Suppose that each component of a certain vector is doubled. (a) by what multiplicative factor does the magnitude of the vector change? (b) by what multiplicative factor does the direction angle of the vector change?

a. The magnitude of the vector is doubled as well. Let's say we have a 2-dimensional vector with components x and y. It's magnitude lâ‚ is given by: lâ‚ = âš(x² + y²) If we double the components x and y, the new magnitude lâ‚‚ is: lâ‚‚ = âš((2x)² + (2y²)) With a bit of algebra... lâ‚‚ = âš(4x² + 4y²) lâ‚‚ = âš4(x² + y²) lâ‚‚ = 2âš(x² + y²) We can write the new magnitude lâ‚‚ in terms of the old magnitude lâ‚. lâ‚‚ = 2lâ‚ Therefore, the new magnitude is double the old one. It should be clear that this relationship applies to 3D (and 1D) vectors as well. b. The direction angle is unchanged. The direction angle θ₠for a 2-dimensional vector is given by: θ₠= arctan(y / x) If we double both components, we get: θ₂ = arctan(2y / 2x) θ₂ = arctan(y / x) θ₂ = θ₠The new direction angle is the same as the old one.

34 degrees fahrenheit

Distance covered, s = 8ft - 4ft = 4 ftinitial velocity, u = 34 ft/sthe acrobat is moving upwards and therefore, a = -g = -9.81 m/s^2 = 32.19 ft/s^2applying equation of motion; s = ut -1/2gt^2then,4 = 34t -1/2*32.19t^24 = 34t - 16.095t^2rearranging; -16.095t^2 + 34t - 4 = 0solving the quadratic equation; t = {-34 +/- sqrt [34^2 - 4(-16.)]}/2(-16.095) = 1.06 +/- 0.93 = 0.13 or 1.99the acrobat will take 0.13 seconds to be at 8 ft above the ground. additionally, once at the maximum height, the acrobat will move down and be at the same spot (that is height) after 1.99 seconds moving down.