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Answer:
R = 148.346 N
M₀ = - 237.2792 N-m
Explanation:
Point O is selected as a convenient reference point for the force-couple system which is to represent the given system
We can apply
∑Fx = Rx = - 60N*Cos 45° + 40N + 80*Cos 30° = 66.8556 N
∑Fy = Ry = 60N*Sin 45° + 50N + 80*Sin 30° = 132.4264 N
Then
R = √(Rx²+Ry²) ⇒ R = √((66.8556 N)²+(132.4264 N)²)
⇒ R = 148.346 N
Now, we obtain the moment about the origin as follows
M₀ = (0 m*40 N)-(7 m*60 N*Sin 45°)+(4 m*60 N*Cos 45°)-(5 m*50 N)+ 140 N-m + (0 m*80 N*Cos 30°) + (0 m*80 N*Sin 30°) = - 237.2792 N-m (clockwise)
We can see the pic shown in order to understand the question.
Answer:
98,614.82 W/m²
Explanation:
Where;
Q = the amount of heat loss from the pipe
h = the heat transfer coefficient of the pipe = 50 W/m².K
T₁ = the ambient temperature of the pipe = 30⁰C
T₂ = the outside temperature of the pipe = 100⁰C
L= the length of pipe
r₁ = inner radius of the pipe = 20mm
r₂ = outer radius of the pipe = 25mm
To determine the amount of heat loss from the pipe per unit length
From the equation above
= 98,614.82 W/m²