Respan Dataset Explorer

Five pulses are generated every 0.116 s in a tank of water. If the wavelength of the surface wave is 1.8 m, the speed of the wave is _______ m/s.

First, to find the frequency of the wave, we need to know how many pulses are generated per unit time. If five pulses are generated every 0.116 seconds, we can calculate the frequency of the pulses (which is equivalent to the frequency of the wave) using the formula:

\[ f = \frac{\text{number of pulses}}{\text{time}} \]

So in this case:

\[ f = \frac{5 \text{ pulses}}{0.116 \text{ s}} \]

Calculating this gives:

\[ f \approx 43.1034 \text{ Hz} \]

Now that we have the frequency, we can calculate the speed of the wave using the relationship between speed (\( v \)), frequency (\( f \)), and wavelength (\( \lambda \)):

\[ v = f \times \lambda \]

Given the wavelength (\( \lambda \)) is 1.8 meters:

\[ v = 43.1034 \text{ Hz} \times 1.8 \text{ m} \]

Calculating the speed:

\[ v \approx 77.5861 \text{ m/s} \]

Therefore, the speed of the wave is approximately 77.59 meters per second (rounded to two decimal places).

All light waves travel at the same speed, which is 3.00 x 108 m/s. If a light wave has a wavelength of 492 nanometers, the frequency of this light wave is ____ Hz.

To find the frequency of a light wave, you can use the equation that relates the speed of light (\( c \)), wavelength (\( \lambda \)), and frequency (\( f \)):

\[ c = f \times \lambda \]

First, convert the wavelength from nanometers to meters since the speed of light is given in meters per second and we want the frequency in hertz. There are \( 1 \times 10^9 \) nanometers in a meter, so:

\[ \lambda = 492 \text{ nm} = 492 \times 10^{-9} \text{ m} \]

We have the speed of light (\( c \)):

\[ c = 3.00 \times 10^8 \text{ m/s} \]

Now, solve for frequency (\( f \)):

\[ f = \frac{c}{\lambda} \]

\[ f = \frac{3.00 \times 10^8 \text{ m/s}}{492 \times 10^{-9} \text{ m}} \]

\[ f = \frac{3.00 \times 10^8}{492 \times 10^{-9}} \]

\[ f = \frac{3.00}{492} \times 10^{8 + 9} \]

\[ f = \frac{3.00}{492} \times 10^{17} \]

Reducing the fraction we get:

\[ f = \frac{300}{492} \times 10^{17} \]

\[ f = \frac{25}{41} \times 10^{17} \]

\[ f = 0.60975609756 \times 10^{17} \]

\[ f \approx 6.0975609756 \times 10^{14} \] Hz

Rounded to a reasonable number of significant figures, the frequency is approximately:

\[ f \approx 6.10 \times 10^{14} \] Hz

Therefore, the frequency of a light wave with a wavelength of 492 nanometers is approximately \( 6.10 \times 10^{14} \) hertz.

A wave has a frequency of 98 Hz. The period of the wave is ____ seconds.

The period (T) of a wave is the reciprocal of the frequency (f). It can be calculated using the formula:

\[ T = \frac{1}{f} \]

For a wave with a frequency of 98 Hz:

\[ T = \frac{1}{98} \]

Performing the calculation:

\[ T \approx 0.01020408 \text{ seconds} \]

To several decimal places, the period is 0.01020408 seconds. You can round this to an appropriate number of significant figures depending on the desired precision; typically, the period might be given as 0.0102 seconds when rounded to four significant figures.

Therefore, the period of the wave is approximately 0.0102 seconds.

A wave has a period of 10.8 seconds. The frequency of the wave is ___ Hz.

The frequency (f) of a wave is the reciprocal of its period (T). The formula to find the frequency is:

\[ f = \frac{1}{T} \]

Given that the period T is 10.8 seconds:

\[ f = \frac{1}{10.8 \text{ s}} \]

\[ f \approx 0.09259259 \text{ Hz} \]

This value can be rounded to an appropriate number of significant figures, depending on the context. If we round it to three significant figures, it would be approximately:

\[ f \approx 0.0926 \text{ Hz} \]

Therefore, the frequency of the wave is approximately 0.0926 Hz.

A wave is observed to have a frequency of 49 Hz and a wavelength of 0.8 meters. The speed of the wave is ____ m/s.

The speed of a wave (v) is the product of its frequency (f) and wavelength (λ). The formula is:

\[ v = f \times \lambda \]

Given the frequency \( f = 49 \) Hz and the wavelength \( \lambda = 0.8 \) meters, we can calculate the speed of the wave as follows:

\[ v = 49 \text{ Hz} \times 0.8 \text{ m} \]
\[ v = 39.2 \text{ m/s} \]

So the speed of the wave is 39.2 meters per second.

A wave is observed to have a period of 31 seconds and a wavelength of 0.69 meters. The speed of the wave is ____ m/s.

