Chapter 4, Problem 4.7P

Solve the differential equation dy-xdx = 0 if the curve passes through (1,0).

find m, if k=460 and v=5.0 k=1/2mv2

The force acting on a particle varies as in the figure below. (The x axis is marked in increments of 10.0 m.) A coordinate plane has a horizontal axis labeled x (m) and a vertical axis labeled Fx (N). There are five points, connected by straight lines in order from A to E. Point A is at (0,0).  Point B is at (40,6).  Point C is at (80,0).  Point D is at (100,−3).  Point E is at (120,0). Find the work done by the force as the particle moves across the following distances. (a) from x = 0 m to x = 80.0 m(b) from x = 80.0 m to x = 120.0 m(c) from x = 0 m to x = 120.0 m Step 1 The work done on the particle by the force F as the particle moves from x = xi to x = xf is the area under the curve from xi to xf.(a) For x = 0 to x = 80 m, WAC =   =   =

solve the Differential Equations

Find the particular solution of the following differential equation:  100 V' = 200 - 10 V  where V is the dependent and t is independent. Given that when t= 0 , V = 0. Find V at 2 seconds.

Determine the magnitude of A-B A= 10i+3k+7k, B=6i+5k

(The complete question is in the picture) The force acting on an object is F = 3.50 [N]ˆi−5.20 [N]ˆj−2.30 [N]kˆ. The vector from the origin to the point where the force is applied is r = 3.50 [m]ˆi + 2.50 [N]ˆj. What is the vector torque with respect to the origin produced by this force?A. 10.9 [N.m]ˆi + 11.5 [N.m]ˆj − 27.0 [N.m]kˆB. −18.2 [N.m]ˆi + 8.75 [N.m]ˆjC. −5.75 [N.m]ˆi + 8.05 [N.m]ˆj − 27.0 [N.m]kˆD. −5.75 [N.m]ˆi − 8.05 [N.m]ˆj + 9.45 [N.m]k

Number 6 part a b and c Explain the physically unacceptable solution of the quadratic equation in t that you obtain

Given the figure, AB=5m, BC=8m and AG=7m and pt 0 is origin. Determine the following: • What is the magnitude of R and Cr in pt G? (Unit must be in N and kN-m) • What is the magnitude of R and Cr in pt 0? (Unit must be in N and kN-m) • Unit Vector of R in the system. • Direct Cosines of Cr in pt 0.