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laila [671]
2 years ago
10

A wave on a string is described by

Physics
1 answer:
liberstina [14]2 years ago
5 0

Explanation:

A wave on a string is described is given by :

D(x,t)=2\ cm\ sin[(12.57\ rad/m)-(638\ rad/s)t]

The linear density of the string is 5 g/m.

Where

x is in meters and t is in seconds

The general equation of a wave is given by :

y=A\ sin(kx-\omega t)

(2) The speed of the wave in terms of tension is given by :

v=\sqrt{\dfrac{T}{\mu}}

Also, v=\dfrac{\omega}{k}

So, \dfrac{\omega}{k}=\sqrt{\dfrac{T}{\mu}}

T=\dfrac{\mu \omega^2}{k^2}

T=\dfrac{5\times 10^{-3}\times (638)^2}{(12.57)^2}

T = 12.88 N

(3) The maximum displacement of a point on the string is equal to the amplitude of the wave. So, the maximum displacement is 2 cm.

(4) The maximum speed of a point on the string is given by :

v=A\omega

v=0.02\times 638

v = 12.76 m/s

Hence, this is the required solution.

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Unlike acceleration and velocity, speed does not need to specify
frosja888 [35]

Unlike acceleration and velocity, speed does not need to specify the direction of motion. Speed is a scalar quality.

4 0
2 years ago
Find an expression for the torsional constant k in terms of the moment of inertia I of the disk and the angular frequency ω of s
sasho [114]

Answer:

Explanation:

The general equation for the disk with moment of inertia I when given small angular displacement  \theta is given by

I\frac{\mathrm{d^2} \theta }{\mathrm{d} t^2}=-k\theta

\frac{\mathrm{d^2} \theta }{\mathrm{d} t^2}+\frac{k\theta }{I}=0

Replacing

\frac{k\theta }{I}=\omega ^2

where \omega is the angular frequency of oscillation

General solution for this Equation is given by

\theta =\theta _{max}\sin \left ( \omega t+\phi \right )

where \theta _{max}=maximum\ angular\ displacement

\phi =Phase\ difference

Thus K can be written as

k=I\omega ^2

5 0
2 years ago
Odległość między kolejnymi grzbietami fal na morzu wynosi 20 m. Łódź opada z grzbietu fali, unosi się i osiąga ponownie najwyższ
Veronika [31]

Answer:

Explanation:

The distance between successive wave crests at sea is 20 m. The boat descends from the crest of the wave, rises and reaches the highest position again within 5 s. Calculate the wave propagation speed.

Given that,

The distance between two successive crest is 20m

Wavelength is the distance between two successive crest or trough

Then, it's wavelength is λ = 20m

The time to reached the maximum height is 5seconds, then it will take (5×4) to complete one period

Then,

Period T = 20seconds

From wave equation

v = fλ

Where

v is speed

f is frequency and

λ is wavelength

The frequency is related to the period

f =  1 / T

Then,

v = λ / T

So, v = 20 / 20

v = 1 m/s

The speed of propagation of the wave is 1m/s

To Polish

Jeśli się uwzględni,

Odległość między dwoma kolejnymi grzebieniami wynosi 20 m

Długość fali to odległość między dwoma kolejnymi grzebieniami lub dolinami

Zatem jego długość fali wynosi λ = 20 m

Czas do osiągnięcia maksymalnej wysokości wynosi 5 sekund, a następnie ukończenie jednego okresu zajmie (5 × 4)

Następnie,

Okres T = 20 sekund

Z równania falowego

v = fλ

Gdzie

v to prędkość

f oznacza częstotliwość, a

λ jest długością fali

Częstotliwość jest związana z okresem

f = 1 / T

Następnie,

v = λ / T

Zatem v = 20/20

v = 1 m / s

Prędkość propagacji fali wynosi 1m/s

6 0
2 years ago
Suppose an isolated box of volume 2V is divided into two equal compartments. An ideal gas occupies half of the container and the
SpyIntel [72]

Answer:

A. the internal energy stays the same

Explanation:

From the first law of thermodynamics, "energy can neither be created nor destroyed but can be transformed from one form to another.

Based on this first law of thermodynamic, the new internal energy of the gas is the same as the internal energy of the original system.

Therefore, when the partition separating the two halves of the box is removed and the system reaches equilibrium again, the internal energy stays the same.

7 0
2 years ago
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