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zvonat [6]
3 years ago
11

A ship

Physics
1 answer:
Juli2301 [7.4K]3 years ago
4 0

Answer:

<em>The sea bed is 1125 m deep.</em>

Explanation:

<u>Speed</u>

It's the rate of change of the distance (d) traveled by an object over time (t).

The speed can be calculated by the formula:

\displaystyle v=\frac{d}{t}

It's known the speed of sound in water is v=1500 m/s. A sound wave is sent to the sea bed and bounces back to the ship 1.5 seconds later.

That is the time the wave takes to travel twice the distance from the ship to the sea bed, thus the required distance uses half of that time or t=1.5/2 = 0.75 seconds.

Solving the equation for d:

d = v.t

d = 1500 m/s * 0.75 s

d = 1125 m

The sea bed is 1125 m deep.

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Approximately 25\; {\rm N} (assuming that this spring is ideal.)

Explanation:

The displacement of a spring is the new length of the spring relative to the original length.

For example:

  • When the 6.0\; {\rm cm}-spring in this question is stretched to 10\; {\rm cm}, the displacement is x = (10\; {\rm cm} - 6.0\; {\rm cm}).
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If this spring is ideal, the force on the spring would be proportional to the displacement of the spring. In other words, if a force of F_{\text{a}} displaces this spring by x_{\text{a}}, while a force of F_{\text{b}} displaces this spring by x_{\text{b}}, then:

\displaystyle \frac{F_{\text{a}}}{x_{\text{a}}} = \frac{F_{\text{b}}}{x_{\text{b}}}.

In this question, it is given that a force of F_{\text{a}} = 7.0 \; {\rm N} would stretch this spring by x_{\text{a}} = (10\; {\rm cm} - 6.0\; {\rm cm}). Thus, the force F_{\text{b}} required to stretch this spring by x_{\text{a}} = (20\; {\rm cm} - 6.0\; {\rm cm}) would satisfy:

\displaystyle \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}}= \frac{F_{\text{b}}}{20\; {\rm cm} - 6.0\; {\rm cm}}.

Rearrange and solve for F_{\text{b}}:

\begin{aligned} F_{\text{b}} &= \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}} \, (20\; {\rm cm} - 6.0\; {\rm cm}) \\ &\approx 25\; {\rm N}\end{aligned}.

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