Answer:
The answer to your question is: total energy = 30100.4 J
Explanation:
Kinetic energy (KE) is the energy due to the movement of and object, its units are joules (J)
Data
mass = 1280 kg
speed = 4.92 m/s
Force = 509 N
distance = 28.7 m
Formula

Work = Fd
Process
- Calculate Kinetic energy
- Calculate work
- Add both results
KE = 
KE = 15492.1 J
Work = (509)(28.7)
Work = 14608.3 J
Total = 15492.1 + 14608.3
Total energy = 30100.4 J
<span>Scientists follow a set order of steps when carrying out a scientific investigation to make sure that the method, interpretation and results that they have obtained are repeatable and reliable. This kind of information can be truly said that their data is true and valid. </span>
Answer:
0.68 m
Explanation:
We know that the speed of sound in air is a product of frequency and wavelength. Taking speed of sound in air as 340 m/s
V=frequency*wavelength
Then wavelength is given by 350/500=0.68 m
Therefore, to repeat constructive interference at the listener's ear, a distance of 0.68 m is needed
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
This gots to be the answer, average, seawater in the world's oceans has a salinity of approximately 3.5%, or 35 parts per thousand.