Answer:
(a) 561.12 W/ m² (b) 196.39 MW
Explanation:
Solution
(a) Determine the energy and power of the wave per unit area
The energy per unit are of the wave is defined as:
E = 1 /16ρgH²
= 1/16 * 1025 kg/ m3* 9.81 m/s² * (2.5 m )²
=3927. 83 J/m²
Thus,
The power of the wave per unit area is,
P = E/ t
= 3927. 83 J/m² / 7 s = 561.12 W/ m²
(b) The average and work power output of a wave power plant
W = E * л * A
= 3927. 83 J/m² * 0.35 * 1 *10^6 m²
= 1374.74 MJ
Then,
The power produced by the wave for one km²
P = P * л * A
= 5612.12 W/m² * 0.35 * 1* 10^6 m²
=196.39 MW
Answer: the standard deviation STD of machine B is s (Lb) = 0.4557
Explanation:
from the given data, machine A and machine B produce half of the rods
Lt = 0.5La + 0.5Lb
so
s² (Lt) = 0.5²s²(La) + 0.5²s²(Lb) + 0.5²(2)Cov (La, Lb)
but Cov (La, Lb) = Corr(La, Lb) s(La) s(Lb) = 0.4s (La) s(Lb)
so we substitute
s²(Lt) = 0.25s² (La) + 0.25s² (Lb) + 0.4s (La) s(Lb)
0.4² = 0.25 (0.5²) + 0.25s² (Lb) + (0.5)0.4(0.5) s(Lb)
0.64 = 0.25 + s²(Lb) + 0.4s(Lb)
s²(Lb) + 0.4s(Lb) - 0.39 = 0
s(Lb) = { -0.4 ± √(0.16 + (4*0.39)) } / 2
s (Lb) = 0.4557
therefore the standard deviation STD of machine B is s (Lb) = 0.4557
Answer:
To run. The machine one at a time
Explanation:
