I’m not sure if I’m correct but I think it is 2^21
Answer:
1/65 is the answer. the main part behind it is using subtraction and divison to find the answer
Answer:
Power generated = 4.315 kW
Step-by-step explanation:
Given,
- speed of wind when enters into the turbine, V = 12 m/s
- speed of wind when exits from the turbine, U = 9 m/s
- mass flow rate of the wind, m = 137 kg/s
According to the law of conservation of energy
Energy generated = change in kinetic energy
Since, the air is exiting at same elevation. So, we will consider only kinetic energy.
Energy generated in one second will be given by,
=4315.5 J
= 4.315 kJ
So, energy is generated in one second = 4.315 kJ
Power generated can be given by,
And the energy is generated is already in per second so power generated will be 4.315 kW.
If z represents the variable of the standard normal distribution curve N(0,1), which is symmetrical about z=0, then we look for
P(z)-P(-z)=0.9642=1-2P(-z) => P(-z)=(1-0.9642)/2=0.0179
=>
-z=-2.099192
=> z=2.099192
Check:
P(2.099192)-P(-2.099192)=0.9642 ok
Answer:
.27
Step-by-step explanation:
4.5*6/100
=.27