area is 9 miles.
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The rate of disappearance of O2(g) under the same conditions is 2.5 × 10⁻⁵ m s⁻¹.
<h3>What is the rate law of a chemical equation? </h3>
The rate law of a chemical reaction equation is usually dependent on the concentration of the reactant species in the equation.
The chemical reaction given is;
The rate law for this reaction can be expressed as:
![\mathbf{= -\dfrac{1}{2}\dfrac{d[NO]}{dt} = -\dfrac{1}{1}\dfrac{d[O_2]}{dt}= +\dfrac{1}{2}\dfrac{d[NO_2]}{dt}}](https://tex.z-dn.net/?f=%5Cmathbf%7B%3D%20-%5Cdfrac%7B1%7D%7B2%7D%5Cdfrac%7Bd%5BNO%5D%7D%7Bdt%7D%20%3D%20-%5Cdfrac%7B1%7D%7B1%7D%5Cdfrac%7Bd%5BO_2%5D%7D%7Bdt%7D%3D%20%2B%5Cdfrac%7B1%7D%7B2%7D%5Cdfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D%7D)
Recall that:
- The rate of disappearance of NO(g) = 5.0× 10⁻⁵ m s⁻¹.
- Since both NO and O2 are the reacting species;
Then:
- The rate of disappearance of NO(g) is equal to the rate of disappearance of O2(g)
![\mathbf{= -\dfrac{1}{2}\dfrac{d[NO]}{dt} = -\dfrac{1}{1}\dfrac{d[O_2]}{dt}}](https://tex.z-dn.net/?f=%5Cmathbf%7B%3D%20-%5Cdfrac%7B1%7D%7B2%7D%5Cdfrac%7Bd%5BNO%5D%7D%7Bdt%7D%20%3D%20-%5Cdfrac%7B1%7D%7B1%7D%5Cdfrac%7Bd%5BO_2%5D%7D%7Bdt%7D%7D)
Thus;
The rate of disappearance of O2 = 2.5 × 10⁻⁵ m s⁻¹.
Therefore, we can conclude that two molecules of NO are consumed per one molecule of O2.
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