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.
Answer:
what is it on? like name one of the questions
Explanation:
Answer:
B. Ecosystem B, because its high species diversity could have resulted from increased competition among its members.
Explanation:
This is because, in the ecosystem with varying level of biodiversity, Ecosystem B has medium level of species diversity found in them with High medium level of habitat diversity which causes increasing competitions among them.
Answer:
Frequency, f = 3.73Hz
Explanation:
The frequency of a simple harmonic 6is given by:
f = w/2pi
But w= Sqrt( k/m)
Where k is the spring constant
And m is the mass
Given:
Mass=0.20kg
Spring constant, k=130N/m
w= Sqrt(130/0.20)
w= Sqrt(650)
w= 25.50 m
Frequency, f = w/2pi
f = 25.50/(2×3.142)
f = 3.73Hz
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