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3 years ago

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A 64.0 cm long cord is vibrating in such a manner that it forms a standing wave with two antinodes. (The cord is fixed at both e

nds.) Which harmonic does this wave represent

1 answer:

xxMikexx [17]3 years ago

5 0

Answer:

the wave represents the second harmonic.

Explanation:

Given;

length of the cord, L = 64 cm

The first harmonic of a cord fixed at both ends is given as;

$f_o = \frac{V}{2L}$

The wavelength of a standing wave with two antinodes is calculated as follows;

L = N---> A -----> N + N ----> A -----> N

Where;

N is node

A is antinode

L = N---> A -----> N + N ----> A -----> N = λ/2 + λ/2

L = λ

The harmonic is calculated as;

$f = \frac{V}{\lambda} \\\\f = \frac{V}{L} = 2(\frac{V}{2L} ) = 2(f_o) = 2^{nd} \ harmonic$

Therefore, the wave represents the second harmonic.

L = λ

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Answer:

Speed is a scalar quantity it has only magnitude. So in the given scenario the student actually described about the velocity.

Explanation:

In this situation, student is describing that the speed is -10 m/s. The speed with the negative sign is wrong because speed is a scalar quantity and scalar quantities have only magnitude, they do not have direction. So, his statement is wrong because he is describing the speed with the negative sign.

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3 years ago

A6 kg bag of rice is on the top shelf at a grocery store.

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Answer:

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Hope this helps you

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3 years ago

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Assume that, when we walk, in addition to a fluctuating vertical force, we exert a periodic lateral force of amplitude 25 NN at

dexar [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From the question we are told

The amplitude of the lateral force is $F = 25 \ N$

The frequency is $f = 1 \ Hz$

The mass of the bridge per unit length is $\mu = 2000 \ kg /m$

The length of the central span is $d = 144 m$

The oscillation amplitude of the section considered at the time considered is $A = 75 \ mm = 0.075 \ m$

The time taken for the undriven oscillation to decay to $\frac{1}{e}$ of its original value is t = 6T

Generally the mass of the section considered is mathematically represented as

$m = \mu * d$

=> $m = 2000 * 144$

=> $m = 288000 \ kg$

Generally the oscillation amplitude of the section after a time period t is mathematically represented as

$A(t) = A_o e^{-\frac{bt}{2m} }$

Here b is the damping constant and the $A_o$ is the amplitude of the section when it was undriven

So from the question

$\frac{A_o}{e} = A_o e^{-\frac{b6T}{2m} }$

=> $\frac{1}{e} =e^{-\frac{b6T}{2m} }$

=> $e^{-1} =e^{-\frac{b6T}{2m} }$

=> $-\frac{3T b}{m} = -1$

=> $b = \frac{m}{3T}$

Generally the amplitude of the section considered is mathematically represented as

$A = \frac{n * F }{ b * 2 \pi }$

=> $A = \frac{n * F }{ \frac{m}{3T} * 2 \pi }$

=> $n = A * \frac{m}{3} * \frac{2\pi}{25}$

=> $n = 0.075 * \frac{288000}{3} * \frac{2* 3.142 }{25}$

=> $n = 1810 \ people$

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3 years ago

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stira [4]

Answer:

0.09 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity.

The S.I unit of acceleration is m/s².

From the question, expression for acceleration is given as

F' = ma

Using Pythagoras Theory,

√(F₁²+F₂²) = ma................... Equation 1

Where F₁ = Force of the First person on the boulder, F₂ = Force of the Second person on the boulder, F' = resultant force acting on the boulder, m = mass of the boulder, a = acceleration of the boulder.

make a the subject of the equation

a = √(F₁²+F₂²) /m................ Equation 2

Given: m = 825 kg, F₁ = 64 N, F₂ = 38 N,

Substitute into equation 2

a = [√(64²+38²)]/825

a = {√(5540)}/825

a = 74.43/825

a = 0.09 m/s²

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Electric field charge E = kq / r^2
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3 years ago

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