In CERN, some work is carried out for developing an accurate
and fast numerical method that can calculate natural gas flow in
a pipeline under non-isothermal steady-state conditions.
The cross section of the pipeline is shown below:
In Diagram 1, O is the centre of the circle of radius 6 cm and P is
the mid-point of the chord AB. The length OP is 3 cm.
Answers
Answer:
A=||| 120°
angle POB=Cos thetha=OP/OB
angle POB=3/6=1/2
angle POB=60°
angle AOB=2(POB)=120°
Given : O is the centre of the circle of radius 6 cm and P is the mid-point of the chord AB. The length OP is 3 cm.
To Find : ∠AOB
Area of ΔAOB in sq cm
Area of bigger sector AOB
Volume of liquid in vessel
Solution:
Cos ∠POB = OP/OB = 3/6 = 1/2 = Cos 60°
=> ∠POB = 60°
∠AOB = 2 ∠POB = 2 × 60° = 120°
∠AOB = 120°
Area of ΔAOB = (1/2) OA * OB Sin ∠AOB
= (1/2) * 6 * 6 * sin 120°
= 18 * √3/2
= 9√3
= 15.588
= 15.6 cm²
∠AOB = 120°
=> bigger angle = 360° - 120° = 240°
Area of bigger sector AOB = ( 240° / 360° ) π * 6²
= 24π
Volume of liquid in vessel = ( area of bigger sector AOB + area of ΔAOB ) * length
= ( 24π + 15.6 ) * 20
= 1820 cm²
Total surface area of cylindrical vassal = 2πrh + 2πr²
= 2πr(r + h)
= 2π * 6 ( 6 + 20)
= 312π cm²
Learn More:
find the curved surface area total surface area and volume of a ...
brainly.in/question/2473335
the circumference of the base of a right circular cylinder is 176cm. if ...
brainly.in/question/1992655
