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- A differential equation encountered in the vibration of beams is d4ydx4=2D where x = distance measured along the beam; [x]=[L] y = displacement of the beam; [y]=[L] = circular frequency of vibration; []=[T1] = mass of the beam per unit length; []=[ML1] D = bending rigidity of beam; [D]=[FL2] Show that the equation is dimensionally homogeneous. Note that [F]=[MLT2] see Eq. (1.21:).Principle of Linear Impulse and Momentum for a System of Particles To apply the principle of linear impulse and momentum to a system of particles. Integrating the equation of motion, as applied to all particles in a system, yields ∑mi(vi)1+∑∫t2t1Fidt=∑mi(vi)2 where mi is the ith particle's mass, vi is the ith particle's velocity, and Fi is the external force that acts on the ith particle. This relationship states that the sum of the initial linear momenta, at time t1, and the impulses of all the external forces acting between times t1 and t2 is equal to the sum of the linear momenta of the system, at time t2. If the system has a mass center, G, the expression becomes m(vG)1+∑∫t2t1Fidt=m(vG)2 This expression allows the principle of linear impulse and momentum to be applied to a system of particles that is represented as a single particle. Two blocks, each of mass m = 6.90 kg , are connected by a massless rope and start sliding down a slope of incline θ = 36.0 ∘ at t=0.000 s. The…Hydrostatic pressure at any given depth of a liquid depends on the_____ of the container.a)Volumeb)ShapeC)Depthd)All of the above .As the temperature of oil increases, the viscosity of oil willa) increaseb) decreaseC)not change Which of the following units would NOT be used for volume flow rate?a) gal/minb) Ib/sc)m⅗d)ft^3/s 2.Dynamic viscosity is defined as,a) weight per unit volume of a fluidb) volume per unit weight of a fluidc) the velocity gradient divided by the shearing stress of a fluidd)the shearing stress divided by the velocity gradient of a fluid
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- A lightweight drone (1.00 kg) is launched at 800 m high and moves upward at a constant velocity (while ignoring the effects of gravity only on the drone). The balloon, when measured at a horizontal distance from you, is about 1600 m away from you. At the moment when the drone moves, you shoot a bullet (weight =180 g) with an initial velocity of 1009 m/s at a fixed angle α, where sin α=3/5 and cos α= 4/5. (g = 9.8 m/s2) c. What is the upward velocity of the drone?The gravitational acceleration on Mars is 3.72 m/s2. The density of water is 1000 kg/m3. Starting from these assumptions, estimate the specific weight of water on Mars in lbf/ft3. Note that when converting lbm (pounds mass) to lbf (pounds force), a unit conversion called gc is required (1 lbf = 32.2 lbm ft s-2). The conversion factor ensures that (under normal gravity), a 100 lbm load generates 100 lbf of gravitational forceAngular Impulse and Momentum Principles To apply the principle of angular impulse and momentum to describe a particle's motion. The moment of a force about a point O, fixed in an inertial coordinate system, MO, and the angular momentum about the same point, HO, are related as follows: ∑MO=H˙O where H˙O is the time derivative of the angular momentum, HO=r×mv. Integrating this equation with respect to time yields the following equation: ∑∫t2t1MO dt=(HO)2−(HO)1 This equation is the principle of angular impulse and momentum, and it is often rearranged to its more familiar form (HO)1+∑∫t2t1MO dt=(HO)2 A centrifugal governor consists of a central rotating shaft that has two thin, pin-connected rods attached to it; a heavy sphere caps the end of each rod. (Figure 1) A centrifugal governor mechanically limits an engine's speed. A part of the engine turns the centrifugal governor, and if the speed exceeds a set amount, the height of the spheres decreases the driving force of the engine by…
- Angular Impulse and Momentum Principles To apply the principle of angular impulse and momentum to describe a particle's motion. The moment of a force about a point O, fixed in an inertial coordinate system, MO, and the angular momentum about the same point, HO, are related as follows: ∑MO=H˙O where H˙O is the time derivative of the angular momentum, HO=r×mv. Integrating this equation with respect to time yields the following equation: ∑∫t2t1MO dt=(HO)2−(HO)1 This equation is the principle of angular impulse and momentum, and it is often rearranged to its more familiar form (HO)1+∑∫t2t1MO dt=(HO)2 A centrifugal governor consists of a central rotating shaft that has two thin, pin-connected rods attached to it; a heavy sphere caps the end of each rod. (Figure 1) A centrifugal governor mechanically limits an engine's speed. A part of the engine turns the centrifugal governor, and if the speed exceeds a set amount, the height of the spheres decreases the driving force of the engine by…A rocket is fired at an angle of 30° above the horizontal. If the x-component of its initial velocity is vix = 326 m/s, what is the y-component of its initial velocity ? ( vy =?) (Take g = 10 m/s²). (A)188.2 m/s (B) 376.4 m/s (C) 330 m/s (D)250 m/s (E) 163 m/sSomewhere DEEP BELOW THE EARTH's surface, at an UNKNOWN displacement from the Earth's center, a particle of mass m is dangled from a long string, length L; the particle oscillates along a small arc according to the differential equation d^2x/dt^2=-(pi^2/36)x. Here x refers to an angular displacement measured from the vertical and t refers to time. The particle's mass is given by m=3kg. The length of the string is given by L=5 meters. Whenever the particle arrives at a location of x=(pi/12) radians from the vertical, the particle has no instantaneous speed. On both sides of the vertical, that is, x=(pi/12) radians is repeatedly observed to be a 'turning point' for the particle's periodic motion. 1. Draw a clear FREE-BODY diagram of this particle at some arbitrary point during oscillation, making sure to label variables and constants described above. 2. Approximating to three significant digits if necessary, what is the angular frequency of this oscillator on a string? 3. Approximating…