Q28P In Problems 17 to 30, for the cu... [FREE SOLUTION]

Chapter 5: Q28P (page 257)

In Problems 17 to 30, for the curve $y = \sqrt{x}$ , between $x = 0$ and $x = 2$ , find:
The mass of a wire bent in the shape of the arc if its density (mass per unit length) is.

Short Answer

Expert verified

The mass of a wire bent in the shape of the arc is $\frac{13}{6}$ .

Step by step solution

Definition of Mass

A physical body's mass is the amount of matter it contains. It is also a measure of the body's inertia, or resistance to acceleration (velocity change) when a net force is applied.

Determination of the Mass of the wire

Calculate the mass of the wire.

$M = 鈭� 位 d s = {鈭�}_{0}^{2} \sqrt{x} \sqrt{1 + y^{c 2}} d x = {鈭�}_{0}^{2} \sqrt{x} \sqrt{1 + \frac{1}{4 x}} d x = {鈭�}_{0}^{2} \sqrt{x + \frac{1}{4} d (x + \frac{1}{4})} = \frac{2}{3} {(x + \frac{1}{4})}^{3 / 2} = \frac{2}{3} [\frac{27}{8} - \frac{1}{8}] = \frac{13}{6}$

Thus, the mass of the bent wire is

\frac{13}{6}

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91影视

In Problems 17 to 30, for the curve $y = \sqrt{x}$ , between $x = 0$ and $x = 2$ , find:
The mass of a wire bent in the shape of the arc if its density (mass per unit length) is.

Short Answer

Step by step solution

Definition of Mass

Determination of the Mass of the wire

One App. One Place for Learning.

Most popular questions from this chapter

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91影视

In Problems 17 to 30, for the curve y=x, betweenx=0 andx=2 , find:The mass of a wire bent in the shape of the arc if its density (mass per unit length) is.

Short Answer

Step by step solution

Definition of Mass

Determination of the Mass of the wire

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Fields in Physics

Nuclear Physics

Radiation

Energy Physics

Space Physics

Turning Points in Physics

Study anywhere. Anytime. Across all devices.

In Problems 17 to 30, for the curve $y = \sqrt{x}$ , between $x = 0$ and $x = 2$ , find:
The mass of a wire bent in the shape of the arc if its density (mass per unit length) is.