Engineering Mechanics Coplanar non-concurrent forces are those forces which __________ at one point, but their lines of action lie on the same plane. None of these Meet Either ‘A’ or ‘B’ Do not meet None of these Meet Either ‘A’ or ‘B’ Do not meet ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The static friction Bears a constant ratio to the normal reaction between the two surfaces Always acts in a direction, opposite to that in which the body tends to move All of these Is independent of the area of contact, between the two surfaces Bears a constant ratio to the normal reaction between the two surfaces Always acts in a direction, opposite to that in which the body tends to move All of these Is independent of the area of contact, between the two surfaces ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics Moment of inertia of a triangular section of base (b) and height (h) about an axis through its base, is bh3/8 bh3/36 bh3/4 bh3/12 bh3/8 bh3/36 bh3/4 bh3/12 ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The law of the machine is (where P = Effort applied to lift the load, m = A constant which is equal to the slope of the line, W = Load lifted, and C = Another constant which represents the machine friction.) P = mW - C P = mW + C P = m/W + C P = C - mW P = mW - C P = mW + C P = m/W + C P = C - mW ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics When two elastic bodies collide with each other, The two bodies tend to compress and deform at the surface of contact The two bodies will momentarily come to rest after collision The two bodies begin to regain their original shape All of these The two bodies tend to compress and deform at the surface of contact The two bodies will momentarily come to rest after collision The two bodies begin to regain their original shape All of these ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics According to parallel axis theorem, the moment of inertia of a section about an axis parallel to the axis through center of gravity (i.e. IP) is given by(where, A = Area of the section, IG = Moment of inertia of the section about an axis passing through its C.G., and h = Distance between C.G. and the parallel axis.) IP = IG - Ah2 IP = IG / Ah2 IP = Ah2 / IG IP = IG + Ah2 IP = IG - Ah2 IP = IG / Ah2 IP = Ah2 / IG IP = IG + Ah2 ANSWER DOWNLOAD EXAMIANS APP
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