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(A) The first wave is given by $y_1 = a \sin(\omega t + kx + 0.57)$.Therefore,the phase of the first wave is $\phi_1 = (\omega t + kx + 0.57)$.The sec

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(A) The first wave is given by $y_1 = a \sin(\omega t + kx + 0.57)$.Therefore,the phase of the first wave is $\phi_1 = (\omega t + kx + 0.57)$.The second wave is given by $y_2 = a \cos(\omega t + kx)$.Using the trigonometric identity $\cos(	heta) = \sin(	heta + \pi/2)$,we can write $y_2 = a \sin(\omega t + kx + \pi/2)$.Therefore,the phase of the second wave is $\phi_2 = (\omega t + kx + \pi/2)$.The phase difference $\Delta \phi$ is given by $\phi_2 - \phi_1$.$\Delta \phi = (\omega t + kx + \pi/2) - (\omega t + kx + 0.57)$.$\Delta \phi = \pi/2 - 0.57$.Since $\pi \approx 3.14$,$\pi/2 \approx 1.57$.$\Delta \phi = 1.57 - 0.57 = 1.0 \ radian$.

Two waves are represented by the equations: $y_1 = a \sin(\omega t + kx + 0.57) \ m$ and $y_2 = a \cos(\omega t + kx) \ m$,where $x$ is in $meters$ and $t$ is in $seconds$. The phase difference between them is ..... $radian$.

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