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2 years ago

A truck slows from a velocity of 25 m/s to a stop in 70 m. What was the truck’s acceleration

Physics

1 answer:

igomit [66]2 years ago

4 0

v^2-u^2=2 x a x d

25^2-0^2=2 x a x 70

625-0=140 x a

625=140a

a=625/140

a=4.46 m/s^2

im not very sure but i think this is how you do this

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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

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

where;

q is the flow rate

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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

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Va = 0.0363 / 0.15

Va = 0.242 m/day

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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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