<span>A hypothesis is testable when you can create an experiment to study the proposition contained within the hypothesis. For example, the hypothesis ‘Santa travels slower than a unicorn’ is testable in theory by measuring the speeds of both, but it is not truly testable because neither exists in reality.</span>
Explanation:
The given data is as follows.
Fluid is water so, density 
Weight flow rate = 500 lbf/s = 2224.11 N/sec
Cross-sectional area (A) = 
= 0.05184 
Hence, weight flow rate will be given as follows.
w = 
2224.11 N/sec = 
V =
m/s
= 4.373 m/s
Thus, we can conclude that average velocity in the given case is 4.373 m/s.

Hi pupil Here's Your answer :::
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Student's justification is not correct. Two equal and opposite force cancel each other if the act on the same body. According to the third law of motion action and reaction forces are equal and opposite but they both act on different bodies. Hence, they cannot cancel each other.
When we push a message track, then the applied force on the truck is not sufficient to overcome the force of friction between the tyres of truck and ground, hence, truck does not move.
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Hope this helps .
We have the following equation for height:
h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
Where,
a: acceleration
vo: initial speed
h0: initial height.
The value of the acceleration is:
a = -g = -9.8 m / s ^ 2
For t = 0 we have:
h (0) = (1/2) * (a) * 0 ^ 2 + vo * 0 + h0
h (0) = h0
h0 = 0 (reference system equal to zero when the ball is hit).
For t = 5.8 we have:
h (5.8) = (1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0
(1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0 = 0
vo = (1/2) * (9.8) * (5.8)
vo = 28.42
Substituting values we have:
h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
h (t) = (1/2) * (- 9.8) * t ^ 2 + 28.42 * t + 0
Rewriting:
h (t) = -4.9 * t ^ 2 + 28.42 * t
The maximum height occurs when:
h '(t) = -9.8 * t + 28.42
-9.8 * t + 28.42 = 0
t = 28.42 / 9.8
t = 2.9 seconds.
Answer:
The ball was at maximum elevation when:
t = 2.9 seconds.
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