practice
1. if the temperature of the air in your region is 32 °c, what is the speed of sound at that temperature? ans: 351 m/s
2. if the speed of sound near you is 333 m/s, what is the ambient temperature? ans: 2.64 °c
3. if the speed of sound near you is 350 m/s, what is the ambient temperature? ans: 31 °c

practice
1. if the local speed of sound is 344 m/s and an aircraft is flying at 910 km/h, what is the mach number? ans: 0.73
2. if the mach number is 0.93 and the local speed of sound is 320 m/s, what is the speed of an airplane in these conditions? ans: 3.0 × 10² m/s = 1100 km/h
3. if the mach number is 0.81 and the speed of an airplane measured by radar is 850 km/h, what is the local speed of sound in kilometres per hour? ans: 290 m/s = 1.0 × 10³ km/h

Question

practice
1. if the temperature of the air in your region is 32 °c, what is the speed of sound at that temperature? ans: 351 m/s
2. if the speed of sound near you is 333 m/s, what is the ambient temperature? ans: 2.64 °c
3. if the speed of sound near you is 350 m/s, what is the ambient temperature? ans: 31 °c

practice
1. if the local speed of sound is 344 m/s and an aircraft is flying at 910 km/h, what is the mach number? ans: 0.73
2. if the mach number is 0.93 and the local speed of sound is 320 m/s, what is the speed of an airplane in these conditions? ans: 3.0 × 10² m/s = 1100 km/h
3. if the mach number is 0.81 and the speed of an airplane measured by radar is 850 km/h, what is the local speed of sound in kilometres per hour? ans: 290 m/s = 1.0 × 10³ km/h

Accepted Answer

### Problem 1 (Top Practice, Question 1)
# Explanation:
## Step1: Identify the formula
The formula for the speed of sound $ v $ is $ v = 331.4 \, \text{m/s} + (0.606 \, \text{m/s}^\circ\text{C}) \times T $, where $ T $ is the temperature in $ ^\circ\text{C} $.
## Step2: Substitute $ T = 32^\circ\text{C} $
Substitute $ T = 32 $ into the formula: $ v = 331.4 + (0.606 \times 32) $.
## Step3: Calculate the product
First, calculate $ 0.606 \times 32 = 19.392 $.
## Step4: Add to the initial speed
Then, add this to 331.4: $ v = 331.4 + 19.392 = 350.792 \approx 351 \, \text{m/s} $.
# Answer:
The speed of sound is $ \boldsymbol{351 \, \text{m/s}} $.

### Problem 2 (Top Practice, Question 2)
# Explanation:
## Step1: Start with the formula
The formula is $ v = 331.4 + 0.606T $. We need to solve for $ T $, so rearrange it: $ T = \frac{v - 331.4}{0.606} $.
## Step2: Substitute $ v = 333 \, \text{m/s} $
Substitute $ v = 333 $ into the rearranged formula: $ T = \frac{333 - 331.4}{0.606} $.
## Step3: Calculate the numerator
$ 333 - 331.4 = 1.6 $.
## Step4: Divide to find $ T $
$ T = \frac{1.6}{0.606} \approx 2.64^\circ\text{C} $.
# Answer:
The ambient temperature is $ \boldsymbol{2.64^\circ\text{C}} $.

### Problem 3 (Top Practice, Question 3)
# Explanation:
## Step1: Use the formula $ v = 331.4 + 0.606T $
We need to solve for $ T $, so rearrange: $ T = \frac{v - 331.4}{0.606} $.
## Step2: Substitute $ v = 350 \, \text{m/s} $
Substitute $ v = 350 $: $ T = \frac{350 - 331.4}{0.606} $.
## Step3: Calculate the numerator
$ 350 - 331.4 = 18.6 $.
## Step4: Divide to find $ T $
$ T = \frac{18.6}{0.606} \approx 30.7 \approx 31^\circ\text{C} $.
# Answer:
The ambient temperature is $ \boldsymbol{31^\circ\text{C}} $.

### Problem 4 (Bottom Practice, Question 1)
# Explanation:
## Step1: Recall Mach number formula
Mach number $ M = \frac{v_{\text{aircraft}}}{v_{\text{sound}}} $. First, convert aircraft speed to m/s: $ 910 \, \text{km/h} = \frac{910 \times 1000}{3600} \approx 252.78 \, \text{m/s} $.
## Step2: Substitute values
$ M = \frac{252.78}{344} \approx 0.73 $.
# Answer:
The Mach number is $ \boldsymbol{0.73} $.

### Problem 5 (Bottom Practice, Question 2)
# Explanation:
## Step1: Use Mach number formula $ M = \frac{v_{\text{airplane}}}{v_{\text{sound}}} $
Rearrange to $ v_{\text{airplane}} = M \times v_{\text{sound}} $.
## Step2: Substitute $ M = 0.93 $ and $ v_{\text{sound}} = 320 \, \text{m/s} $
$ v_{\text{airplane}} = 0.93 \times 320 = 297.6 \approx 3.0 \times 10^2 \, \text{m/s} $. Convert to km/h: $ 297.6 \times \frac{3600}{1000} \approx 1071.36 \approx 1100 \, \text{km/h} $.
# Answer:
The speed of the airplane is $ \boldsymbol{3.0 \times 10^2 \, \text{m/s} (or 1100 \, \text{km/h})} $.

### Problem 6 (Bottom Practice, Question 3)
# Explanation:
## Step1: Use Mach number formula $ M = \frac{v_{\text{airplane}}}{v_{\text{sound}}} $
Rearrange to $ v_{\text{sound}} = \frac{v_{\text{airplane}}}{M} $. First, use the speed in km/h: $ v_{\text{airplane}} = 850 \, \text{km/h} $, $ M = 0.81 $.
## Step2: Calculate $ v_{\text{sound}} $ in km/h
$ v_{\text{sound}} = \frac{850}{0.81} \approx 1049.38 \approx 1.0 \times 10^3 \, \text{km/h} $. Convert to m/s: $ \frac{1049.38 \times 1000}{3600} \approx 291.49 \approx 290 \, \text{m/s} $.
# Answer:
The local speed of sound is $ \boldsymbol{290 \, \text{m/s} (or 1.0 \times 10^3 \, \text{km/h})} $.

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QUESTION IMAGE

Question

Explanation:

Problem 1 (Top Practice, Question 1)

Step1: Identify the formula

Step2: Substitute \( T = 32^\circ\text{C} \)

Step3: Calculate the product

Step4: Add to the initial speed

Step1: Start with the formula

Step2: Substitute \( v = 333 \, \text{m/s} \)

Step3: Calculate the numerator

Step4: Divide to find \( T \)

Step1: Use the formula \( v = 331.4 + 0.606T \)

Step2: Substitute \( v = 350 \, \text{m/s} \)

Step3: Calculate the numerator

Step4: Divide to find \( T \)

Answer:

Problem 2 (Top Practice, Question 2)