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- Question 1: Complete the calculations of the table based on the data given below Test No Weight (N) H, (cm) H2 (cm) P.E. (initial) (J) P.E. (final) (J) Loss in P.E. (J) 1 1.8 60 43.0 1.08 0.774 0.306 2 2.0 60 44.4 1.2 0.888 0.312 3 2.2 60 45.7 1.32 1.005 0.3151) What is the momentum of a car of mass 1850 kg and is moving at 27.75 m/s? p = m vmgh = 1 mv ₁ ² 2 Duph FAST 2 m = 1.00 kg 9 = 9.80 m/s² V₁ = 11 m/s ² 2 08.P Solve for hieght (h) MUN TIE
- 1. Based on the values provided, complete the following table. Test NoWeight (N) H, (cm) H2 (cm) P.E. (initial) (J) P.E. (final) (J) Loss in P.E. (J) 1 0.8 50 23.4 2 1.0 50 25.0 3 1.2 50 26.2 4 1.4 50 27.8 5 1.6 50 29.42) Find the missing values: (a) 3.0 N1.0 kg 5.0N (b) 2.0N- 120 kg 3.0 N 5.0N -5.0 N -2.0N i- 2.0 m/s? Fret =? a = 0 Ft-? 10 N -5N (d) 30 N- (e) 4000g (f) 5.0 kg 1.0N 2.0N 3.0N constant Fnet ? a-1.5 m/s? - 0.5 m/s? 7, = ? Fret = ? F, = ? -SN 20 N- (h) (i) Fz 30 18 N a = 5 m/s? Fnut = ? a = + 0.6 m/s? Frst = 1.8 × 10-2 N 10 N - constant (down] m-? m= ? Fz = ? net1. Based on the values provided, complete the following table. Test Weight H1 H2 P.E. (initial) P.E. (final) Loss in P.E. No (N) (cm) (cm) (J) (J) (J) 1 1.4 42.5 2 1.6 65 43.8 1.8 65 45.0 4 2.0 65 47.4 2.2 65 49.7
- 27. Review. A 5.75-kg object passes through the origin at time t = 0 such that its x component of velocity is 5.00 m/s and its y component of velocity is -3.00 m/s. (a) What is the kinetic energy of the object at this time? (b) At a later time t = 2.00 s, the particle is located at x = 8.50 m and y = 5.00 m. What constant force acted on the object during this time interval? (c) What is the speed of the particle at t = 2.00 s?30. A large number of small projectiles with initial velocity v = 20.22 m/s are launched in all directions. The points at which these projectiles reach their maximum height then form a closed surface with volume. Find this volume.Water leaks from a hole in a water tank which is a 17.9 meters below the surface of the water which is 7.31 meters deep. The velocity of the water just leaving the hole is a. 12 b. 140 c. 19 d. 350
- Lo= 0.09 m , g= 10 m/s? m (kg) L (m) x = L – Lo (m) F (N) 0.02 0.108 0.018 0.2 0.04 0.126 0.036 0.4 0.06 0.145 0.055 0.6 0.08 0.167 0.077 0.8 0.1 0.185 0.095 1 2. Plot a graph between Fg versus slope for the best fit line. And experiment value of K for spring by using the slope of graph? in your graph book. And from graph calculate theThe question: Add the following quantities: 4.05 kg + 567.95 g + 100.1 g My solution: 4.05 + 567.95 + 100.1 = 672.1 There are several things wrong with my solution. Tell me what they are.1. Suppose the position vector for a particle is given as a func- tion of time by 7(1) = x(t)i + y(t) j, with x(t) y(t) = ct² + d, where a = = at + b and = 0.125 m/s², 1.00 m/s, b = 1.00 m, c= 1.00 m. (a) Calculate the average velocity during and d the time interval from t = 2.00 s to t = 4.00 s. (b) Determine the velocity and the speed at t = 2.00 s.