Engineering Mechanics The velocity of a particle moving with simple harmonic motion is _________ at the mean position. Zero Maximum Minimum None of these Zero Maximum Minimum None of these ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The three forces of 100 N, 200 N and 300 N have their lines of action parallel to each other but act in the opposite directions. These forces are known as Coplanar non-concurrent forces Coplanar concurrent forces Like parallel forces Unlike parallel forces Coplanar non-concurrent forces Coplanar concurrent forces Like parallel forces Unlike parallel forces ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics Three forces acting on a rigid body are represented in magnitude, direction and line of action by the three sides of a triangle taken in order. The forces are equivalent to a couple whose moment is equal to Twice the area of the triangle Area of the triangle Half the area of the triangle None of these Twice the area of the triangle Area of the triangle Half the area of the triangle None of these ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The resolved part of the resultant of two forces inclined at an angle 'θ' in a given direction is equal to The difference of the forces multiplied by the cosine of θ The sum of the resolved parts of the forces in the given direction The sum of the forces multiplied by the sine of θ The algebraic sum of the resolved parts of the forces in the given direction The difference of the forces multiplied by the cosine of θ The sum of the resolved parts of the forces in the given direction The sum of the forces multiplied by the sine of θ The algebraic sum of the resolved parts of the forces in the given direction ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics Moment of inertia of a rectangular section having width (b) and depth (d) about an axis passing through its C.G. and parallel to the width (b), is db³/36 db³/12 bd³/12 bd³/36 db³/36 db³/12 bd³/12 bd³/36 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 = Ah2 / IG IP = IG + Ah2 IP = IG - Ah2 IP = IG / Ah2 IP = Ah2 / IG IP = IG + Ah2 IP = IG - Ah2 ANSWER DOWNLOAD EXAMIANS APP
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