Answer:
a) 24 kg
b) 32 kg
Explanation:
The gauge pressure is of the gas is equal to the weight of the piston divided by its area:
p = P / A
p = m * g / (π/4 * d^2)
Rearranging
p * (π/4 * d^2) = m * g
m = p * (π/4 * d^2) / g
m = 1200 * (π/4 * 0.5^2) / 9.81 = 24 kg
After the weight is added the gauge pressure is 2.8kPa
The mass of piston plus addded weight is
m2 = 2800 * (π/4 * 0.5^2) / 9.81 = 56 kg
56 - 24 = 32 kg
The mass of the added weight is 32 kg.
Moisture content is measured in terms of pounds of water per pound of air (lb water/lb air) or grains of water per pound of air (gr. of water/lb air).
Hope this helps❤
Answer:


Explanation:
= Area of section 1 = 
= Velocity of water at section 1 = 100 ft/min
= Specific volume at section 1 = 
= Density of fluid = 
= Area of section 2 = 
Mass flow rate is given by

The mass flow rate through the pipe is 
As the mass flowing through the pipe is conserved we know that the mass flow rate at section 2 will be the same as section 1

The speed at section 2 is
.
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