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
.
Answer:
Final Velocity = 4.9 m/s
Explanation:
We are given;. Initial velocity; u = 2 m/s
Constant Acceleration; a = 0.1 m/s²
Distance; s = 100 m
To find the final velocity(v), we will use one of Newton's equations of motion;
v² = u² + 2as
Plugging in the relevant values to give;
v² = 2² + 2(0.1 × 100)
v² = 4 + 20
v² = 24
v = √24
v = 4.9 m/s
Answer:
v₂ = 70 m / s
Explanation:
For this exercise let's use Bernoulli's equation
where subscript 1 is for the top of the mountain and subscript 2 is for Tuesday's level
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ +1/2 ρ v₂² + ρ g y₂
indicate that the pressure in the two points is the same, y₁ = 250 m, y₂ = 0 m, the water in the upper part, because it is a reservoir, is very large for which the velocity is very small, we will approximate it to 0 (v₁ = 0), we substitute
ρ g y₁ = ½ ρ v₂²
v₂ =
let's calculate
v₂ = √( 2 9.8 250)
v₂ = 70 m / s
Answer:
Heating water to produce steam which drives a turbine
Explanation:
Generation of electricity in coal-burning power plants and nuclear power plants both involve heating water to produce steam which drives a turbine.