λ = 2 m.
The easiest way to solve this problem is using the equation of frecuency of a wave f = v/λ, where v is the velocity of the wave, and λ is the wavelength.
To calculate the wavelength of a microwave light travels through a liquid, it moves at a speed of 2.2 x 10⁸ m/s. If the frecuency of the light wave is 1.1 x 10⁸ Hz, we have to clear λ from the equation f = v/λ:
f = v/λ -------> λ = v/f
λ = 2.2 x 10⁸ m/s / 1.1 x 10⁸ Hz
λ = 2 m (wavelength of the microwave)
Yes I would expect them too
<span>let the fsh jump with initial velocity (u) in direction (angle p) with horizontal
it can cross and reach top of trajectory if its top height h = 1.5m
and horizontal distance d = (1/2) Range
--------------------------------------...
let t be top height time
at top height, vertical component of its velocity =0
vy = 0 = u sin p - gt
t = u sin p/g
h = [u sin p]*t - 0.5 g[t[^2
1.5 = u^2 sin^2 p/g - u^2 sin^2 p/2g
u^2 sin^2 p/2g = 1.5
u^2 sin^2 p = 1.5*2*9.8 = 29.4
u sin p = 5.42 m/s >>>>>>>>>>>>>>> V-component
=====================
t = HALF the time of flight
d = (1/2) Range (R) = (1/2) [2 u^2 sin p cos p/g]
1 = u^2 sin p cos p/g
u sin p * u cos p = 9.8
5.42 * u cos p = 9.8
u cos p = 1.81 m/s >>>>>>>>>>>>> H-component
check>>
u = sqrt[u^2 cos^2 p + u^2 sin^2 p] = 5.71 m/s
u < less than fish's potential jump speed 6.26 m/s
so it will able to cross</span>
