Answer:
The pressure measured at this moment is 0.875 mPa
Explanation:
Given that,
Flow energy = 124 L/min
Boundary to system P= 108.5 kJ/min
We are tasked with finding the pressure here
Applying the pressure formula
Here, 
Where, v refers to velocity
Insert the values into the equation
Therefore, the pressure at this moment is 0.875 mPa
Response:
The population mean is parameter = 65 c
Explanation:
In the analysis of samples and inferring population behavior, two key elements are essential.
To ascertain the population mean, we typically extract various samples and calculate their average. The average of all these means will serve as an estimate for the population mean. According to the central limit theorem, as sample sizes increase, the average of a sample tends to follow a normal distribution with an estimated mean being the sample mean.
A statistic pertains to a sample, while a parameter refers to the whole population.
In this case, 65 degrees C represents the entire population; thus, it constitutes a parameter.
Response: Numerous elements can be found, all situated within the same vertical column as bromine.
Explanation:
Elements are organized by their atomic numbers on the periodic table. Those in the same vertical column (known as groups) exhibit the same valence electron configurations, resulting in similar chemical characteristics. Consequently, there are numerous elements sharing analogous chemical properties grouped with Bromine.
Hypothesis: The liquid will project far.
Independent Variable: Height of the hole.
Dependent Variable: Distance of the squirt.
Constant: All other factors aside from the independent variable, such as the liquid volume.
Control: None that I recognize.
Number of groups: 4
Trials per group: 4
Response:
The ball remained airborne for 3.896 seconds
Explanation:
Given that
g = 9.8 m/s², representing gravitational acceleration,
If the angle of launch is 45°, the horizontal range will be maximized.
Both horizontal and vertical launch velocities are equal, each equating to
v_h = v cos θ
v_h = 27 × cos 45°
= 19.09 m/s.
The duration to reach maximum height is half of the flight time.
v = u + at ∵ v = 0 (at maximum height)
19.09 - 9.8 t₁ = 0
t₁ = 1.948 s
The total time in the air equals twice the time to reach maximum height
2 t₁ = 3.896 s
The horizontal distance covered is
D = v × t
D = 3.896×19.09
= 74.375 m
The ball was in the air for 3.896 seconds
