Applied Mechanics and Graphic Statics A particle moves with a velocity of 2 m/sec in a straight line with a negative acceleration of 0.1 m/sec2. Time required to traverse a distance of 1.5 m, is 20 sec 15 sec 40 sec 30 sec 20 sec 15 sec 40 sec 30 sec ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics The centre of gravity of a triangle is at the point where three None of these Bisectors of the angle of the triangle meet Perpendicular bisectors of the sides of the triangle meet Medians of the triangle meet None of these Bisectors of the angle of the triangle meet Perpendicular bisectors of the sides of the triangle meet Medians of the triangle meet ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics The centre of gravity of a homogeneous body is the point at which the whole All listed here Area of the surface of the body is assumed to be concentrated Volume of the body is assumed to be concentrated Weight of the body is assumed to be concentrated All listed here Area of the surface of the body is assumed to be concentrated Volume of the body is assumed to be concentrated Weight of the body is assumed to be concentrated ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics A projectile is fired with a velocity of 100.3 m/sec. at an elevation of 60°. The velocity attained by the projectile when it is moving at a height of 100 m, is 75 m/sec 90 m/sec 80 m/sec 70 m/sec 75 m/sec 90 m/sec 80 m/sec 70 m/sec ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics If the resultant of two forces ‘P’ and ‘Q’ acting at an angle ‘θ’ makes an angle ‘α’ with ‘P’, then tan α equals P sin θ/P + Q tan θ P sin θ/P - Q cos θ Q sin θ/P + Q cos θ Q sin θ/P + Q sin θ P sin θ/P + Q tan θ P sin θ/P - Q cos θ Q sin θ/P + Q cos θ Q sin θ/P + Q sin θ ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics If a particle moves with a uniform angular velocity ‘ω’ radians/sec along the circumference of a circle of radius ‘r’, the equation for the velocity of the particle, is y = ω √(y - r) v = ω √(r² + y²) v = ω √(y² - r²) v = ω √(r² - y²) y = ω √(y - r) v = ω √(r² + y²) v = ω √(y² - r²) v = ω √(r² - y²) ANSWER DOWNLOAD EXAMIANS APP
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