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kompoz [17]
3 years ago
15

A man dropped a dime in a wishing well , he heard it 5 sec later ..Find distance traveled if moving at 10 m/s 2

Physics
1 answer:
Alenkasestr [34]3 years ago
8 0

Answer:

The distance traveled is 109.58 m

Explanation:

The Speed of sound in air = 344 m/s

Let the time in which the dime dropped by the man reach an impact in the well = t₁

Let the time in which the sound travel from the well to the man = t₂

Then

1/2× 10 × t₁² = 344 × t₂ which gives;

5 × t₁² = 344 × t₂.........................(1)

Also the total time before the man heard the dime = t₁ + t₂ = 5

Therefore;

 t₂ = 5 - t₁

Substituting the value of t₂ in equation (1), we have;

5 × t₁² = 344 × (5 - t₁)

5·t₁² + 344·t₁ - 1720 = 0

Using the quadratic formula, we have;

x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}

Which gives;

t_1 = \dfrac{-344\pm \sqrt{344^{2}-4\times 5\times (-1720)}}{2\times 5}

t₁ = 4.68 s or -73.48 s

Therefore, with the positive value for t₁ = 4.68 s, we have

The distance = 1/2× 10 × 4.68² = 109.58 m

The distance traveled = 109.58 m.

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Explanation:

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If an object is dropped from a tall building and hits the ground 3.0 s later, how tall is the building?
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2 years ago
Read 2 more answers
Two points charges are brought closer together,increasing the force between them by a factor of 25.By what factor wa their separ
Roman55 [17]

Answer:

The separation between the charges was decreased by a factor of 0.2

Explanation:

The Coulomb's force between two charges is given by;

F = \frac{kq^2}{r^2} \\\\let \ kq^2 \ be \ constant\\\\F_1r_1^2 = F_2r_2^2\\\\r_2^2 = \frac{F_1r_1^2}{F_2} \\\\increasing \ the \ force \ between \ them \ by \ factor \ of \ 25\\\\(F_2 = 25F_1)\\\\r_2^2 = \frac{F_1r_1^2}{25F_1}\\\\r_2^2 = \frac{r_1^2}{25}\\\\r_2 = \sqrt{\frac{r_1^2}{25} }\\\\ r_2 = \frac{r_1}{5}

r₂ = 0.2r₁

Therefore, the separation between the charges was decreased by a factor of 0.2.

6 0
3 years ago
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