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3 years ago

10

A stone is dropped into a well. The sound of the splash is heard 6.25 s later. What is the depth of the well? (Take the speed of

sound to be 343 m/s.)

1 answer:

Naya [18.7K]3 years ago

3 0

Answer:

depth of well is 163.30 m

Explanation:

Given data

speed of sound = 343 m/s

timer = 6.25 s

to find out

depth of well

solution

let us consider depth d

so equation will be

depth = 1/2 ×g ×t² ..............1

and

depth = velocity of sound × time .................2

here we have given time 6.25 that is sum of 2 time

when stone reach at bottom that time

another is sound reach us after stone strike on bottom

so time 1 + time 2 = 6.25 s

so from equation 1 and 2 we get

1/2 ×g ×t² = velocity of sound × time

1/2 ×9.8 × t1² = 343 × (6.25 - t1 )

t1 = 5.77376 sec

so height = 1/2 ×g ×t²

height = 1/2 ×9.8 × (5.773)²

height = 163.30 m

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Answer:

interest point:

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4) point within the radius r₂ of the second sphere

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Explanation:

In this case, the total electric field is the vector sum of the electric fields of each sphere, to simplify the calculation on the line that joins the two spheres

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Let's analyze the field for various points of interest.

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2) Point within the radius r₁ of the first sphere.

In this case, according to Gauus' law, the charge is on the surface of the sphere at the point, there is no charge inside so this sphere has no electric field on its inner point

E_ {total} = $-k \frac{Q}{x_2^2} = -k \frac{Q}{((d-x_1)^2}$

this expression holds for the points located at

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This champ is always different from zero

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E_ {total} = $+ k \frac{Q}{(d-x_1)^2}$ + k Q / (d-x1) 2

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