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Minchanka [31]
2 years ago
9

The wavelength of a water wave is 6 cm when the frequency is 10 hertz .

Physics
1 answer:
dedylja [7]2 years ago
4 0

  1. The distance between successive crests (or successive troughs) is defined as the wavelength. The time elapsed between the crossings of two successive crests through a given point is the period of the wave.
  2. The Velocity of propagation (Vp) of a cable is the speed at which an electrical signal can propagate through the cable in comparison to the speed of light. For example, in a vacuum, the velocity of propagation would be 100% or 1 (depending on how it is represented).
  3. The wave period is the time it takes to complete one cycle. The standard unit of a wave period is in seconds, and it is inversely proportional to the frequency of a wave, which is the number of cycles of waves that occur in one second.

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What determines how the plates interact at their boundaries
swat32

Answer:

Tectonic plate interactions are of three different basic types: Divergent boundaries are areas where plates move away from each other, forming either mid-oceanic ridges or rift valleys. These are also known as constructive boundaries. Convergent boundaries are areas where plates move toward each other and collide.

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3 years ago
A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What
AleksandrR [38]

Answer:

The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

Explanation:

We are given that

Angular acceleration, \alpha=3.3 rad/s^2

Diameter of the wheel, d=21 cm

Radius of wheel, r=\frac{d}{2}=\frac{21}{2} cm

Radius of wheel, r=\frac{21\times 10^{-2}}{2} m

1m=100 cm

Magnitude of total linear acceleration, a=1.7 m/s^2

We have to find the linear speed  of a  at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.

Tangential acceleration,a_t=\alpha r

a_t=3.3\times \frac{21\times 10^{-2}}{2}

a_t=34.65\times 10^{-2}m/s^2

Radial acceleration,a_r=\frac{v^2}{r}

We know that

a=\sqrt{a^2_t+a^2_r}

Using the formula

1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}

Squaring on both sides

we get

2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}

\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}

v^4=r^2\times 2.7699

v^4=(10.5\times 10^{-2})^2\times 2.7699

v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}

v=0.418 m/s

Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

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A toy rocket fires its engines to launch it straight up from rest. After a short time, the engine turns off and the
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Answer:

at 10 seconds.. b/c that's when the velocity stops increasing..

Explanation:

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