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1 year ago

A transverse wave on a string is described by the wave function

Physics

1 answer:

svp [43]1 year ago

5 0

The period of the transverse wave from what we have here is 0.5

<h3>How to find the period of the transverse wave</h3>

The period of a wave can be defined as the time that it would take for the wave to complete one complete vibrational cycle.

The formula with which to get the period is

w = 4π

where w = 4 x 22/7

2π/T = 4π

6.2857/T = 12.57

From here we would have to cross multiply

6.2857 = 12.57T

divide through by 12.57

6.2857/12.57 = T

0.500 = T

Hence we can conclude that the value of T that can determine the period based on the question is 0.500.

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The answer would be in the chart or graph A is 1 B is 2

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3 years ago

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Devon was watching an old television show about people stranded on an island. One of the characters wanted to make an electric g

navik [9.2K]

D. a and c

Explanation:

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The magnets provide magnetic field for the process.
The coil of wire is placed in between the magnetic field generated by the magnets.
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3 years ago

Four traveling waves are described by the following equations, where all quantities are measured in SI units and y represents th

agasfer [191]

Answer:

$T_1=T_3=\dfrac{2\pi}{21}$

$T_2=T_4=\dfrac{2\pi}{42}$

Explanation:

Wave 1, $y_1=0.12\ cos(3x-21t)$

Wave 2, $y_2=0.15\ sin(6x+42t)$

Wave 3, $y_3=0.13\ cos(6x+21t)$

Wave 4, $y_4=-0.27\ sin(3x-42t)$

The general equation of travelling wave is given by :

$y=A\ cos(kx\pm \omega t)$

The value of $\omega$ will remain the same if we take phase difference into account.

For first wave,

$\omega_1=21$

$\dfrac{2\pi }{T_1}=21$

$T_1=\dfrac{2\pi}{21}$

For second wave,

$\omega_2=42$

$\dfrac{2\pi }{T_2}=42$

$T_2=\dfrac{2\pi}{42}$

For the third wave,

$\omega_3=21$

$\dfrac{2\pi }{T_3}=21$

$T_3=\dfrac{2\pi}{21}$

For the fourth wave,

$\omega_4=42$

$\dfrac{2\pi }{T_4}=42$

$T_4=\dfrac{2\pi}{42}$

It is clear from above calculations that waves 1 and 3 have same time period. Also, wave 2 and 4 have same time period. Hence, this is the required solution.

3 0

3 years ago

A long solenoid that has 1 200 turns uniformly distributed over a length of 0.420 m produces a magnetic field of magnitude 1.00

Fantom [35]

Answer:

<h2>The current required winding is $2.65*10^-^2 mA$ </h2>

Explanation:

We can use the expression B=μ₀*n*I-------1 for the magnetic field that enters a coil and

n= N/L (number of turns per unit length)

Given data

The number of turns n= 1200 turns

length L= 0.42 m

magnetic field B= 1*10^-4 T

μ₀= $4\pi*10^-^7 T.m/A$

Applying the equation B=μ₀*n*I

I= B/μ₀*n

I= B*L/μ₀*n

$I= \frac{1*10^-^4*0.42}{4\pi*10^-^7*1.2*10^3 }$

$I= 2.65*10^-^2 mA$

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3 years ago

Describe the importance of SI units?

alex41 [277]

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Explanation:

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3 years ago