An aggregate blend is composed of 55% aggregate A (Sp. Gr. 2.631), 25% aggregate B (Sp. Gr. 2.331) and 20% sand (Sp. Gr. 2.609).
1 answer:
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Answer:
<em>866.1 N</em>
Explanation:
The torque on the flywheel = 300 N-m
The force from the hydraulic cylinder will generate a moment on CA about point A.
The part of this moment that will be at point B about A must be proportional to the torque on the cylinder which is 300 N-m
we know that moment = F x d
where F is the force, and
d is the perpendicular distance from the turning point = 1 m
Equating, we have
300 = F x 1
F = 300 N this is the frictional force that stops the flywheel
From F = μN
where F is the frictional force
μ is the coefficient of static friction = 0.4
N is the normal force from the hydraulic cylinder
substituting, we have
300 = 0.4 x N
N = 300/0.4 = 750 N
This normal force calculated is perpendicular to CA. This actual force, is at 30° from the horizontal. To get the force from the hydraulic cylinder R, we use the relationship
N = R sin (90 - 30)
750 = R sin 60°
750 = 0.866R
R = 750/0.866 = <em>866.1 N</em>
Answer:
a) 1.16 m/s
b) 1/216000
c) (√15)/6480000
Explanation:
The parameters given are;
Length of boat prototype, lp = 30 m
Speed of boat prototype = 9 m/s
Length of boat model, lm= 0.5 m
a) lm/lp = 0.5/30 = 1/60 = ∝
(vm/vp) = ∝^(1/2) = √∝ = (1/60)^(1/2)
vm = 9 × (1/60)^(1/2) = 1.16 m/s
b) The ratio of the model to prototype drag, Fm/Fp, is given as follows;
Fm/Fp = (vm/vp)²×(lm/lp)² = ∝³
Fm/Fp = (1/60)³ = 1/216000
c) The ratio of the model to prototype power pm/p = (Fm/Fp) × (vm/vp) = ∝³×√∝
The ratio of the model to prototype power pm/p = √(1/60) × (1/60)³
pm/p = √(1/60) × (1/60)³ = (√15)/6480000