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- 1) 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 TIE1. 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 29.4
- 2) 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 = ? net1.67 Ka into ma 3.296X10° Km into mm. 4.967 Kg into da S8961m into (m 6.3.45X lo? m/s into Km/hr 7.234x 63Kmlhre into ms 8.233 um into m 9,432x10" Kq into9The 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. 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 49.730. 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.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 the
- 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. 3501. 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.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?