Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration

Q8PE

Expert-verified
College Physics (Urone)
Found in: Page 33

Answers without the blur.

Just sign up for free and you're in.

Illustration

Short Answer

The speed of sound is measured to be \({\rm{342 }}{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}\) on a certain day. What is this in km/h?

Hence, the speed of sound is \(1231.2{\rm{ }}{{km} \mathord{\left/ {\vphantom {{km} {hr}}} \right. \\} {hr}}\).

See the step by step solution

Step by Step Solution

Step 1:

Meter is a SI unit of distance. It is abbreviated as a meter.

The relation between kilometers and meters is given as

\(1{\rm{ }}kilometer = 1000{\rm{ }}meters\)

Given data

Consider the given data as below.

The speed of sound =\(342{\rm{ }}{m \mathord{\left/ {\vphantom {m s}} \right. \\} s}\)

Step 2: Conversion of units

The value of speed of sound in km/h can be expressed through conversion.

Converting 1 sec to 1 hour

\({1 \mathord{\left/ {\vphantom {1 {3600}}} \right.\\} {3600}} = 0.000277778{\rm{ }}hours\)

Therefore, we can imply 1m/sec

\(\begin{aligned}{}1{\rm{ }}{m \mathord{\left/ {\vphantom {m s}} \right. \\} s} = {\raise0.7ex\hbox{${{{10}^{ - 3}}{\rm{ }}km}$} \!\mathord{\left/ {\vphantom {{{{10}^{ - 3}}{\rm{ }}km} {0.00027778{\rm{ }}hour}}}\right.\\}\!\lower0.7ex\hbox{${0.00027778{\rm{ }}hour}$}}\\ = 599.99\times {10^{ - 3}}{\rm{ }}{{km} \mathord{\left/ {\vphantom {{km} {hr}}} \right. \\} {hr}}\end{aligned}\)\(342{\rm{ }}{m \mathord{\left/ {\vphantom {m s}} \right. \\} s} = 3599.99 \times {10^{ - 3}}{\rm{ }}{{km} \mathord{\left/ {\vphantom{{km} {hr}}} \right. \\} {hr}}\)

Therefore,\(1{\rm{ }}{m \mathord{\left/ {\vphantom {m s}} \right. \\} s} = 3.6{\rm{ }}{{km} \mathord{\left/ {\vphantom {{km} {hr}}} \right. \\} {hr}}\)

So, the velocity will be,

\(3.6{\rm{ }}x{\rm{ }}342 = 1231.2{\rm{ }}{{km} \mathord{\left/ {\vphantom {{km} {hr}}} \right. \\} {hr}}\)

Step 3: Conclusion:

Hence, the speed of sound in \({{km} \mathord{\left/ {\vphantom {{km} {hr}}} \right. \\} {hr}}\) is \(1231.2{\rm{ }}{{km} \mathord{\left/ {\vphantom {{km} {hr}}} \right. \\} {hr}}\).

Recommended explanations on Physics Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.