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irinina [24]
3 years ago
5

The vertical displacement of an ocean wave is described by the function, y = A sin(ωt - kx). k is called the wave number (k = 2π

/λ) and has a value of k = 18 rad/m. The remaining values are A = 9.5 m and ω = 14.5 rad/s.
a) Using y = A sin(B), input an expression for B where the wave would be traveling in the -x-direction. sig.gif?tid=7M79-31-9F-4E-8624-20536

b) What is the wave's velocity in m/s?

c) What is the wave's amplitude in m?
Physics
1 answer:
MrMuchimi3 years ago
4 0

Answer:

Explanation:

a ) y = A sin(B)  ; here B is the phase of the wave which moves so that it remains constant

ωt - kx = constant

differentiating on both sides

ωdt - kdx =0

ωdt =   kdx

dx / dt = ω / k

wave velocity = ω / k

b ) ω = 14.5 rad / s ,

k = 18 rad / m

wave velocity = ω / k

= 14.5 / 18

= .805 m /s

80.5 cm / s

c )

Amplitude = A

= 9.5 m

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Answer:

a)v=1.77\times 10^6\ m/s

b)v=3.872\times 10^6\ m/s

c)v=5.5\times 10^6\ m/s

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e)v=7.7\times 10^6\ m/s

Explanation:

Given that

d = 2 cm

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u=4\times 10^5\ m/s

We know that

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F = m a

E = V/d

So

m a =  q .V/d b            

a=\dfrac{q.V}{m.d}                 ---------1

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m=9.1\times 10^{-31}\ kg

The charge on electron

q=1.6\times 10^{-19}\ C

Now by putting the all values in equation 1

a=\dfrac{1.6\times 10^{-19}\times 200}{9.1\times 10^{-31}\times 0.02}\ m/s^2

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v^2=u^2+2as

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s = 0.1 cm

v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 0.1\times 10^{-2}

v=\sqrt{3.16\times 10^{12}}\ m/s

v=1.77\times 10^6\ m/s

b)

s = 0.5 cm

v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 0.5\times 10^{-2}

v=\sqrt{1.5\times 10^{13}}\ m/s

v=3.872\times 10^6\ m/s

c)

s = 1 cm

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s = 1.5 cm

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v=\sqrt{4.5\times 10^{13}}\ m/s

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e)

s = 2 cm

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v=\sqrt{6.06\times 10^{13}}\ m/s

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