Answer:
The answer is "
"
Explanation:
Air flowing into the
Flow rate of the mass 
inlet temperature 
Pipeline
Its air is modelled as an ideal gas Apply the ideum gas rule to the air to calcule the basic volume v:




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
Complete question:
A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of 98.9 MPa root m (90 ksi root in.) and a yield strength of 860 MPa (125,000 psi). The flaw size resolution limit of the flaw detection apparatus is 3.0 mm (0.12 in.). If the design stress is one-half of the yield strength and the value of Y is 1.0, determine whether or not a critical flaw for this plate is subject to detection.
Answer:
Since the flaw 17mm is greater than 3 mm the critical flaw for this plate is subject to detection
so that critical flow is subject to detection
Explanation:
We are given:
Plane strain fracture toughness K 
Yield strength Y = 860 MPa
Flaw detection apparatus = 3.0mm (12in)
y = 1.0
Let's use the expression:

We already know
K= design
a = length of surface creak
Since we are to find the length of surface creak, we will make "a" subject of the formula in the expression above.
Therefore
![a= \frac{1}{pi} * [\frac{k}{y*a}]^2](https://tex.z-dn.net/?f=%20a%3D%20%5Cfrac%7B1%7D%7Bpi%7D%20%2A%20%5B%5Cfrac%7Bk%7D%7By%2Aa%7D%5D%5E2%20)
Substituting figures in the expression above, we have:
![= \frac{1}{pi} * [\frac{98.9 MPa \sqrt{m}} {10 * \frac{860MPa}{2}}]^2](https://tex.z-dn.net/?f=%20%3D%20%5Cfrac%7B1%7D%7Bpi%7D%20%2A%20%5B%5Cfrac%7B98.9%20MPa%20%5Csqrt%7Bm%7D%7D%20%7B10%20%2A%20%5Cfrac%7B860MPa%7D%7B2%7D%7D%5D%5E2)
= 0.0168 m
= 17mm
Therefore, since the flaw 17mm > 3 mm the critical flow is subject to detection
The true statement about the dot plot is 1 has 4 and 0 dots.
Explanation:
- after creating a Bar chart 4 is the right answer.
- The highest among st all the other plots is 1 but 4 shows 3.
- Taking an average from all the data points 3 comes to the right answer.
- Median the central Mid-point is 3.
- Mode also comes to 3.
- It is skewed on the right due to the 1st one.
- Skewed shows the data point either increasing or in decreasing.
- There a to Bi- histograms which has two ups and downs.
- The true statement has to be as Mean=Median=Mode is 3.