To find the speed of a wave, you can use the relationship between its frequency, wavelength, and speed. The speed (v) of a wave is the product of its frequency (f) and its wavelength (λ):

\[ v = f \times \lambda \]

The frequency of a wave is the reciprocal of its period (T):

\[ f = \frac{1}{T} \]

Given that the period T is 31 seconds, the frequency f is:

\[ f = \frac{1}{31 \text{ s}} \]

Now, multiply the frequency by the wavelength (λ = 0.69 meters) to find the speed:

\[ v = f \times \lambda = \frac{1}{31} \times 0.69 \]

Calculate the speed:

\[ v \approx \frac{0.69}{31} \]
\[ v \approx 0.0222580645 \text{ m/s} \]

Rounded to more practical decimal places, the speed of the wave is approximately:

\[ v \approx 0.0223 \text{ m/s} \]

Therefore, the speed of the wave is approximately 0.0223 meters per second.

All light waves travel at the same speed, which is 3.00 x 108 m/s. If a light wave has a wavelength of 580 nanometers, the frequency of this light wave is ____ Hz.

The speed of light (\( c \)) in a vacuum is \( 3.00 \times 10^8 \) meters per second (m/s). To find the frequency (\( f \)) of a light wave when given its wavelength (\( \lambda \)) in nanometers, we use the relationship:

\[ c = f \times \lambda \]

First, convert the wavelength from nanometers to meters, since the speed of light is in meters per second. There are \( 1 \times 10^9 \) nanometers in a meter, so the wavelength in meters is:

\[ \lambda = 580 \text{ nm} = 580 \times 10^{-9} \text{ m} \]

Now, using the speed of light, calculate the frequency:

\[ f = \frac{c}{\lambda} \]

\[ f = \frac{3.00 \times 10^8 \text{ m/s}}{580 \times 10^{-9} \text{ m}} \]

\[ f = \frac{3.00 \times 10^8}{580 \times 10^{-9}} \]

\[ f = \frac{3.00}{580} \times 10^{8 + 9} \]

\[ f = \frac{3.00}{580} \times 10^{17} \]

\[ f \approx 5.1724137931 \times 10^{14} \]

Rounded to a reasonable number of significant figures, the frequency of the light wave is approximately:

\[ f \approx 5.17 \times 10^{14} \] Hz

Therefore, the frequency of a light wave with a wavelength of 580 nanometers is approximately \( 5.17 \times 10^{14} \) hertz.

At standard temperature and pressure, sound waves travel through air at 343 m/s. If a sound wave has a frequency of 1,997 Hz, its wavelength must be _____ m.

To find the wavelength (\( \lambda \)) of a sound wave, use the formula that relates the wave's speed (\( v \)), frequency (\( f \)), and wavelength:

\[ v = f \times \lambda \]

First, solve the equation for wavelength:

\[ \lambda = \frac{v}{f} \]

Given that the speed of sound at standard temperature and pressure is \( v = 343 \) m/s, and the frequency of the sound wave is \( f = 1,997 \) Hz, plug in the values to calculate the wavelength:

\[ \lambda = \frac{343 \text{ m/s}}{1,997 \text{ Hz}} \]

\[ \lambda \approx \frac{343}{1,997} \]

\[ \lambda \approx 0.171757136 \text{ m} \]

Rounded to appropriate significant figures, the wavelength is:

\[ \lambda \approx 0.172 \text{ m} \]

Therefore, the wavelength of the sound wave with a frequency of 1,997 Hz is approximately 0.172 meters.

Sound waves travel through the air at 343 m/s at standard temperature and pressure. If a sound wave has a frequency of 8.23 kHz, the wavelength must be ____ m.

To calculate the wavelength of a sound wave, use the formula that relates the speed of sound (v), the frequency (f), and the wavelength (λ):

\[ v = f \times \lambda \]

To calculate the wavelength:

\[ \lambda = \frac{v}{f} \]

First, convert the frequency from kilohertz to hertz:

\[ f = 8.23 \text{ kHz} = 8.23 \times 10^3 \text{ Hz} \]

Given the speed of sound (v) is 343 m/s:

\[ \lambda = \frac{343 \text{ m/s}}{8.23 \times 10^3 \text{ Hz}} \]

\[ \lambda = \frac{343}{8.23 \times 10^3} \]

\[ \lambda = 0.04167631 \text{ m} \]

To a few significant figures, the wavelength is:

\[ \lambda \approx 0.0417 \text{ m} \]

Therefore, the wavelength of the sound wave with a frequency of 8.23 kHz is approximately 0.0417 meters.

At standard temperature and pressure, sound waves travel through air at 343 m/s. If a sound wave has a frequency of 1,426 Hz, its period must be _____ s.

The period (T) of a wave, which is the time for one complete cycle, is the reciprocal of its frequency (f). The formula to find the period is:

\[ T = \frac{1}{f} \]

Given that the frequency of the sound wave is \(f = 1,426\) Hz, we can calculate the period as follows:

\[ T = \frac{1}{1,426 \text{ Hz}} \]

Performing the calculation:

\[ T \approx 0.00070126 \text{ seconds} \]

To a reasonable number of significant digits, the period of the sound wave is approximately \(0.000701\) seconds (rounded to six significant digits).

Therefore, the period of the sound wave with a frequency of 1,426 Hz is approximately 0.000701 seconds.