Wouldn't it be the employee? Because the employee has to adjust to the needs of his/her supervisor. If the supervisor wants 100 boxes the employee has to make those 100 boxes and so on and so forth.
A graph of real speed can have a section that's as steep as you want,
but it can never be a perfectly vertical section.
Any vertical line on a graph, even it it's only a tiny tiny section, means
that at that moment in time, the speed had many different values.
It also means that the speed took no time to change from one value to
another, and THAT would mean infinite acceleration.
<h3>Question -:</h3>
The Earth orbits around the sun because the gravitational force that the sun
exerts on the Earth:
O A. causes Earth's acceleration toward the sun.
O B. is very small because the sun is so far from the Earth.
O c. is smaller than the force the Earth exerts on the sun.
O D. pushes the Earth away from the sun.
<h3>Answer -:</h3>
O A. causes Earth's acceleration toward the sun.
<em>I </em><em>hope </em><em>this</em><em> </em><em>helps</em><em>,</em><em> </em><em>have </em><em>a </em><em>nice </em><em>time </em><em>ahead!</em>
Answer:
The upper limit on the flow rate = 39.46 ft³/hr
Explanation:
Using Ergun Equation to calculate the pressure drop across packed bed;
we have:

where;
L = length of the bed
= viscosity
U = superficial velocity
= void fraction
dp = equivalent spherical diameter of bed material (m)
= liquid density (kg/m³)
However, since U ∝ Q and all parameters are constant ; we can write our equation to be :
ΔP = AQ + BQ²
where;
ΔP = pressure drop
Q = flow rate
Given that:
9.6 = A12 + B12²
Then
12A + 144B = 9.6 -------------- equation (1)
24A + 576B = 24.1 --------------- equation (2)
Using elimination methos; from equation (1); we first multiply it by 2 and then subtract it from equation 2 afterwards ; So
288 B = 4.9
B = 0.017014
From equation (1)
12A + 144B = 9.6
12A + 144(0.017014) = 9.6
12 A = 9.6 - 144(0.017014)

A = 0.5958
Thus;
ΔP = AQ + BQ²
Given that ΔP = 50 psi
Then
50 = 0.5958 Q + 0.017014 Q²
Dividing by the smallest value and then rearranging to a form of quadratic equation; we have;
Q² + 35.02Q - 2938.8 = 0
Solving the quadratic equation and taking consideration of the positive value for the upper limit of the flow rate ;
Q = 39.46 ft³/hr