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Eduardwww [97]
2 years ago
12

A 6.00-m long string sustains a three-loop standing wave pattern as shown. The wave speed is 2.00 × 102 m/s.What is the lowest p

ossible frequency for standing waves on this string?

Physics
1 answer:
KATRIN_1 [288]2 years ago
4 0

Given:

The length of the string is l = 6 m

The speed of the wave is

v=2\times10^2\text{ m/s}

Required: Lowest possible frequency for the standing wave.

Explanation:

The lowest possible frequency is the fundamental frequency.

The fundamental frequency can be calculated by the formula

f=\frac{v}{2l}

On substituting the values, the fundamental frequency will be

\begin{gathered} f=\frac{2\times10^2}{2\times6} \\ =16.67\text{ Hz} \end{gathered}

Final Answer: The lowest possible frequency for standing waves on this string is 16.67 Hz

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A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate
Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by b) is at 36 ft. from the wall is the following:

\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s

Explanation:

The Area of the triangle is given by A=h\times b where h=\sqrt{l^2-b^2} (by using the Pythagoras' Theorem) and b is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

A=\sqrt{l^2-b^2}b

The rate of change of the area is given by its time derivative

\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)

\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}

\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Product rule

\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Chain rule

\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}

\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)

In here we can identify b=36\, ft, l=39 and \frac{db}{dt}=8\,ft/s.

The result is then

\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s

3 0
3 years ago
What is happening in this picture?
Alexandra [31]
The answer is b no problem
7 0
3 years ago
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nasty-shy [4]

Answer:

a) Ws = 2.548 J

b) Wf = 1.153 J

c) v = 1.923 m / s

Explanation:

a) The work done by the spring force  

Ws =  ½ * k * x²

Ws =  ½ * 260 N/m *0.14² m  

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b) The increase in thermal energy can by find using  

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EK = Ws + Et

Ek = ( 2.548 + 1.153 ) J = 3.7 J

Ek = ½ * m * v²

v² = 2* Ek / m

v = √[2 * 3.7 J / 2.0 kg]

v = 1.923 m / s

8 0
3 years ago
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Illusion [34]

Answer:

I think the answers March 21

6 0
3 years ago
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Technician A says the charging system recharges the battery by supplying a constant and relatively low charge to the battery. Te
brilliants [131]

Answer:

Technician B is correct

Explanation:

Charging systems works on the principle of magnetism. Charging system uses electromagnetic induction to generate electrical energy.

Electromagnetic principles states that voltage will be produced when there is a motion between conductor and the magnetic field.

5 0
3 years ago
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