Answer:
When the imposter is sus : O
Explanation:
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:
both
Explanation:
Both the technician are correct, ac generator output can be tested in both ways. The two ways are current output test to check ac generator output. and voltage output test to check output.
Answer:
The 1st one:Your natural ability
Answer:
11/8
Explanation:
7/8+4/8=11/8 if you want it on a ruler you need to go online and type fraction to decimal converter and put in 11/8 into a decimal
