Physics, 04.07.2019 23:00 seider7997

Four 8.5 kg spheres are located at the corners of a square of side 0.52 m. calculate the magnitude and direction of the gravitational force exerted on one sphere by the other three. magnitude

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3.4x10^-8 N of force directly towards the sphere in the opposite corner of the square. The gravitational attraction of two masses towards each other is F = G(m1m2/r^2) where G = gravitational constant m1,m2 = masses r = distance between the centers of the masses. So the force being exerted by the masses alone an edge of the square would be F = 6.674x10^-11 N(m/kg)^2 * ((8.5 kg)^2)/((0.52m)^2) F = 6.674x10^-11 N(m/kg)^2 * (72.25 kg^2)/(0.2704 m^2) F = 6.674x10^-11 N(m/kg)^2 * 267.1967 (kg/m)^2 F = 1.78327x10^-8 N There are 2 masses that affect a mass on a corner so that the sum of their vectors will result in a 3rd vector aiming towards the mass in the diagonal corner. So sqrt(2(1.78327x10^-8 N)^2) = 2.52193x10^-8 N The mass in the diagonal corner will also be attracting. The distance to that mass is sqrt(2*(0.52m)^2) = 0.735391052 m F = 6.674x10^-11 N(m/kg)^2 * ((8.5 kg)^2)/(2*(0.52m)^2) F = 6.674x10^-11 N(m/kg)^2 * (72.25 kg^2)/(0.5408 m^2) F = 6.674x10^-11 N(m/kg)^2 * 133.5984 (kg/m)^2 F = 8.916355x10^-9 N This vector will be along the same line as the combined vector from the other 2 masses, so they'll add directly. F = 8.916355x10^-9 N + 2.52193x10^-8 N = 3.4136x10^-8 N Since we have 2 significant figures in our data, the result rounded to 2 significant figures is 3.4x10^-8 N

Answer from: Quest

cant without the picture! ?

Answer from: Quest

we have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is $[2,\infty)$ .

if we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

let us know find its inverse.

$y=(x-5)^{2}+2$

upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

let us now solve this function for y.

$(y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\$

hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .

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Four 8.5 kg spheres are located at the corners of a square of side 0.52 m. calculate the magnitude a...