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 diameter is 50mm
Explanation:
The answer is in two stages. At first the torque (or twisting moment) acting on the shaft and needed to transmit the power needs to be calculated. Then the diameter of the shaft can be obtained using another equation that involves the torque obtained above.
T=(P×60)/(2×pi×N)
T is the Torque 
P is the the power to be transmitted by the shaft; 40kW or 40×10³W
pi=3.142
N is the speed of the shaft; 250rpm
T=(40×10³×60)/(2×3.142×250)
T=1527.689Nm
Diameter of a shaft can be obtained from the formula
T=(pi × SS ×d³)/16
Where 
SS is the allowable shear stress; 70MPa or 70×10⁶Pa
d is the diameter of the shaft
Making d the subject of the formula
d= cubroot[(T×16)/(pi×SS)]
d=cubroot[(1527.689×16)/(3.142×70×10⁶)]
d=0.04808m or 48.1mm approx 50mm
 
        
             
        
        
        
Tell me why i got this question got it right and now won’t remember but i’ll get back at you when i remember
        
             
        
        
        
What standard feature is on all circular saws?
4. All of the above
 
        
             
        
        
        
Answer:
Absolute Pressure=315.06256 kPa
Explanation:
Gauge pressure= 31 psi
Atmospheric Pressure at Sea level= 1 atm=101.325 kPa
