For the answer to the question above,
I'll write down my solution to your problem
sin(A+B) = sinAcosB + cosAsinB
<span>sin(2A) = 2sinAcosA </span>
<span>cos(2A) = 1-2sin^2A </span>
<span>sin(3x) = sin(2x+x) </span>
<span>sin(3x) = sin(2x)cos(x) + cos(2x)sin(x) </span>
<span>= 2sin(x)cos(x)cos(x) + (1-2sin^2(x))sin(x) </span>
<span>= 2sin(x)cos^2(x) + sin(x) - 2sin^3(x) </span>
<span>= 2sin(x)(1-sin^2(x)) + sin(x) - 2sin^3(x) </span>
<span>= 2sin(x) - 2sin^3(x) + sin(x) - 2sin^3(x) </span>
<span>= 3sin(x) - 4sin^3(x)
</span>My closest answer is multiple choice letter D.
Answer:
The force (thrust) is equal to the exit mass flow rate times the exit velocity minus the free stream mass flow rate times the free stream velocity.
Answer:
5 sq. root 3
Explanation:
theta= 60°
=> u sin theta = 10 × sin 60
= 10× sq. root 3/2
= 5 sq. root 3