Answer:
True
Explanation:
When trying to solve a frame problem in Engineering or Physics, it will typically be necessary to draw more than one body diagram.
When we have several parts of the frame or a set of frames, we have the anchor point, as well as the intersections of frames. Besides that, usually, there is a particle or rigid body together with the frame system. In this sense, usually, it is required to analyze a body diagram for the particle or rigid body suspended, as well as the intersections of the frames. So, usually, it will be required a minimum of two body diagrams.
If the system is more complex, or there are many intersections points, it will be required more than two body diagrams.
Finally, indeed, it will typically be necessary to draw many-body diagrams.
Answer:
The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.
Explanation:
The Upper Surface Cp is given as

The Lower Surface Cp is given as

The difference of the Cp over the airfoil is given as

Now the Lift Coefficient is given as

Now the coefficient of moment about the leading edge is given as

So the lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.
Answer:
The correct response will be "0.992". The further explanation to the following question is given below.
Explanation:
The probability that paging would be beneficial becomes 0.8
Effective paging at the very first attempted is 0.8
On the second attempt the success probability will be:
⇒ 
⇒ 
On the third attempt the success probability will be:
⇒ 
⇒ 
So that the success probability will be:
⇒ 
⇒ 
Answer: a) 0.948 b) 117.5µf
Explanation:
Given the load, a total of 2.4kw and 0.8pf
V= 120V, 60 Hz
P= 2.4 kw, cos θ= 80
P= S sin θ - (p/cos θ) sin θ
= P tan θ(cos^-1 (0.8)
=2.4 tan(36.87)= 1.8KVAR
S= 2.4 + j1. 8KVA
1 load absorbs 1.5 kW at 0.707 pf lagging
P= 1.5 kW, cos θ= 0.707 and θ=45 degree
Q= Ptan θ= tan 45°
Q=P=1.5kw
S1= 1.5 +1.5j KVA
S1 + S2= S
2.4+j1.8= 1.5+1.5j + S2
S2= 0.9 + 0.3j KVA
S2= 0.949= 18.43 °
Pf= cos(18.43°) = 0.948
b.) pf to 0.9, a capacitor is needed.
Pf = 0.9
Cos θ= 0.9
θ= 25.84 °
(WC) V^2= P (tan θ1 - tan θ2)
C= 2400 ( tan (36. 87°) - tan (25.84°)) /2 πf × 120^2
f=60, π=22/7
C= 117.5µf