I can think of two possible and logical questions for the problem given. First, you can calculate for the maximum height reached by the blue ball. Second, you can compute the length of time for the two balls to be at the same height. If so, the solution are as follows:
When the object is thrown upwards or when the object is dropped from a height, the only force acting upon it is the gravitational force. Because of this, it simplifies equations of motion.
1. For the maximum height, the equation is
H = v₀²/2g
where
v₀ is the initial speed
g is the acceleration due to gravity equal to 9.81 m/s²
For the blue ball, v₀ = 21.8 m/s. Substituting the values:
H = (21.8 m/s)²/2(9.81m/s²)
H = 24.22 m
The maximum height reached by the blue ball is 24.22 m + 0.9 = 25.12 m.
2. For this, you equate the y values of both balls:
y for red ball = y for blue ball
v₀t + 0.5gt² = v₀t + 0.5gt²
(10.4 m/s)t + 0.5(9.81 m/s²)(t²) + 26.6 m = (21.8 m/s)t + 0.5(9.81 m/s²)(t²) + 0.9 m
Solving for t,
t = 2.25 seconds
Thus, the two balls would be at the same height after 2.25 seconds.
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.
Aristotle created and it’s credited as the creator.
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