Answer:
I'd say C is the answer they want, though my pedantic side wants to argue for B being true as well.
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
Answer:
The applied torque is 3.84 N-m.
Explanation:
Given that,
Moment of inertia of the wheel is 
Initial speed of the wheel is 0 (at rest)
Final angular speed is 25 rad/s
Time, t = 13 s
The relation between moment of inertia and torque is given by :

So, the applied torque is 3.84 N-m.
Answer:
acceleration = v-u /t
30- 20/5
= 10/5 = 2m/sec²
Force = mass * acceleration
Force = 0.1 * 2
Force = 0.2 Newton
Answer:
a = Δv/t = (vf - vi)/t = (0 - 5)/4 = -1.25 m/s²
Explanation:
You may or may not need the negative sign, depending on how the question designer was thinking about the problem.