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Crazy boy [7]
2 years ago
14

A rope is attached to a hook in the ceiling and is hanging straight down. The rope has a mass m, and nothing is attached to the

free end of the rope. As a transverse wave travels down the rope from the top, ________________.
Physics
1 answer:
Andru [333]2 years ago
6 0

As a transverse wave travels down the rope from the top, the speed of the wave decreases. In a transverse wave, movement occurs at right angles.

<h3>What is a transverse wave?</h3>

A transverse wave can be defined as a type of wave where movement occurs at right angles to the direction of the wave.

Some examples of types of transverse waves include, among others, seismic waves, radio waves, and light waves.

The ripples on the surface of water move at right angles to the direction of the wave, thereby they represent transverse waves.

Learn more about transverse waves here:

brainly.com/question/13761336

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3 years ago
An engineer is designing a runway. She knows that a plane, starting at rest, needs to reach a speed of 180mph at take-off. If th
kvv77 [185]

Answer:

The plane would need to travel at least 8,\!580\; {\rm ft} (8.58 \times 10^{3}\; {\rm ft}.)

The 10,\!000\; {\rm ft} runway should be sufficient.

Explanation:

Convert unit of the the take-off velocity of this plane to \rm ft\cdot s^{-1}:

\begin{aligned}v &= 180\; {\rm mph} \\ &= 180\; {\rm mi \cdot hrs^{-1}} \times \frac{1\; {\rm hrs}}{3600\; {\rm s}} \times \frac{5280\; {\rm ft}}{1\; {\rm mi}} \\ &= 264\; {\rm ft \cdot s^{-1}}\end{aligned}.

Initial velocity of the plane: u = 0\; {\rm ft \cdot s^{-1}}.

Take-off velocity of the plane v =264\; {\rm ft\cdot s^{-1}}.

Let x denote the distance that the plane travelled along the runway. Since acceleration is constant but unknown, make use of the SUVAT equation x = ((u + v) / 2) \, t.

Notice that this equation does not require the value of acceleration. Rather, this equation make use of the fact that the distance travelled (under constant acceleration) is equal to duration t times average velocity (u + v) / 2.

The distance that the plane need to cover would be:

\begin{aligned}x &= \left(\frac{u + v}{2}\right)\, t \\ &= \frac{0\; {\rm ft \cdot s^{-1}} + 264\; {\rm ft \cdot s^{-1}}}{2} \times 65.0\; {\rm s} \\ &= 8.58\times 10^{3}\; {\rm ft}\end{aligned}.

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2 years ago
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