Technician A says that one planetary gear set can provide gear reduction, overdrive, and reverse. Technician B says that most tr
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:
The current at t= 0 sec, is 0 A
The current at t= 0.5 sec, is 2.2 A
The current at t= 1 sec, is 4.4 A
Explanation:
Given that
q(t) = 2.2 t²
We know that:- the change in the charge w.r.t. time is known as current. So,
q(t) = 2.2 t²
I= 4.4 t
1.
t = 0 s :
I = 4.4 x 0 = 0 A
<u>Therefore, the current at t= 0 sec, is 0 A</u>
2 .
t= 0.5 s :
I = 4.4 x 0.5 = 2.2 A
<u>Therefore, the current at t= 0.5 sec, is 2.2 A</u>
3.
t= 1 s
I = 4.4 x 1 =4.4 A
<u>Therefore, the current at t= 1 sec, is 4.4 A</u>