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ycow [4]
3 years ago
13

A vertical spring stretches 3.8 cm when a 13-g object is hung from it. The object is replaced with a block of mass 20 g that osc

illates in simple harmonic motion. Calculate the period of motion.
Physics
1 answer:
Alex777 [14]3 years ago
3 0

Answer:

The period of motion is 0.5 second.

Explanation:

Given;

extension of the spring, x = 3.8 cm = 0.038 m

mass of the object, m = 13 g = 0.013 kg

Determine the force constant of the spring, k;

F = kx

k = F / x

k = mg / x

k = (0.013 x 9.8) / 0.038

k = 3.353 N/m

When the object is replaced with a block of mass 20 g, the period of motion is calculated as;

T = 2\pi\sqrt{\frac{m}{k} } \\\\T =  2\pi\sqrt{\frac{0.02}{3.353} } \\\\T = 0.5 \ second

Therefore, the period of motion is 0.5 second.

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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
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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;

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

where;

q is the flow rate

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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

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