Question

Tolong bantu jawab soal fisika nomor 9 dan 11​

Answer 1

Jawaban:

9. D

11. B

Penjelasan:

9.

y = A sin ( 0,2π.t - 3x )

v = dy/dt  = 0,2π.A cos ( 0,2π.t - 3x )

11.

y = 2 sin (0,5πx) cos (8π.t)

ω = 8π

2π/T = 8π

T = 1/4 s

k = 0,5π

2π/λ = 0,5π

λ = 4 m

Cepat Rambat

v = λ/T = 4/(1/4)

v = 16 m/s

Answer 2

9.

v = dy/dt

= d(Asin(0,2πt - 3x))/dt

= 0,2π * Acos(0,2πt- 3x)

= 0,2πAcos(0,2πt - 3x)

11.

v = λ f

y = A sin kx cos wt

y = 2 sin 0,5πx cos 8πt

k = 0,5π

k = 2π/λ

λ = 2π/k =2π/0,5π = 4 m

w = 8π

w = 2πf

f = w/2π = 8π/2π = 4

maka

v = λf = 4*4=16 m/s