Question
A buoy bobs up and down in the ocean. The waves have a wavelength of 2.5 meters, and they pass the buoy at a speed of 4.0 meters/second. What is the frequency of the waves?
1.6 Hertz.
10 Hertz.
7.5 Hertz.
4 Hertz.
1.6 Hertz.
10 Hertz.
7.5 Hertz.
4 Hertz.
Answers
Answered by
GPT-4o mini
To find the frequency of the waves, we can use the formula:
\[
\text{Frequency} (f) = \frac{\text{Speed} (v)}{\text{Wavelength} (\lambda)}
\]
Given:
- Speed (\(v\)) = 4.0 meters/second
- Wavelength (\(\lambda\)) = 2.5 meters
Substituting the values into the formula:
\[
f = \frac{4.0 \, \text{m/s}}{2.5 \, \text{m}} = 1.6 \, \text{Hertz}
\]
Thus, the frequency of the waves is **1.6 Hertz**.
\[
\text{Frequency} (f) = \frac{\text{Speed} (v)}{\text{Wavelength} (\lambda)}
\]
Given:
- Speed (\(v\)) = 4.0 meters/second
- Wavelength (\(\lambda\)) = 2.5 meters
Substituting the values into the formula:
\[
f = \frac{4.0 \, \text{m/s}}{2.5 \, \text{m}} = 1.6 \, \text{Hertz}
\]
Thus, the frequency of the waves is **1.6 Hertz**.
Answer
Are you sure?
Answered by
GPT-4o mini
Let's double-check the calculation step by step.
We have:
- Wave speed (\(v\)) = 4.0 m/s
- Wavelength (\(\lambda\)) = 2.5 m
Using the formula for frequency:
\[
f = \frac{v}{\lambda}
\]
Let's substitute the values in:
\[
f = \frac{4.0 \, \text{m/s}}{2.5 \, \text{m}}
\]
Calculating that gives:
\[
f = \frac{4.0}{2.5} = 1.6 \, \text{Hertz}
\]
So, I can confirm that the frequency is indeed **1.6 Hertz**. Thank you for your patience!
We have:
- Wave speed (\(v\)) = 4.0 m/s
- Wavelength (\(\lambda\)) = 2.5 m
Using the formula for frequency:
\[
f = \frac{v}{\lambda}
\]
Let's substitute the values in:
\[
f = \frac{4.0 \, \text{m/s}}{2.5 \, \text{m}}
\]
Calculating that gives:
\[
f = \frac{4.0}{2.5} = 1.6 \, \text{Hertz}
\]
So, I can confirm that the frequency is indeed **1.6 Hertz**. Thank you for your patience!
Answer
You are at a pool party and notice that 2 waves pass by a floating ball in 4 seconds. If the distance between each wave crest is 2 meters, what is the speed of the waves?
1 m/s.
6 m/s.
2 m/s.
1.5 m/s.
1 m/s.
6 m/s.
2 m/s.
1.5 m/s.
Answered by
GPT-4o mini
To find the speed of the waves, we can use the formula:
\[
\text{Speed} (v) = \text{Frequency} (f) \times \text{Wavelength} (\lambda)
\]
First, we need to determine the frequency of the waves.
**Step 1: Calculate the frequency.**
You observe that 2 waves pass by in 4 seconds, so the frequency \(f\) can be calculated as follows:
\[
f = \frac{\text{Number of waves}}{\text{Time (seconds)}}
\]
\[
f = \frac{2 \, \text{waves}}{4 \, \text{seconds}} = 0.5 \, \text{Hertz}
\]
**Step 2: Use the wavelength.**
We are given that the distance between each wave crest (wavelength) is:
\[
\lambda = 2 \, \text{meters}
\]
**Step 3: Calculate the speed.**
Now we can calculate the speed of the waves:
\[
v = f \times \lambda
\]
\[
v = 0.5 \, \text{Hz} \times 2 \, \text{m} = 1 \, \text{m/s}
\]
Thus, the speed of the waves is **1 m/s**.
\[
\text{Speed} (v) = \text{Frequency} (f) \times \text{Wavelength} (\lambda)
\]
First, we need to determine the frequency of the waves.
**Step 1: Calculate the frequency.**
You observe that 2 waves pass by in 4 seconds, so the frequency \(f\) can be calculated as follows:
\[
f = \frac{\text{Number of waves}}{\text{Time (seconds)}}
\]
\[
f = \frac{2 \, \text{waves}}{4 \, \text{seconds}} = 0.5 \, \text{Hertz}
\]
**Step 2: Use the wavelength.**
We are given that the distance between each wave crest (wavelength) is:
\[
\lambda = 2 \, \text{meters}
\]
**Step 3: Calculate the speed.**
Now we can calculate the speed of the waves:
\[
v = f \times \lambda
\]
\[
v = 0.5 \, \text{Hz} \times 2 \, \text{m} = 1 \, \text{m/s}
\]
Thus, the speed of the waves is **1 m/s**.
Answer
A student moves the end of a compressed coiled spring up and down to demonstrate wave motion. She then moves the compressed coiled spring faster but keeps everything else the same. Which characteristic will increase?
Answered by
GPT-4o mini
When the student moves the end of the compressed coiled spring faster while keeping everything else the same, the frequency of the wave will increase.
**Explanation:**
- **Frequency** is the number of waves that pass a point in a given amount of time. If the student moves the spring faster, more waves are created in the same period, resulting in a higher frequency.
- **Wavelength** will typically decrease as frequency increases (for a given wave speed), maintaining the wave speed constant.
Therefore, the characteristic that will increase is the **frequency** of the waves produced in the spring.
**Explanation:**
- **Frequency** is the number of waves that pass a point in a given amount of time. If the student moves the spring faster, more waves are created in the same period, resulting in a higher frequency.
- **Wavelength** will typically decrease as frequency increases (for a given wave speed), maintaining the wave speed constant.
Therefore, the characteristic that will increase is the **frequency** of the waves produced in the spring.