Axis Components

from the position at t = 4 s to find the displacement Δx = 12 m. (f) The position at t = 2 s is subtracted from the position at t = 4 s to give the displacement Δx = 14 m. Eq. 2-2, then, leads to vavg = Δx 14 m = = 7 m/s. Δt 2s (g) The horizontal axis is 0 ≤ t ≤ 4 with SI units understood. Not shown is a straight line drawn from the point at (t, x) = (2, –2) to the highest point shown (at t = 4 s) which would represent the answer for part (f). 6. (a) Using the fact that time =…

from the position at t = 4 s to find the displacement Δx = 12 m. (f) The position at t = 2 s is subtracted from the position at t = 4 s to give the displacement Δx = 14 m. Eq. 2-2, then, leads to vavg = Δx 14 m = = 7 m/s. Δt 2s (g) The horizontal axis is 0 ≤ t ≤ 4 with SI units understood. Not shown is a straight line drawn from the point at (t, x) = (2, –2) to the highest point shown (at t = 4 s) which would represent the answer for part (f). 6. (a) Using the fact that time =…

from the position at t = 4 s to find the displacement Δx = 12 m. (f) The position at t = 2 s is subtracted from the position at t = 4 s to give the displacement Δx = 14 m. Eq. 2-2, then, leads to vavg = Δx 14 m = = 7 m/s. Δt 2s (g) The horizontal axis is 0 ≤ t ≤ 4 with SI units understood. Not shown is a straight line drawn from the point at (t, x) = (2, –2) to the highest point shown (at t = 4 s) which would represent the answer for part (f). 6. (a) Using the fact that time =…

position at t = 4 s to find the displacement Δx = 12 m. (f) The position at t = 2 s is subtracted from the position at t = 4 s to give the displacement Δx = 14 m. Eq. 2-2, then, leads to vavg = Δx 14 m = = 7 m/s. Δt 2s (g) The horizontal axis is 0 ≤ t ≤ 4 with SI units understood. Not shown is a straight line drawn from the point at (t, x) = (2, –2) to the highest point shown (at t = 4 s) which would represent the answer for part (f). 6. (a) Using the fact that time =…

we assume Q > 0 and q < 0 (although the final result does not depend on this particular choice). (a) The x-component of the force experienced by q1 = Q is 1 4πε 0 F1x = − ( Q )( Q ) cos 45° + (| q |) ( Q ) ( 2a ) 2 a 2 = Q|q| Q /|q| − +1 2 4 πε 0 a 2 2 which (upon requiring F1x = 0) leads to Q / | q |= 2 2 , or Q / q = −2 2 = −2.83. (b) The y-component of the net force on q2 = q is 1 4πε 0 | q |2 F2 y = ( 2a ) sin 45° − 2 (|…

we assume Q > 0 and q < 0 (although the final result does not depend on this particular choice). (a) The x-component of the force experienced by q1 = Q is 1 4πε 0 F1x = − ( Q )( Q ) cos 45° + (| q |) ( Q ) ( 2a ) 2 a 2 = Q|q| Q /|q| − +1 2 4 πε 0 a 2 2 which (upon requiring F1x = 0) leads to Q / | q |= 2 2 , or Q / q = −2 2 = −2.83. (b) The y-component of the net force on q2 = q is 1 4πε 0 | q |2 F2 y = ( 2a ) sin 45° − 2 (|…

coordinates and the invariant line element 1.5 Geometry and groups 1.6 Vectors 1.7 Quaternions 1.8 3-vector analysis 1.9 Linear algebra and n-vectors 1.10 The geometry of vectors 1.11 Linear operators and matrices 1.12 Rotation operators 1.13 Components of a vector under coordinate rotations 1 2 5 7 10 13 13 16 18 21 24 25 27 2 KINEMATICS: THE GEOMETRY OF MOTION 2.1 2.2 2.3 2.4 Velocity and acceleration Differential equations of kinematics Velocity in…

rod’s total mass m. y l Iy = M L x2 dm A l = = 0 L x 2 x (r A dx) 1 r A l3 3 m = rAl Thus, Iy = 1 m l2 3 Ans. 17–2. The right circular cone is formed by revolving the shaded area around the x axis. Determine the moment of inertia Ix and express the result in terms of the total mass m of the cone. The cone has a constant density r. y y r –x h r x 2 dm = r dV = r(p y dx) r(p) ¢ h m= dIx = L 0…

........................... 12 2.3.1 Define the Problem........................................................................ 12 2.3.2 Starting Triflex Windows .............................................................. 14 2.3.3 Coding the Components ................................................................ 17 2.3.3.1 Anchor Data Point 1000…………………………………………... 2.3.3.2 Pipe Data Point 1010……………………………………………… 2.3.3.3 Branch Connection Data Point 1020……………………………… 2.3.3.4 Pipe &…

which becomes, upon finding its components (and using the fact that cos 45◦ = 1/ 2), q (2q ) ˆ q (2q ) ˆ i+k √ j. F2 = k √ 2 2 2a 2 2 a2 Finally, the force exerted by the charge −2q on +2q is F3 = k (2q )(2q ) ˆ i. a2 (a) Therefore, the horizontal component of the resultant force on +2q is Fx = = F 1 x + F 2 x + F3 x = k 8.99 × 109 q2 a2 1 √ +4 2 1.0 × 10−7 0.0502 2 1 √ +4 2 = 0.17 N . (b) The vertical component of the net force is F y = F1 y