Answer:
1.2 m/s
0.31 m
0.15 m
Explanation:
Time period is
[tex]T=2.5\times 2\\\Rightarrow T=5\ s[/tex]
Frequency is
[tex]f=\dfrac{1}{T}\\\Rightarrow f=\dfrac{1}{5}\\\Rightarrow f=0.2\ Hz[/tex]
Velocity is given by
[tex]v=f\lambda\\\Rightarrow v=0.2\times 6\\\Rightarrow v=1.2\ m/s[/tex]
The waves are traveling at 1.2 m/s
Amplitude is given by
[tex]A=\dfrac{d}{2}\\\Rightarrow A=\dfrac{0.62}{2}\\\Rightarrow A=0.31\ m[/tex]
Amplitude is 0.31 m
If d = 0.3 m
[tex]A=\dfrac{0.3}{2}=0.15\ m[/tex]
The amplitude would be 0.15 m. The speed would remain the same.
A) The speed at which the waves are travelling is; v = 1.2 m/s
B) The amplitude of each wave is; A = 0.31 m
C) If the total vertical distance travelled by the boat were 0.3 m;
- The speed will be the same but the amplitude will be lesser than it was in answer B.
We are gjven;
Distance from highest to lowest point; 2A = 0.62 m
Thus, A = 0.62/2
A = 0.31 m
Wavelength; λ = 6m
Half of time period; T/2 = 2.5
Thus; period; T = 2.5 × 2
T = 5 s
A) Formula for velocity is;
v = fλ
Where;
f is frequency = 1/T = 1/5
λ is wavelength
Thus;
v = (1/5) × 6
v = 1.2 m/s
B) From earlier we saw that A = 0.31 m
Thus, amplitude = 0.31 m
C) We are told that the vertical distance travelled by the boat is now 0.3 m. This is the highest to lowest point. Thus;
Amplitude = 0.3/2
A = 1.5 m
Since only amplitude changes, the speed would remain the same but the amplitude will be lesser than that in answer B.
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