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allochka39001 [22]
4 years ago
8

Jim runs off a diving board and lands in the water 3 meters from the end of the board. When he runs at the same speed on a highe

r diving board, he lands in the water 5 meters from the end of the board. Why is this? A. His horizontal velocity was constant. B. Only his vertical velocity affected the time he stayed airborne. C. Running off the higher board he was in the air for a longer time. D. All of these are true.
Physics
1 answer:
satela [25.4K]4 years ago
4 0

Answer:

C.

Explanation:

If we look at the equation x(final) = v(initial)*t + x(initial), where v(initial in the x dir.) in the same both times, and x(initial) is 0, we can conclude that t is responsible for the change in x(final).

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An object is given an initial velocity. What will happen to the object if no other forces act on it?
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The object will travel at a constant rate in along a straight line.

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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

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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

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Change in market price is m<span>ovement along the demand curve. </span>

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