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question:In a tennis serve, a 0.079 kg ball is accelerated from rest to 33 m/s over a distance of 0.80 m. Determine the magnitude of the average force exerted by the racket on the ball during this serve.
answer:Using the work-energy theorem, we can find the average force: {eq}begin{align*} W &= Delta KE F cdot d &= frac{1}{2} m (v_f^2 - v_o^2) F (0.80) &= frac{1}{2} (0.079) (33^2 - 0^2) F &= frac{1}{2} (0.079) (1089) F &= 43.77 , N end{align*} {/eq} Thus, the magnitude of the average force exerted by the racket on the ball is 43.77 N.
question:Find the estimated interior angles in radians, the area, and the perimeter of a convex polygon with 7 vertices. The given coordinates of the vertices are: left( begin{array}{cc} -0.822 & 0.56 -0.807 & 0.003 -0.543 & 0.8 -0.304 & 0.819 -0.054 & 0.412 -0.043 & 0.372 0. & 0. end{array} right) Also, classify the polygon as 'Simple' or 'Convex'.
answer:The polygon is classified as Convex. The estimated perimeter is 2.87. The interior angles (in radians) are estimated as follows: {2.26, 1.6, 1.45, 2.99, 2.86, 2.05, 2.51}. The estimated area of the polygon is 0.54 square units.
question:In the proof of the division algorithm for polynomials with integer coefficients, why is the restriction of being "monic" imposed on the divisor element? Where does the division algorithm fail if the above restriction is removed?
answer:The divisor has to be monic because otherwise the result is not necessarily true. It is easy to find two polynomials f, g in Z[x] such that g | f in C[x] but whose quotient f/g is not in Z[x]. For instance, as @ancientmathematician mentions in a comment, f = 1 and g = 2. The normal division algorithm cancels the highest term of f by subtracting a suitable multiple of g (and then repeating this until the degree of f drops below the degree of g). This suitable multiple is (lc(f) / lc(g)) g. If lc(g) = 1 (or -1), this is guaranteed to be in Z[x].
question:Find the Jacobian matrix of the vector-valued function mathbf{r}(x, y, z) = langle f(x, y, z), g(x, y, z), h(x, y, z) rangle, where f(x, y, z) = cos^{-1}(y + z), g(x, y, z) = ln(y), and h(x, y, z) = e^{xz}.
answer:The Jacobian matrix of mathbf{r}(x, y, z) is given by: J(mathbf{r}(x, y, z)) = begin{bmatrix} frac{partial f}{partial x} & frac{partial f}{partial y} & frac{partial f}{partial z} frac{partial g}{partial x} & frac{partial g}{partial y} & frac{partial g}{partial z} frac{partial h}{partial x} & frac{partial h}{partial y} & frac{partial h}{partial z} end{bmatrix} Calculating each partial derivative, we get: frac{partial f}{partial x} = 0, quad frac{partial f}{partial y} = -frac{1}{sqrt{1-(y+z)^2}}, quad frac{partial f}{partial z} = -frac{1}{sqrt{1-(y+z)^2}} frac{partial g}{partial x} = 0, quad frac{partial g}{partial y} = frac{1}{y}, quad frac{partial g}{partial z} = 0 frac{partial h}{partial x} = ze^{xz}, quad frac{partial h}{partial y} = 0, quad frac{partial h}{partial z} = xe^{xz} Therefore, the Jacobian matrix of mathbf{r}(x, y, z) is: J(mathbf{r}(x, y, z)) = begin{bmatrix} 0 & -frac{1}{sqrt{1-(y+z)^2}} & -frac{1}{sqrt{1-(y+z)^2}} 0 & frac{1}{y} & 0 ze^{xz} & 0 & xe^{xz} end{bmatrix}