The question is incomplete. The complete question is :
A plate of uniform areal density 
is bounded by the four curves:
where x and y are in meters. Point 
has coordinates 
and 
. What is the moment of inertia 
of the plate about the point ?
Solution :
Given :
and , 
, 
.
So,
, 
![$I=2 \int_1^2 \left( \left[ (x-1)^2y+\frac{(y+2)^3}{3}\right]_{-x^2+4x-5}^{x^2+4x+6}\right) \ dx$](https://tex.z-dn.net/?f=%24I%3D2%20%5Cint_1%5E2%20%5Cleft%28%20%5Cleft%5B%20%28x-1%29%5E2y%2B%5Cfrac%7B%28y%2B2%29%5E3%7D%7B3%7D%5Cright%5D_%7B-x%5E2%2B4x-5%7D%5E%7Bx%5E2%2B4x%2B6%7D%5Cright%29%20%5C%20dx%24)
So the moment of inertia is 
.
Correct answer choice is :
B) Petroleum
Explanation:
Petroleum is commonly happening, yellow to the black liquid obtained in geological structures below the Earth's surface. It is usually filtered into different types of fuels. Petroleum outcomes involve transport fuels, fuel oils for heating and power generation, pavement and road oil, and feedstocks for making the compounds, plastics, and plastic elements that are in almost everything we use.
<span>b. They share their experimental results with other researchers</span>
At the initial state: v1 = vf = 0.001053 m
3
/kg, h1 = hf = 467.11 kJ/kg, and s1 = sf = 1.4336 kJ/kgK.
The mass of the water is: m = V/v1 = 0.005/0.001053 = 4.7483 kg.
To find the final state, we will use the First Law:
Q12 = m(h2 - h1) for closed system undergoing a constant pressure process.
h2 = 1Q2/m + h1 = 2200/4.7483 + 467.11 = 930.43 kJ/kg.
At P2 = P1 = 150 kPa, this is a saturated mixture.
hf = 467.11 kJ/kg, hfg = 2226.5 kJ/kg, sf = 1.4336 kJ/kgK, and sfg = 5.7897 kJ/kgK
s2 = sf + sfg (h2 – hf )/hfg = 1.4336 + 5.7897(930.43 – 467.11)/2226.5 = 2.6384 kJ/kgK.
The entropy change of water is:
Delta Ssys= m(s2 – s1) = 4.7483(2.6384 – 1.4336) = 5.72 kJ/K.
