#### Answer

(a) The amplitude is $2.5~cm$
(b) The wavelength of the wave is $14~m$
(c) The speed of the wave is $20~m/s$
(d) $f = 1.4~Hz$
(e) $T = 0.70~s$

#### Work Step by Step

(a) Each point oscillates between $y = -2.5~cm$ and $y = 2.5~cm$. Therefore, the amplitude is $2.5~cm$
(b) From the graph for $t=0$, we can see that one cycle starts at $x = 2.0~m$ and ends at $x = 16~m$. Therefore, the wavelength of the wave is $14~m$
(c) On the graph, we can see that the wave moves $2.0~m$ to the right in a time of $0.1~s$. Therefore, the speed of the wave is $20~m/s$
(d) We can find the frequency:
$f = \frac{v}{\lambda} = \frac{20~m/s}{14~m} = 1.4~Hz$
(e) We can find the period:
$T = \frac{1}{f} = \frac{14~m}{20~m/s} = 0.70~s$