Engineering Mechanics A cable with a uniformly distributed load per horizontal meter run will take the following shape Straight line Elliptical Parabola Hyperbola Straight line Elliptical Parabola Hyperbola ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics In order to determine the effects of a force, acting on a body, we must know Magnitude of the force All of these Line of action of the force Nature of the force i.e. whether the force is push or pull Magnitude of the force All of these Line of action of the force Nature of the force i.e. whether the force is push or pull 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 bh3/36 bh3/8 bh3/4 bh3/12 bh3/36 bh3/8 bh3/4 bh3/12 ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The velocity of a particle (v) moving with simple harmonic motion, at any instant is given by (where, r = Amplitude of motion, and y = Displacement of the particle from mean position.) ω².√(r² - y²) ω.√(r² - y²) ω.√(y² - r²) ω².√(y² - r²) ω².√(r² - y²) ω.√(r² - y²) ω.√(y² - r²) ω².√(y² - r²) ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics A particle moves along a straight line such that distance (x) traversed in 't' seconds is given by x = t² (t - 4), the acceleration of the particle will be given by the equation 6t² - 8t 6f - 8 6f - 4 3t² + 2t 6t² - 8t 6f - 8 6f - 4 3t² + 2t 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 = IG / Ah2 IP = Ah2 / IG IP = IG + Ah2 IP = IG - Ah2 IP = IG / Ah2 IP = Ah2 / IG ANSWER DOWNLOAD EXAMIANS APP
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