006 (part 1 of 2) 10.0 points
1 answer:
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Answer:
Part (a) The flow rate per unit width of the aquifer is 1.0875 m³/day
Part (b) The specific discharge of the flow is 0.0363 m/day
Part (c) The average linear velocity of the flow is 0.242 m/day
Part (d) The time taken for a tracer to travel the distance between the observation wells is 4132.23 days = 99173.52 hours
Explanation:
Part (a) the flow rate per unit width of the aquifer
From Darcy's law;
where;
q is the flow rate
K is the permeability or conductivity of the aquifer = 25 m/day
b is the aquifer thickness
dh is the change in th vertical hight = 50.9m - 52.35m = -1.45 m
dl is the change in the horizontal hight = 1000 m
q = -(25*30)*(-1.45/1000)
q = 1.0875 m³/day
Part (b) the specific discharge of the flow
V = 0.0363 m/day
Part (c) the average linear velocity of the flow assuming steady unidirectional flow
Va = V/Φ
Φ is the porosity = 0.15
Va = 0.0363 / 0.15
Va = 0.242 m/day
Part (d) the time taken for a tracer to travel the distance between the observation wells
The distance between the two wells = 1000 m
average linear velocity = 0.242 m/day
Time = distance / speed
Time = (1000 m) / (0.242 m/day)
Time = 4132.23 days
Answer:
Fr = 48 [N] forward.
Explanation:
Suppose the movement is on the X axis, in this way we have the force of the engine that produces the movement to the right, while the force produced by the brake causes the vehicle to decrease its speed in this way the sign must be negative.
∑F = Fr
The movement remains forward, since the force produced by the movement is greater than the braking force.
The frequency of middle C on a string is
f = 261.6 Hz.
The given linear density is
ρ = 0.02 g/cm = (0.02 x 10⁻³ kg)/(10⁻² m)
= 0.002 kg/m
The length of the string is L = 1 m.
Let T = the tension in the string (N).
The velocity of the standing wave is
In the fundamental mode, the wavelength, λ, is equal to the length, L.
That is
Because v = fλ, therefore
From given information, obtain
T = (0.002 kg/m)*(261.6 1/s)²*(1 m)²
= 136.87 N
Answer: 136.9 N (nearest tenth)
f = 261.6 Hz.
The given linear density is
ρ = 0.02 g/cm = (0.02 x 10⁻³ kg)/(10⁻² m)
= 0.002 kg/m
The length of the string is L = 1 m.
Let T = the tension in the string (N).
The velocity of the standing wave is
In the fundamental mode, the wavelength, λ, is equal to the length, L.
That is
Because v = fλ, therefore
From given information, obtain
T = (0.002 kg/m)*(261.6 1/s)²*(1 m)²
= 136.87 N
Answer: 136.9 N (nearest tenth)