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
Because sound waves don't travel through the vaccume of space. Hope this helped
0.4 N-s is the "impulse" acted on the "beach ball".
Option: C
Explanation:
Given that,
Mass of the "beach ball" is 0.1 kg.
The speed of the ball hits is 4 m/s.
We know that,
Whenever an object is collide with other object then an impulse is acted on object, this "impulse" causes "change in momentum".
Impulse acted on the beach ball is "mass" times "velocity".
Impulse = mass × velocity
Impulse = 0.1 × 4
Impulse = 0.4 kg m/s
Impulse = 0.4 N-s
Therefore, the "impulse" acted on the ball is 0.4 N-s.
Answer: 258.3 s
Explanation:
The speed
is given by the following equation:

Where:
is the speed of light in vacuum
is the double of the distance between Earth and Moon, since the beam of light travels from Earth to the Moon and back to Earth again.
is the time it takes to the beam of light to travel the mentioned distance
Isolating
and solving with the given information:


Finally:
