With increases in magnification, which of the following occur? a. The field of view decreases. b. The ambient illumination decre
1 answer:
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Answer: 3/2mg
Explanation:
Express the moment equation about point B
MB = (M K)B
-mg cosθ (L/6) = m[α(L/6)](L/6) – (1/12mL^2 )α
α = 3g/2L cosθ
express the force equation along n and t axes.
Ft = m (aG)t
mg cosθ – Bt = m [(3g/2L cos) (L/6)]
Bt = ¾ mg cosθ
Fn = m (aG)n
Bn -mgsinθ = m[ω^2 (L/6)]
Bn =1/6 mω^2 L + mgsinθ
Calculate the angular velocity of the rod
ω = √(3g/L sinθ)
when θ = 90°, calculate the values of Bt and Bn
Bt =3/4 mg cos90°
= 0
Bn =1/6m (3g/L)(L) + mg sin (9o°)
= 3/2mg
Hence, the reactive force at A is,
FA = √(02 +(3/2mg)^2
= 3/2 mg
The magnitude of the reactive force exerted on it by pin B when θ = 90° is 3/2mg
Answer:
v = 1.076 m /s
Explanation:
Initial volume of balloon = 4/3 x 3.14 x (9.905/2)³
=508.56 m³
Final volume of balloon = 4/3 x 3.14 x (16.502/2)³
= 2351.73 m³
Increase in volume = 1843.17 m³
Cross sectional area of inlet A = 3.14 x( 1.458/2)²
A = 1.6687 m²
Volume rate of flow of air = cross sectional area x velocity of inflow
= 1 .6687 V [ V is velocity of inflow ]
Total time taken = Increase in volume / rate of flow of air
17.108 X 60 = 1843.17 / 1.6687 V
V =
v = 1.076 m /s