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dlinn [17]
3 years ago
11

Ocean waves are traveling to the east at 3.2 m/s with a distance of 19 m between crests. (a) With what frequency do the waves hi

t the front of a boat when the boat is at anchor? Hz (b) With what frequency do the waves hit the front of a boat when the boat is moving westward at 1.5 m/s?
Physics
1 answer:
rjkz [21]3 years ago
8 0

Answer:

(a) Frequency will be 0.168 Hz

(b) Frequency will be 0.0789 Hz

Explanation:

(a) We have given velocity of the ocean waves in east direction v = 3.2 m /sec

Distance between the crest \Delta x=\lambda =19m

We know that v=\lambda f

So 3.2=19\times  f

f=0.168Hz

(b)Velocity of the ocean waves in westward direction v = 1.5 m /sec

Distance between the crest \Delta x=\lambda =19m

We know that v=\lambda f

So 1.5=19\times  f

f=0.0789Hz

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The speed of a boat undergoing constant acceleration increases from 5 meters per second to 15 meters per second over a period of
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Should be 250m. check with your teacher and let me know

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statuscvo [17]

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it becomes a gas

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Proper design of automobile braking systems must account for heat buildup under heavy braking. Part A Calculate the thermal ener
AVprozaik [17]

Answer:

1838216 J

Explanation:

95 km/h = 26.39 m/s

40 km/h = 11.11 m/s

Initial kinetic energy

= .5 x 1600 x(26.39)²

= 557145.67 J

Final kinetic energy

= .5 x 1600 x ( 11.11)²

= 98745.68 J

Loss of kinetic energy

= 458400 J

Loss of potential energy

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= 1600 x 9.8 x 340 sin 15

= 1379816 J

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8 0
3 years ago
A 2-m long string is stretched between two supports with a tension that produces a wave speed equal to vw=50.00m/s. What are the
svetoff [14.1K]

Answer

given,

Length of the string, L = 2 m

speed of the wave , v = 50 m/s

string is stretched between two string

For the waves the nodes must be between the strings

the wavelength  is given by

           \lambda = \dfrac{2L}{n}

where n is the number of antinodes; n = 1,2,3,...

the frequency expression is given by

            f = n\dfrac{v}{2L}

now, wavelength calculation

      n = 1

           \lambda_1 = \dfrac{2\times 2}{1}

                    λ₁ = 4 m

      n = 2

           \lambda_2 = \dfrac{2\times 2}{2}

                   λ₂ = 2 m

      n =3

           \lambda_3 = \dfrac{2\times 2}{3}

                    λ₃ = 1.333 m

now, frequency calculation

      n = 1

            f = n\dfrac{v}{2L}

            f_1 =1\times \dfrac{50}{2\times 2}

                    f₁ = 12.5 Hz

      n = 2

            f = n\dfrac{v}{2L}

            f_2 =2\times \dfrac{50}{2\times 2}

                    f₂= 25 Hz

      n = 3

            f = n\dfrac{v}{2L}

            f_3 =3\times \dfrac{50}{2\times 2}

                    f₃ = 37.5 Hz

8 0
3 years ago
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