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MissTica
3 years ago
5

A river flows from south to north at 5.4 km/hr. on the west bank of this river, a boat launches and travels perpendicular to the

current with a velocity of 7.6 km/hr due east. if the river is 1.4 km wide at this point, how far downstream does the boat land on the east bank of the river relative to the point it started at on the west bank?
Physics
1 answer:
Aleks [24]3 years ago
7 0
The motion of the boat is basically a uniform motion on both directions, north-south (NS) and east-west (EW), with two constant velocities: v_{SN}=5.4~km/h and v_{EW}=7.6~km/h. 
From the law of motion on the EW direction we can find the total time of the motion:
S_{EW}(t) = v_{EW} t
since we know S_{EW}=1.4~km, we find
t= \frac{S_{EW}}{v_{EW}} =0.18~h
and then, we can find how far the boat went on the NS direction during this time:
S_{NS}=v_{NS} t =5.4~km/h \cdot 0.18~h = 1~km
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\begin{gathered} W=(17)(12)\cos0+(36)(12)\cos30 \\ W=578.123 \end{gathered}

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Read 2 more answers
A confined aquifer with a porosity of 0.15 is 30 m thick. The potentiometric surface elevation at two observation wells 1000 m a
AlekseyPX

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;

q = -Kb\frac{dh}{dl}

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 = \frac{Q}{A} = \frac{q}{b} = -K\frac{dh}{dl}\\\\V = -(25 m/d).(\frac{-1.45 m}{1000 m}) = 0.0363 m/day

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

        = 4132.23 days *\frac{24 .hrs}{1.day} = 99173.52, hours

4 0
2 years ago
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