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OLga [1]
4 years ago
13

A standing wave is established on a string that is fixed at both ends. If the string is vibrating at its fundamental frequency,

how is the length of the string related to the wavelength of the standing wave?
Physics
1 answer:
Alisiya [41]4 years ago
7 0

The concept related to solving the problem is called Standing Waves: This property relates to how, through a chain that is fixed at both ends, it has a fundamental frequency that relates the wavelength of a standing wave to the length of rope.

In other words, if we take a wave that travels in a direction on the rope from one direction, it will immediately tend to return but in the opposite direction. When having two waves traveling in the opposite direction, it is necessary to take into account that during the vibration two reflections are generated on each side. This concept sometimes called constructive interference results in a wave reflection of the type,

\frac {n} {\lambda 2} = L

Where n is a positive integer,

λ is the wavelength

L is the length of the chain.

In this way it is possible to appreciate that <em>the length of the string is equal to half of a wavelength.</em>

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The polar coordinates of the collar A are given as functions of time in seconds by r = 2+ 0.7 t2 ft and ????= 3.5t rad. What are
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Answer with explanation:

Part a)

v_{radial}=\frac{dr}{dt}=\frac{d(2+0.7t^{2})}{dt}\\\\v_{radial}=1.4t\\\\\therefore v_{radial}|_{t=4}=1.4\times 4=5.6ft/s\\\\v_{angular}=r|_{t=4}\times \frac{d\theta }{dt}=13.2\frac{3.5t}{dt}=46.2fts^{-1}\\\\\therefore v=\sqrt{v_{radial}^{2}+v_{angular}^{2}}\\\\v=46.53ft/s

Part b)

a_{radial}=\frac{d^{2}r}{dt^{2}}=\frac{d^{2}(2+0.7t^{2})}{dt^{2}}\\\\a_{radial}=1.4ft/s^{2}\\\\a_{angular}=r\times \frac{d^{2}\theta }{dt^{2}}\\\\a_{angular}=r\times \frac{d^{2}(3.5t) }{dt^{2}}\\\\\therefore a_{angular}=0\\\\\therefore Accleration=1.4ft/s^{2}

8 0
3 years ago
The junction rule describes the conservation of which quantity? Note that this rule applies only to circuits that are in a stead
denis-greek [22]

Explanation:

The junction rule says that the sum of the currents going into a junction must equal the sum of the currents leaving a junction.  This describes the conservation of current.

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A man pushes a 10 kg box a distance of 5 m for 3 hours. How much work<br> did the man complete?*
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alot

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As part of a study of the effects of urbanization, local scientists have taken many data samples over the course of 20 years. Th
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A boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s .
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Answer:

Force exerted, F = 1.5 N

Explanation:

It is given that, a boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s.

i.e. u = 0

v = 30 m/s

Time taken, t = 0.06 s

Mass of the paper, m = 0.003 kg

We need to find the force the boxer exert on it. The force can be calculated using second law of motion as :

F=m\times a

F=m\times (\dfrac{v-u}{t})

F=0.003\times (\dfrac{30}{0.06})

F = 1.5 N

So, the force the boxer exert on the paper is 1.5 N. Hence, this is the required solution.

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