In deep water, the speed s (in meters per second) of a series of waves and the wavelength L (in meters) of the waves are related by the equation 2(pi)s^2=9.8L.

a. Find the speed the the nearest hundredth of a meter per second of a series of waves with the following wavelengths: 6 meters, 10 meters and 25 meters. (Use 3.14 for pi)

b. Does the speed of a series of waves increase or decrease as the wavelength of the waves increases? Explain

turtle doveas

To find the speed of a series of waves, we need to solve the given equation: 2(pi)s^2 = 9.8L.

a. Let's substitute the given wavelengths into the equation and solve for speed.

For a wavelength of 6 meters:
2(pi)s^2 = 9.8 * 6
2(pi)s^2 = 58.8
s^2 = 58.8 / (2(pi))
s^2 ≈ 9.34
s ≈ √9.34
s ≈ 3.05

The speed of waves with a wavelength of 6 meters is approximately 3.05 meters per second.

For a wavelength of 10 meters:
2(pi)s^2 = 9.8 * 10
2(pi)s^2 = 98
s^2 = 98 / (2(pi))
s^2 ≈ 15.57
s ≈ √15.57
s ≈ 3.95

The speed of waves with a wavelength of 10 meters is approximately 3.95 meters per second.

For a wavelength of 25 meters:
2(pi)s^2 = 9.8 * 25
2(pi)s^2 = 245
s^2 = 245 / (2(pi))
s^2 ≈ 39.01
s ≈ √39.01
s ≈ 6.24

The speed of waves with a wavelength of 25 meters is approximately 6.24 meters per second.

b. From the calculations above, we can observe that as the wavelength of the waves increases, the speed of the series of waves also increases. This can be inferred from the equation 2(pi)s^2 = 9.8L. Since L is directly proportional to the wavelength, an increase in wavelength results in a larger value for L. To maintain the equation's validity, the speed, denoted as s, must increase as well.

What is your difficulty with this? You plug in the wavelengths for L, and compute s

s=sqrt(1/2pi * 9.8L)

obviously, from the equation, as L increases, s increases