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Kamila [148]
3 years ago
10

A water wave has a speed of 23.0 meters/second. If the wave’s frequency is 0.0680 hertz, what is the wavelength?

Physics
2 answers:
Nitella [24]3 years ago
8 0

Answer: Option (B) is the correct answer.

Explanation:

Given data is as follows.

Frequency = 0.0680 hertz = 0.0680 per second

Speed or velocity = 23.0 m/s

We can calculate the wavelength with the help of formula as follows.

                            v = \lambda f

                            \lambda = \frac{v}{f}

                                    = \frac{23.0 m/s}{0.0680 s^{-1}}

                                    = 338.23 meters

or,                                 = 338 meters (approx)

Thus, we can conclude that wavelength is 338 meters (approx).

tankabanditka [31]3 years ago
3 0
     Using the Fundamental Equation of Wave, we have:

v=\lambda f \\ \lambda= \frac{23}{0.068}  \\ \boxed {\lambda \approx 338m}

Letter B
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A sled is moving down a steep hill. The mass of the sled is 50 kg and the net force acting on it is 20 N. What must be done to f
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You need to first measure the angle of descent, i.e. the angle the hill makes with the ground. Then identify the forces acting on the sled, split them up into horizontal and vertical components, or into components that are parallel and perpendicular to the hill, and use Newton's second law to determine the components of the sled's acceleration vector.

There are at least 2 forces acting on the sled:

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• the normal force, pointing perpendicular to the hill and away from the ground with mag. <em>N</em>

The question doesn't specify, but there might also be friction to consider, indicated in the attachment by the vector <em>F</em> pointing parallel to the slope of the hill and opposing the direction of the sled's motion with mag. <em>F</em>.

Splitting up the forces into parallel/perpendicular components is less work. By Newton's second law, the net force (denoted with ∑ or "sigma" here) in a particular direction is equal to the mass of the sled times its acceleration in that direction:

∑ (//) = <em>W</em> (//) = <em>m</em> <em>a</em> (//)

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If the hill makes an angle of <em>θ</em> with flat ground, then <em>W</em> makes the same angle with the hill so that

<em>W</em> (//) = -<em>m g </em>sin(<em>θ</em>)

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Then the acceleration vector is

<em>a</em> = <em>a</em> (//)

with magnitude

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