# Need Thursday By 8:30pm Est Physics Math Problems.

**Physics and Math Assignment**

**Problem 1: Wavelength of Sound at the Upper Limit of Human Hearing**

The upper limit of human hearing is around 20,000 Hz. Calculate the corresponding wavelength at an air temperature of 30°C.

**Solution:**

To find the wavelength, we first need to determine the speed of sound in air at 30°C. The formula for the speed of sound in air is:

$v=331.3+0.6⋅T$

where $T$ is the air temperature in degrees Celsius. For $T=30°C$:

$v=331.3+0.6×30=349.3m/s$

The wavelength $λ$ is calculated using the wave equation:

$λ=fv $

where $v=349.3m/s$ is the speed of sound, and $f=20,000Hz$ is the frequency of the sound. Substituting the values:

$λ=, =0.017465m≈1.75cm$

**Question:** How does temperature affect the speed of sound, and what happens to the wavelength as temperature increases?

**Answer:** The corresponding wavelength of sound at 20,000 Hz is approximately **1.75 cm**.

**Problem 2: Distance to a Cliff from an Echo**

An echo is heard from a cliff 5.2 seconds after a rifle is fired. How far away is the cliff if the air temperature is 46°F?

**Solution:**

First, convert the air temperature from Fahrenheit to Celsius using the formula:

$T_{C}=95 (T_{F}−32)$

For $T_{F}=46°F$:

$T_{C}=95 (46−32)=95 ×14=7.78°C$

Now, calculate the speed of sound at 7.78°C:

$v=331.3+0.6×7.78=331.3+4.67=335.97m/s$

The time for the sound to travel to the cliff and back is 5.2 seconds, so the one-way time is:

$t=25.2 =2.6seconds$

Using the speed of sound, the distance to the cliff is:

$d=v×t=335.97×2.6=873.52m$

Convert the distance to feet (1 meter = 3.28084 feet):

$d=873.52×3.28084=2865.18feet$

**Question:** If the temperature were lower, would the distance calculated be greater or smaller? Why?

**Answer:** The distance to the cliff is approximately **2865 feet**.