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VikaD [51]
3 years ago
6

A water skier notices that the distance between successive wave crest is 12 meters. The wave crests pass his boat every 4.8 seco

nds. How fast are the waves traveling?a.456Hzb. 638Hzc. 368Hzd. 213Hz
Physics
1 answer:
Alenkinab [10]3 years ago
6 0

Answer:

The velocity of wave=2.5 m/s

Explanation:

We are given that

The distance between successive wave crest=\lambda=12m

Time =2.8 s

We have to find the the velocity of wave.

Frequency of wave=\nu=\frac{1}{T}=\frac{1}{4.8} s=0.2083 Hz

The relation between frequency, wavelength and velocity is given by

\lambda=\frac{v}{\nu}

12=\frac{v}{\frac{1}{4.8}}

v=12\times \frac{1}{4.8}=2.5 m/s

Hence, the velocity of wave= 2.5 m/s

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DanielleElmas [232]

Answer:

A

Explanation:

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3 years ago
A 1.8-kg object is attached to a spring and placed on frictionless, horizontal surface. A force of 40 N stretches a spring 20 cm
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Answer:

a) k = 200 N/m

b) E = 4 J

c) Δx = 6.3 cm

Explanation:

a)

  • In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:

       F = - k * \Delta x (1)

  • where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
  • Solving for k:

       k =- \frac{F}{\Delta x} =-\frac{40 N}{-0.2m} = 200 N/m (2)

b)

  • Assuming no friction present, total mechanical energy mus keep constant.
  • When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:

        U = \frac{1}{2}* k* (\Delta x)^{2} (3)

  • Replacing k and Δx by their values, we get:

       U = \frac{1}{2}* k* (\Delta x)^{2} = \frac{1}{2}* 200 N/m* (0.2m)^{2} = 4 J (4)

c)

  • When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
  • We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
  • So, we can write the following expression:

        \frac{1}{2}* k* \Delta x_{1} ^{2} + \frac{1}{2} * m* v^{2}  = \frac{1}{2}*k*\Delta x^{2}   (5)

  • Replacing the right side of (5) with (4), k, m, and v by the givens, and simplifying, we can solve for Δx₁, as follows:

        \frac{1}{2}* 200N/m* \Delta x_{1} ^{2} + \frac{1}{2} * 1.8kg* (-2.0m/s)^{2}  = 4J   (6)

⇒      \frac{1}{2}* 200N/m* \Delta x_{1} ^{2}   = 4J  - 3.6 J = 0.4 J (7)

⇒     \Delta x_{1}   = \sqrt{\frac{0.8J}{200N/m} } = 6.3 cm (8)

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