Engineering Mechanics The resolved part of the resultant of two forces inclined at an angle 'θ' in a given direction is equal to The algebraic sum of the resolved parts of the forces in the given direction The sum of the forces multiplied by the sine of θ The sum of the resolved parts of the forces in the given direction The difference of the forces multiplied by the cosine of θ The algebraic sum of the resolved parts of the forces in the given direction The sum of the forces multiplied by the sine of θ The sum of the resolved parts of the forces in the given direction The difference of the forces multiplied by the cosine of θ ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics Non-coplanar concurrent forces are those forces which Meet at one point, but their lines of action do not lie on the same plane Meet at one point and their lines of action also lie on the same plane Do not meet at one point, but their lines of action lie on the same plane Do not meet at one point and their lines of action do not lie on the same plane Meet at one point, but their lines of action do not lie on the same plane Meet at one point and their lines of action also lie on the same plane Do not meet at one point, but their lines of action lie on the same plane Do not meet at one point and their lines of action do not lie on the same plane ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics A redundant frame is also called __________ frame. Deficient None of these Imperfect Perfect Deficient None of these Imperfect Perfect ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics If n = number of members and y = number of joints, then for a perfect frame, n = ? 43892 43864 43832 43865 43892 43864 43832 43865 ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The time of flight (t) of a projectile on a horizontal plane is given by t = 2u. sinα/g t = 2u/g.sinα t = 2u. cosα/g t = 2u. tanα/g t = 2u. sinα/g t = 2u/g.sinα t = 2u. cosα/g t = 2u. tanα/g 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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