A steel ball is traveling through water with a speed of x meters per second, where x is positive. The drag force, F, in newtons (N) is: F=0.5+0.004(x+50)(x-2 At what speed in meters per second does the ball have a force of 0.5N on it?

Given Equation: We are given the drag force equation F = 0.5 + 0.004(x+50)(x-2) and we need to find the value of x when F = 0.5N. First, we set the force F to 0.5N in the equation and solve for x. 0.5 = 0.5 + 0.004(x+50)(x-2)Set Force to 0.5N: Subtract 0.5 from both sides of the equation to isolate the term with x. 0 = 0.004(x+50)(x-2)Subtract 0.5: Divide both sides of the equation by 0.004 to simplify the equation. 0 = (x+50)(x-2)Divide by 0.004: Now we have a quadratic equation in the form of (x+50)(x-2) = 0. We can find the values of x by setting each factor equal to zero. x+50 = 0 or x-2 = 0Quadratic Equation: Solve for x in both equations. x = -50 or x = 2Solve for x: Since x represents a speed and speed cannot be negative, we discard the negative value. Therefore, the speed at which the ball has a force of 0.5N on it is x = 2 meters per second.

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A steel ball is traveling through water with a speed of
x meters per second, where
x is positive. The drag force,
F, in newtons (N) is:

F=0.5+0.004(x+50)(x-2
At what speed in meters per second does the ball have a force of
0.5N on it?

A steel ball is traveling through water with a speed of $x$ meters per second, where $x$ is positive. The drag force, $F$ , in newtons $(\mathrm{N})$ is: $\newline$ $F=0.5+0.004(x+50)(x-2$ $\newline$ At what speed in meters per second does the ball have a force of $0.5 \mathrm{~N}$ on it?

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Q. A steel ball is traveling through water with a speed of

x

meters per second, where

x

is positive. The drag force,

F

, in newtons

(\mathrm{N})

is:

\newline

F=0.5+0.004(x+50)(x-2

\newline

At what speed in meters per second does the ball have a force of

0.5 \mathrm{~N}

on it?

Given Equation: We are given the drag force equation $F = 0.5 + 0.004(x+50)(x-2)$ and we need to find the value of $x$ when $F = 0.5N$ . First, we set the force $F$ to $0.5N$ in the equation and solve for $x$ . $0.5 = 0.5 + 0.004(x+50)(x-2)$
Set Force to $0.5\,\text{N}$ : Subtract $0.5$ from both sides of the equation to isolate the term with $x$ . $\newline$ $0 = 0.004(x+50)(x-2)$
Subtract $0.5$ : Divide both sides of the equation by $0.004$ to simplify the equation. $\newline$ $0 = (x+50)(x-2)$
Divide by $0.004$ : Now we have a quadratic equation in the form of $(x+50)(x-2) = 0$ . We can find the values of $x$ by setting each factor equal to zero. $\newline$ $x+50 = 0$ or $x-2 = 0$
Quadratic Equation: Solve for $x$ in both equations. $x = -50$ or $x = 2$
Solve for $x$ : Since $x$ represents a speed and speed cannot be negative, we discard the negative value. $\newline$ Therefore, the speed at which the ball has a force of $0.5\,\text{N}$ on it is $x = 2$ meters per second.