Chapter 1.10, Problem 68E

(a)

To determine

To find:Thestretch in the spring.

Answer to Problem 68E

The stretch is $0.0513$ meters.

Explanation of Solution

The model has the form,

  $y=kx$ ...... (1)

Substitute $0.15$ for $y$ and $265$ for $x$ in equation (1).

  $\begin{array}{c}0.15=265k\\ k=\frac{0.15}{265}\\ \approx 0.000566\end{array}$

Substitute $0.000566$ for $k$ and $90$ for $x$ in equation (1).

  $\begin{array}{c}y=0.000566\left(90\right)\\ \approx 0.0513\end{array}$

(b)

To determine

To find:The value of $x$ .

Answer to Problem 68E

The value is $\frac{530}{3}$ .

Explanation of Solution

Substitute $0.15$ for $y$ and $265$ for $x$ in equation (1).

  $\begin{array}{c}0.15=265k\\ k=\frac{3}{5300}\end{array}$

Substitute $\frac{3}{5300}$ for $k$ in equation (1).

  $y=\frac{3}{5300}x$ ...... (2)

Substitute $0.1$ for $y$ in equation (2).

  $\begin{array}{c}0.1=\frac{3}{5300}x\\ x=0.1\left(\frac{5300}{3}\right)\\ =\frac{530}{3}\end{array}$