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3 years ago

Read the scenario.

Physics

2 answers:

Setler [38]3 years ago

6 0

Distance=6m
scalar quantity shows only magnitude

tankabanditka [31]3 years ago

4 0

Answer:

distance = 6 m

Explanation:

- Distance is a scalar quantity (so, only magnitude, no direction), and it is calculated as the scalar sum of all the distances travelled by an object during its motion, regardless of the direction. So, in this problem, the distance covered by the pinecone is

d = 4 m + 2 m = 6 m

- Displacement is a vector quantity (magnitude+direction), and its magnitude is calculate as the distance in a straight line between the final position and the initial position of the object. In this case, the final position is 2 m west and the initial position is 0 m, so the displacement of the pinecone is

d = 2 m west - 0 m = 2 m west

So, a scalar quantity from this scenario is

distance = 6 m

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It will take 3 hours

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4 years ago

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podryga [215]

Answer:

8F_i = 3F_f

Explanation:

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Let us denote that the initial charge of A and B are Q. Then after C is touched to A, their respective charges are Q/2.

Then, C is touched to B, and they share the total charge of Q + Q/2 = 3Q/2. Their respective charges afterwards is 3Q/4 each.

The electrostatic force, Fi, in the initial configuration can be calculated as follows.

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4 years ago

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

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3 years ago

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3 years ago

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ryzh [129]

Answer:

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