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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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The energy per unit volume in an electromagnetic wave is: ________
Marianna [84]

Answer:

c.

Explanation:

Electromagnetic waves are waves that are generated when an electric field and a magnetic field vibrate together. They are usually created whenever an electric field collides with a magnetic field.

In more generic terms for an electric field, the energy per unit volume is given by:

Energy density = \dfrac{1}{2}\varepsilon _oE^2

For magnetic field:

Energy density =\dfrac{\beta^2}{2 \mu _o}

∴

For the electromagnetic wave (u):

Total energy density is:

u = \dfrac{1}{2}\varepsilon_oE^2 + \dfrac{\beta ^2}{2 \mu_o}

Due to the fact that the energy related with both fields is equivalent:

Then:

E = cB

6 0
3 years ago
A parallel circuit has three branches. Each branch has a resistance of 37.5 ohms. The total current is 7.5 amps. What is the cur
Sophie [7]

Answer:

Current in each branch will be 2.5 A

Explanation:

We have given that a circuit has three branch each has a resistance of 37.5 ohm

We have also given that a total current of 7.5 A is flows through the circuit

As we know that in parallel circuit total current is the sum of current flowing through each branch

As the resistances of each branch is same so current will also be same

As there is 3 branches in circuit so current in each branch will be \frac{7.5}{3}=2.5A

4 0
4 years ago
An object of mass M oscillates on the end of a spring. To double the period, replace the object with one of mass: (a) 2M. (b)M/2
harkovskaia [24]

Answer:

(c) 4M

Explanation:

The system is a loaded spring. The period of a loaded spring is given by

T = 2\pi\sqrt{\dfrac{m}{k}}

<em>m</em> is the mass and <em>k</em> is the spring constant.

It follows that, since <em>k</em> is constant,

T\propto\sqrt{m}

\dfrac{T}{\sqrt{m}} = C where <em>C</em>  represents a constant.

\dfrac{T_1}{\sqrt{m_1}} = \dfrac{T_2}{\sqrt{m_2}}

m_2 = m_1\left(\dfrac{T_2}{T_1}\right)^2

When the period is doubled, T_2 = 2T_1.

m_2 = m_1\left(\dfrac{2T_1}{T_1}\right)^2 = 4m_1

Hence, the mass is replaced by 4M.

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