/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 43 A cylindrical tube \(9.5 \mathrm... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 43

A cylindrical tube \(9.5 \mathrm{~cm}\) high and \(0.85 \mathrm{~cm}\) in diameter is used to collect blood samples. How many cubic decimeters (dm \(^{3}\) ) of blood can it hold ( \(V\) of a cylinder \(=\pi r^{2} h\) )?

Short Answer

Expert verified
0.0036 cubic decimeters

Step by step solution

01

- Convert Diameter to Radius

The diameter of the cylindrical tube is 0.85 cm. The radius (r) is half of the diameter. Therefore, calculate the radius as follows:\(r = \frac{0.85 \, \text{cm}}{2} = 0.425 \, \text{cm}\)
02

- Calculate the Volume of the Cylinder in Cubic Centimeters

The formula to calculate the volume \(V\) of a cylinder is given by \(V = \pi r^{2} h\). Substitute the values of r and h:\(V = \pi (0.425 \, \text{cm})^{2} (9.5 \, \text{cm})\)First, calculate \(r^2\):\(0.425^{2} = 0.180625 \, \text{cm}^{2}\)Then calculate the entire volume:\(V = \pi (0.180625) (9.5) \approx 3.6 \, \text{cm}^{3}\)
03

- Convert Cubic Centimeters to Cubic Decimeters

Since there are 1000 cubic centimeters in a cubic decimeter, convert the volume from cubic centimeters to cubic decimeters:\(V = \frac{3.6 \, \text{cm}^{3}}{1000} = 0.0036 \, \text{dm}^{3}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cylindrical Volume Formula
To calculate the volume of a cylinder, we use the formula: \( V = \pi r^{2} h \). This formula tells us that the volume is determined by multiplying the area of the base (which is a circle) by the height of the cylinder.

Here's a simple breakdown:
  • \( \pi \footnote\) is a mathematical constant, approximately equal to 3.14159, representing the ratio of a circle's circumference to its diameter.
  • \( r \footnote\) is the radius of the cylinder's base.
  • \( r^2 \footnote\) is the radius squared, which gives the area of the base circle.
  • \( h \footnote\) is the height of the cylinder.
By substituting these values into our formula, we can easily determine the volume of any cylindrical object.

This approach is crucial in many practical applications, from calculating how much liquid a container can hold to determining material usage for manufacturing. Knowing this formula helps solve many real-world problems involving cylindrical shapes.
Unit Conversion
Understanding how to convert between different units is essential for solving many math and science problems. In this exercise, we converted cubic centimeters (cm鲁) to cubic decimeters (dm鲁).

Why is this important? Because units must be consistent in calculations.

Steps for conversion in our example:
  • The cylinder's volume was first calculated in cubic centimeters.
  • Since 1 cubic decimeter (dm鲁) is equivalent to 1000 cubic centimeters (cm鲁), we divide the volume in cubic centimeters by 1000 to get the volume in cubic decimeters.
So, our final volume in cubic decimeters is \( \frac{3.6 \text{cm}^3}{1000} = 0.0036 \text{dm}^3 \footnote\).

By mastering unit conversion, students can handle diverse problems more effectively, ensuring that the units they work with are consistent across different contexts.
Radius and Diameter in Geometry
In geometry, understanding the difference between radius and diameter is fundamental.

Here's a quick rundown:
  • The diameter (\footnote\text{d\footnote\}) of a circle is the distance across the circle, passing through its center.
  • The radius (\footnote\text{r\footnote\}) is half the diameter, extending from the center of the circle to any point on its perimeter.
  • Mathematically, this relationship is expressed as r = \(\frac{\text{d}}{2} \footnote\).
In our exercise, we convert the diameter of the cylinder's base to its radius:

The given diameter is 0.85 cm, so the radius is \(r = \frac{0.85 \text{cm}}{2} = 0.425 \text{cm} \footnote\).

Knowing how to switch between these two measurements helps in applying the cylindrical volume formula accurately.

Grasping these fundamental concepts in geometry not only simplifies volume calculations but also aids in understanding the properties and relationships of different shapes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Asbestos is a fibrous silicate mineral with remarkably high tensile strength. But it is no longer used because airborne asbestos particles can cause lung cancer. Grunerite, a type of asbestos, has a tensile strength of \(3.5 \times 10^{2} \mathrm{~kg} / \mathrm{mm}^{2}\) (thus, a strand of grunerite with a \(1-\mathrm{mm}^{2}\) cross-sectional area can hold up to \(3.5 \times 10^{2} \mathrm{~kg}\) ). The tensile strengths of aluminum and Steel No. 5137 are \(2.5 \times 10^{4} \mathrm{lb} / \mathrm{in}^{2}\) and \(5.0 \times 10^{4} \mathrm{lb} / \mathrm{in}^{2},\) respectively. Calculate the cross-sectional areas (in \(\mathrm{mm}^{2}\) ) of wires of aluminum and of Steel No. 5137 that have the same tensile strength as a fiber of grunerite with a crosssectional area of \(1.0 \mu \mathrm{m}^{2}\).

An empty Erlenmeyer flask weighs 241.3 g. When filled with water \(\left(d=1.00 \mathrm{~g} / \mathrm{cm}^{3}\right),\) the flask and its contents weigh \(489.1 \mathrm{~g} .\) (a) What is the volume of water in the flask? (b) How much does the flask weigh when filled with the same volume of chloroform \(\left(d=1.48 \mathrm{~g} / \mathrm{cm}^{3}\right) ?\)

Liquid nitrogen is obtained from liquefied air and is used industrially to prepare frozen foods. It boils at \(77.36 \mathrm{~K}\). (a) What is this temperature in \({ }^{\circ} \mathrm{C} ?\) (b) What is this temperature in \({ }^{\circ} \mathrm{F}\) ? (c) At the boiling point, the density of the liquid is \(809 \mathrm{~g} / \mathrm{L}\) and that of the gas is \(4.566 \mathrm{~g} / \mathrm{L}\). How many liters of liquid nitrogen are produced when \(895.0 \mathrm{~L}\) of nitrogen gas is liquefied at \(77.36 \mathrm{~K} ?\)

At room temperature \(\left(20^{\circ} \mathrm{C}\right)\) and pressure, the density of air is \(1.189 \mathrm{~g} / \mathrm{L}\). An object will float in air if its density is less than that of air. In a buoyancy experiment with a new plastic, a chemist creates a rigid, thin-walled ball that weighs \(0.12 \mathrm{~g}\) and has a volume of \(560 \mathrm{~cm}^{3}\). (a) Will the ball float if it is evacuated? (b) Will it float if filled with carbon dioxide \((d=1.830 \mathrm{~g} / \mathrm{L}) ?\) (c) Will it float if filled with hydrogen \((d=0.0899 \mathrm{~g} / \mathrm{L}) ?\) (d) Will it float if filled with oxygen \((d=1.330 \mathrm{~g} / \mathrm{L}) ?\) (e) Will it float if filled with nitrogen \((d=1.165 \mathrm{~g} / \mathrm{L}) ?\) (f) For any case in which the ball will float, how much weight must be added to make it sink?

he radius of a barium atom is \(2.22 \times 10^{-10} \mathrm{~m} .\) What is its radius in angstroms (脌)?
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Making Measurements

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.