The tides are considered an example of shallow-water waves because the tidal bulges have a wavelength that is on the order of __
1 answer:
You might be interested in
Explanation:
The frequency of radio waves is 1.667 GHz
One portion of the same wave front travels 1.260 mm farther than the other before the two signals are combined.
There are two conditions for interference either constructive or destructive.
For constructive interference , the path difference is n times of wavelength and for destructive interference, the path difference is (n+1/2) times of wavelength
We can find wavelength in this case as follows :
If we divide path difference by wavelength,
It means that the path difference is 7 times of the wavelength. it means the two waves combine constructively and the value of m for the path difference between the two signals is 7.
(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.
(b) The maximum height above the ground reached by the ball is 8.6 m.
(c) The distance off course the ball would be carried is 0.38 m.
(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.
<h3>Horizontal and vertical components of the ball's velocity</h3>Vx = Vcosθ
Vx = 39.7 x cos(17.8)
Vx = 37.8 m/s
Vy = Vsin(θ)
Vy = 39.7 x sin(17.8)
Vy = 12.14 m/s
<h3>Maximum height reached by the ball</h3>Maximum height above ground = 7.51 + 1.09 = 8.6 m
<h3>Distance off course after 2 second </h3>Upward speed of the ball after 2 seconds, V = V₀y - gt
Vy = 12.14 - (2x 9.8)
Vy = - 7.46 m/s
Horizontal velocity will be constant = 37.8 m/s
Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)
d = Δvt = (38.72 - 38.53) x 2
d = 0.38 m
The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.
Learn more about projectiles here: brainly.com/question/11049671