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grin007 [14]
3 years ago
12

You send a traveling wave along a particular string by oscillating one end. If you increase the frequency of oscillations, does

the speed of the wave increase, decrease, or remain the same?
Physics
1 answer:
eimsori [14]3 years ago
4 0

Answer:

The speed of the wave remains the same

Explanation:

Since the speed of the wave v = √(T/μ) where T is the tension in the string and μ is the linear density of the string.

We observed that the speed, v is independent of the frequency of the  wave in the string. So, increasing the frequency of the wave has no effect on the speed of the wave in the string, since the speed of the wave in the string is only dependent on the properties of the string.

<u>So, If you increase the frequency of oscillations, the speed of the wave remains the same.</u>

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The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be
kap26 [50]

Answer:

v = 14.86 m/s

Explanation:

As we know that the force equation at the top is given as

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now we know that

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3 years ago
An old light bulb draws only 54.3 W, rather than its original 60.0 W, due to evaporative thinning of its filament. By what facto
Lemur [1.5K]

Answer:

The factor of the diameter is 0.95.

Explanation:

Given that,

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We know that,

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The power is inversely proportional to the resistance.

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Using relation of power and diameter

\dfrac{P_{i}}{P_{f}}=\dfrac{D_{i}^2}{D_{f}^2}

Put the value into the formula

\dfrac{D_{i}^2}{D_{f}^2}=\dfrac{54.3}{60}

\dfrac{D_{i}}{D_{f}}=0.95

D_{i}=0.95 D_{f}

Hence, The factor of the diameter is 0.95.

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If we continue to warm a body that is "white hot", it would emit in the ultraviolet spectrum, with what would become ... black! then we would not see it emits light in the visible spectrum (well, we would see a very faint bluish light corresponding to the tail of the distribution of the spectrum it emits, but the peak of that spectrum would be in the ultraviolet).

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