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konstantin123 [22]
3 years ago
11

A swimming duck paddles the water with its feet once per time interval of 1.6 s, producing surface waves with this period. The d

uck is moving at constant speed in a pond where the speed of surface waves is 0.40 m/s, and the crests of the waves ahead of the duck have a spacing of 0.20 m.
Part A: What is the duck's speed?
Part B: How far apart are the crests behind the duck?
Physics
1 answer:
soldier1979 [14.2K]3 years ago
4 0

Answer:

a. 0.275m/s

b. 1.08m

Explanation:

Since the duck the paddle the water at an interval of 1.6sec, we can determine the frequency of the wave formed using the equation

f=1/T

Where T is the period.

f=1/1.6

f=0.625Hz.

Also from the equation used in determining the speed of a wave

V=fλ,

v=0.625*0.2

v=0.125m/s

in the question it was stated that that the duck produce a wave moving at a speed of 0.40m/s.

Hence the speed of the duck is

v(duck)=0.40-0.125

v(duck)=0.275m/s

b. The distance between the crest behind the duck is the wavelength of the waves.

To determine this, the wavelength is expressed as

λ=v/f

but the speed in this case is the speed of the duck and the surface wave,as this account for the wave speed behind the duck,

Hence we have

λ=(0.40+0.275)/0.625

λ=1.08m.

The wavelength behind the duck is 1.08m

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The mass of water is

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The volume of heptane will be

\begin{gathered} V_h=\frac{m_h}{d_h} \\ =\frac{31}{0.684} \\ =45.32\text{ mL} \end{gathered}

The volume of water will be

\begin{gathered} V_w=\frac{m_w}{d_w} \\ =\frac{37}{1} \\ =37\text{ mL} \end{gathered}

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What is the peak emf generated by rotating a 940-turn, 24 cm diameter coil in the Earth’s 5·10−5 T magnetic field, given th
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Answer:

The peak emf generated by the coil is 2.67 V

Explanation:

Given;

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diameter, d = 24 cm = 0.24 m

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time, t = 5 ms = 5 x 10⁻³ s

peak emf, V₀ = ?

V₀ = NABω

Where;

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A is the area

B is the magnetic field strength

ω is the angular velocity

V₀ = NABω and ω = 2πf = 2π/t

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V₀ = N x (πd²/4) x B x (2π/t)

V₀ = 940 x (π x 0.24²/4) x 5 x 10⁻⁵ x (2π/0.005)

V₀ = 940 x 0.04524 x  5 x 10⁻⁵ x 1256.8

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