The value of cos(4M) where M is the magnitude of the vector field with potential ƒ = e² sin(лy) cos(π²) at x = 1, y = 1/4, z = 1/3 is 0.602 -0.323 0.712 -0.816 0.781 0.102 0.075 0.013
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- A sodium ion (Na+) moves in the xy-plane with a speed of 2.90 ✕ 103 m/s. If a constant magnetic field is directed along the z-axis with a magnitude of 3.25 ✕ 10−5 T, find the magnitude of the magnetic force acting on the ion and the magnitude of the ion's acceleration. HINT (a) the magnitude (in N) of the magnetic force acting on the ion N (b) the magnitude (in m/s2) of the ion's acceleration m/s2In the images below, the vector field F is shown by the blue arrows and the curve Cis shown in red. Indicate the value of F ·dT. 1. /, F ·dÌ = 0 2. ,F ·dT > 0 3. F dT < 0 4. The value of F ·dT cannot be determined with only this information. XFind the curl of the vector field F (6x sin(y), 2y cos(x)). curl F =
- Let be a potential for the vector field F = (−24 x³ z + 6 xyz², 3 x² z² + 15 y¹, −6 x¹ + 6 x²yz – 10 %). Then the value of sin((-1.63, 2.06, 0.57) - (0,0, 0)) isShow that (vector)F = <ln(y) + y/x, ln(x) + x/y > is a gradient field. Then find a potential function f for (vector) F.Use the equation giving the flux of the vector field across the curve to calculate the flux of F(x, y) = x+1 y ( (x + 1)² + y²' (x + 1)² + y² across C, the segment 7≤ y ≤ 9 along the y-axis, oriented upwards. (Use symbolic notation and fractions where needed.) F. dr = 2 (1.9085 - 1.6987)
- fellas is it controversial to existQuestion A1 Consider a vector field F= 2x° yị +x'j., and a curve C defined by the equation y = 2x². Calculate the line integral (F•dr along the curve C starting at x = 0 and finishing at x = 1. 2 1.8 1.4 12 1 0.8 F 0.6 0.4 0.2 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 X-axisAnswer both remaining part in 10 minutes in the order to get positive feedback
- Sketch the vector field on the real line and find the stability of the fixed points.The motion of a point on the circumference of a rolling wheel of radius 4 feet is described by the vector function r(t) = 4(12t - sin(12t))i + 4(1 − cos(12t))j Find the velocity vector of the point. v(t) = Find the acceleration vector of the point. a(t) = Find the speed of the point. s(t) = =We are given a vector field and a closed path.