In the figure shown, Given: VB =1.3 m/s 20° Solve for: a) velocity of A (m/s) b) angular velocity of AB (rad/s) B 60° 500 mm A
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- IN THE FIGURE, THE WIDTH OF A RIVER BEING OBSERVED IS 63m. ASSUMING THE TRANSIT BEING USED HAS AN EXTERNAL FOCUSING TELESCOPE AND K = 95, DETERMINE THE FOLLOWING IF C=0. *Determine the horizontal distance DR if the stadia intercept at R is 0.88. *Determine the horizontal distance DL. *Determine the stadia intercept at rod L.The circular disk of radius r = 0.40m rotates about a fixed axis through point O with the angular properties omega w= 2.2 rad/s and alpha a= 3.3 rad/s 2 with directions as shown in the figure. Determine the instantaneous values of the velocity and acceleration of point A.A wheel 0.70 m in dianeter starts from rest and accelerates uniformly to an angular velocity of 80 rad/sec in 20 seconds. Find the angular acceleration in rad/s^2
- Situation 2: The displacement (in meter) of a particle moving along x - axis is givenby x=At^2 + B , where A = 2m and B = 3m. Calculate average velocity between t =3sand t = 5s.Situation 2: Calculate instantaneous velocity at t = 5sSituation 2: Calculate instantaneous acceleration.A 500 N block is resting on an inclined plane and is subjected to a constant force of 600N acting parallel to the inclined plane. After the block has moved 3 m from rest along the inclined plane the force 600N is removed. The inclined plane has a slope of three vertical to four horizontal. Coefficient of friction is 0.20 a. Which is the following gives the distance that the black will move further along the inclined plane until it will stop? b. Which of the following gives the velocity of the block when this for 600 Newton was removed? c. Which of the following gives the velocity of the block when it returns to its initial position?With the line of sight horizontal, solve for the stadia constant given the following? Station Occupied: A Station Observed: B Distance from A to B: [7+7](10) m Stadia Intercept = [10+7](10) em Station Occupied: A Station Observed: C Distance from A to C: [7](10) m Stadia Intercept = [10](10) cm Fundamental of surveying
- Given the acceleration-time diagram sketch the velocity-time diagram and determine the displacement at time=9s.The position coordinate of a particle which is confined to move along a straight line is given by s = 2t³ - 24t + 6, where s is measured in meters from a convenient origin and t is in seconds. Determine (a) the time required for the particle to reach a velocity of 72 m/s from its initial condition at t = 0Consider a velocity field where the x and y components of velocity aregiven by u = cx/(x2 + y2) and v = cy/(x2 + y2) where c is a constant. For source flow, calculate: a. The time rate of change of the volume of a fluid element per unitvolume.b. The vorticity.Hint: It is simpler to convert the velocity components to polar coordinatesand deal with a polar coordinate system.
- The elevations of the ski jumping hill shown below are as follows: hA = 44 mhB = 1.6 mhC = 4.9 m If the velocity of a ski jumper is measured to be 27.9 m/s at position B, what will be the launch velocity (in m/s) at position Ck = D10*(γw/μ)*(C*e3)/(1+e) Where does this formula come from, it is not one that I could find in our class notes. Also how was the dynamic velocity calculated?The angular velocity of the disk is defined by ω = (5t2 + 2) rad/s, where t is in seconds. Determine the magnitudes of the velocity and acceleration of point A on the disk when t = 2.5 s.