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allsm [11]
1 year ago
7

If a piece of ribbon were tied to a stretched string carrying a transverse wave, then how is the ribbon observed to oscillate?a.

)perpendicular to wave directionb.)both perpendicular to and parallel to wave direction
Physics
1 answer:
noname [10]1 year ago
5 0

The ribbon was observed to oscillate perpendicular to the wave direction. <u>Option A.</u>

<u />

In transverse waves, particles move perpendicular to the direction of wave propagation. Examples of transverse waves are string vibrations and water waves. By moving the Slinky vertically up and down, you can create horizontal transverse waves. Her one type of mechanical wave is the transverse wave.

where the motion of the medium is perpendicular to the direction of energy propagation. A transverse wave is a wave whose vibration is perpendicular to the direction of wave propagation. This is in contrast to longitudinal waves, which propagate in the direction of oscillation. Water waves are an example of transverse waves.

Learn more about The ribbon here:-brainly.com/question/11685530

#SPJ9

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mihalych1998 [28]

Explanation:

By the second law of Newton we get the relation

F = ma

5 0
3 years ago
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g In 1956, Frank Lloyd Wright proposed the construction of a mile-high building in Chicago. Suppose the building had been constr
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To solve this problem it is necessary to apply the concepts related to acceleration due to gravity, as well as Newton's second law that describes the weight based on its mass and the acceleration of the celestial body on which it depends.

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a = \frac{GM}{r^2}

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This equation can be differentiated with respect to the radius of change, that is

\frac{da}{dr} = -2\frac{GM}{r^3}

da = -2\frac{GM}{r^3}dr

At the same time since Newton's second law we know that:

F_w = ma

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From the previous value given for acceleration we have to

F_W = m (\frac{GM}{r^2} ) = 600N

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dF_W = -2(m\frac{GM}{r^2})(\frac{dr}{r})

dF_W = -2F_W(\frac{dr}{r})

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4 0
3 years ago
The period of a pendulum is measured 16 times. The average value of the period over these 16 trials is calculated to be 1.50 sec
marin [14]

Answer:

The additional trials needed is 48 trials

Explanation:

Given;

initial number of trials, n = 16 trials

the standard deviation, σ = 0.24 s

initial standard error, ε = 0.06 s

The standard error is given by;

\epsilon = \frac{\sigma}{\sqrt{n} }

To reduce the standard error to 0.03 s, let the additional number of trials = x

0.03= \frac{0.24}{\sqrt{n+x} } \\\\0.03= \frac{0.24}{\sqrt{16+x} }\\\\0.03\sqrt{16+x} = 0.24\\\\\sqrt{16+x} = \frac{0.24}{0.03} \\\\\sqrt{16+x} = 8\\\\16+x = 8^2\\\\16+x = 64\\\\x = 64 -16\\\\x = 48 \ trials

Therefore, the additional trials needed is 48 trials.

6 0
3 years ago
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spayn [35]
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Answer:

energy is added to it

Explanation:

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