To solve this exercise it is necessary to use the concepts related to Difference in Phase.
The Difference in phase is given by
Where
Horizontal distance between two points
Wavelength
From our values we have,
The horizontal distance between this two points would be given for
Therefore using the equation we have
Therefore the correct answer is C.
Answer: minimum speed of launch must be 7.45m/s
Explanation:
Given the following:
Height or distance (s) = 2.83m
The final velocity(Vf) at maximum height = 0
Upward motion, acceleration due to gravity(g) us negative = -9.8m/s^2
From the 3rd equation of motion:
V^2 = u^2 - 2gs
Where V = final velocity
u = initial velocity
Therefore, u = Vi
u = √Vf^2 - 2gs
u = √0^2 - 2(-9.8)(2.83)
u = √0 + 55.468
u = √55.468
u = 7.4476 m/s
u = 7.45m/s
Answer:
90 meters
Explanation:
Given:
x₀ = 0 m
v₀ = 0 m/s
v = 30 m/s
t = 6 s
Find:
x
x = x₀ + ½ (v + v₀)t
x = 0 + ½ (30 + 0)(6)
x = 90
The car travels 90 meters.
Explanation:
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