Question

27S 140-1 November 20, 2019 all work in the examination book. A block of mass 1.80 kg is suspended from an ideal spring with force constant k = 0.200 N/m. The mass is pushed 15.0 cm below its equilibrium position and then given an initial speed of 0.300 m/s directed downward (watch signs). a. What is the period of the resulting SHM of the block? Show work. (15 points) b. Taking the upward direction as a positive y-axis, the resulting motion of the block can be written as y = y,cos(cor+o) y in mir in s). Find the values of o, y, and Ⓡ. Show work. (15 points) A pump is filling a large open vat of cross-sectional area 2.00 m' with water. The cross-sectional area of the filling pipe at the base of the vat is 5.00x10"mand the top of the water is exposed to the atmosphere. See Figure 1. p. = 1.00x10 p . -1.01x10 a. Use the equation of continuity to find vif y = 50.0 cm/s. (14 points) b. If the pump is operating with power P - 265 watts, use P=Fvy to find the force F and then the pressure P, the pump is exerting on the water. (14 points) c. Use p+1 pv? + Pgy = constant (Bernoulli's equation) to find the depth h of the water in the tank at the instant shown in Figure 1. Do not round off until you Figure 1 obtain your final answer. Show work. (15 points) d. Assuming the pump maintains constant hydrostatic pressure pı (the value you found in part b) on the water (so that the water is not flowing), find the height of the water in the tank. Show work. (12 points) State Kepler's three laws of planetary motion. (15 points) atra Credit A small mass m sits to the right of a uniform thin rod of mass M and length L. See Figure 2. duide (1. = ML) a. Write a formula for dU, the gravitational potential energy of m and the small piece of the rod dM (5 points) b. Use integration to find the total gravitational potential energ U of m and the rod. (10 points) c. Find the force of gravitational attraction on m from the rod using Fu (10 points) Figure 2

Answer 1

ay Angular dowquency = 8 = V portation = (0.33 sally and time period - I nteger (19.0387 whis b.) Apply energy conservation on - kyla = 1k22 + 1 min Ź XO.2 x 47 = x0.2% (0-15)2 + 1*18x0.32 Toy = 0.837 loengute t=0 D = Hoy Cos CO. E +®] => 0.15= 0.83 205() and angular frequency wath in a for [W= 0:53 vadle ya 0.83 Cos (0-33£ + 73.5) Plz ook long question pastwise