All categories

3 years ago

Wave is traveling at 60 cm/second and has a wavelength of 15 cm, what is the frequency?

Physics

1 answer:

emmasim [6.3K]3 years ago

7 0

The frequency of the wave is 4 Hz

Explanation:

The relationship between wave, frequency and speed of a wave is given by the equation:

$v=f \lambda$

where

v is the speed of the wave

f is the frequency

$\lambda$ is the wavelength

For the wave in this problem, we have:

$v=60 cm/s$ is the speed

$\lambda=15 cm$ is the wavelength

Solving the equation for f, we find its frequency:

$f=\frac{v}{\lambda}=\frac{60 cm/s}{15 s}=4 s^{-1} = 4 Hz$

Learn more about waves and frequency:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

You might be interested in

Which is the best example of the concept of heredity? A. A black cat blends into its environment at night. B. A baby has his mot

Hitman42 [59]

B) A baby has his mother's hair and eye color

4 0

3 years ago

Read 2 more answers

A man has a mass of 66 kg on earth. What’s his weight in pounds?

Vlad1618 [11]

Answer:

646.8 N

Explanation:

3 0

3 years ago

Which sentence is correct?

Alexeev081 [22]

Answer:

pretty sure its B if it isnt im so so sorry

Explanation:

7 0

3 years ago

The force exerted by the wind on the sails of a sailboat is Fsail = 330 N north. The water exerts a force of Fkeel = 210 N east.

Elena L [17]

Answer:

The magnitude of the acceleration is $a_r = 1.50 \ m/s^2$

The direction is $\theta = 32.5 6^o$ north of east

Explanation:

From the question we are told that

The force exerted by the wind is $F_{sail} = (330 ) \ N \ north$

The force exerted by water is $F_{keel} = (210 ) \ N \ east$

The mass of the boat(+ crew) is $m_b = 260 \ kg$

Now Force is mathematically represented as

$F = ma$

Now the acceleration towards the north is mathematically represented as

$a_n = \frac{F_{sail}}{m_b}$

substituting values

$a_n = \frac{330 }{260}$

$a_n = 1.269 \ m/s^2$

Now the acceleration towards the east is mathematically represented as

$a_e = \frac{F_{keel}}{m_b }$

substituting values

$a_e = \frac{210}{260}$

$a_e =0.808 \ m/s^2$

The resultant acceleration is

$a_r = \sqrt{a_e^2 + a_n^2}$

substituting values

$a_r = \sqrt{(0.808)^2 + (1.269)^2}$

$a_r = 1.50 \ m/s^2$

The direction with reference from the north is evaluated as

Apply SOHCAHTOA

$tan \theta = \frac{a_e}{a_n}$

$\theta = tan ^{-1} [\frac{a_e}{a_n } ]$

substituting values

$\theta = tan ^{-1} [\frac{0.808}{1.269 } ]$

$\theta = tan ^{-1} [0.636 ]$

$\theta = 32.5 6^o$

5 0

3 years ago

A 40 kg boy on a skateboard throws a 2 kg ball at 20 m/s to the left. How fast did the boy move to the right on his skateboard?

sergey [27]

According to my examination I have confirmed that I do NOT repeat do NOT know this .

4 0

3 years ago