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kotykmax [81]
3 years ago
8

PSYW - Please Show Your Work

Physics
1 answer:
Ad libitum [116K]3 years ago
7 0

Answer:

9.66E4 J

Explanation:

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A transverse mechanical wave is traveling along a string lying along the x-axis. The displacement of the string as a function of
Wewaii [24]

(1) The wavelength of the wave is 1.164 m.

(2) The  velocity of the wave is 23.7 m/s.

(3) The maximum speed in the y-direction of any piece of the string is 6.14 m/s.

<h3> Wavelength of the wave</h3>

A general wave equation is given as;

y(x, t) = A sin(Kx - ωt)

<h3>Velocity of the wave</h3>

v = ω/K

From the given wave equation, we have,

y(x, t) = 0.048 sin(5.4x - 128t)

v = ω/K

where;

  • ω corresponds to 128
  • k corresponds to 5.4

v = 128/5.4

v = 23.7 m/s

<h3>Wavelength of the wave</h3>

λ = 2π/K

λ = (2π)/(5.4)

λ = 1.164 m

<h3>Maximum speed of the wave</h3>

v(max) = Aω

where;

  • A is amplitude of the wave
  • ω is angular speed of the wave

v(max) = (0.048)(128)

v(max) = 6.14 m/s

Thus, the wavelength of the wave is 1.164 m.

The  velocity of the wave is 23.7 m/s.

The maximum speed in the y-direction of any piece of the string is 6.14 m/s.

Learn more about wavelength here: brainly.com/question/10728818

#SPJ1

3 0
1 year ago
Zad. 2 W marszobiegu chłopiec przebiegł 900 m w ciągu 205 min, a następnie przeszedł 300m w tym danym czasie. Jaka była średnia
USPshnik [31]

Odpowiedź:

0,049 m / s

Wyjaśnienie:

Biorąc pod uwagę, że:

Dystans biegu = 900m

Czas trwania = 205 minut

Długość przejścia = 300 m

Zajęty czas = 205 minut

Średnia prędkość :

(Przebieg + pokonany dystans) / całkowity czas

Średnia prędkość :

(900 m +. 300 m) / 205 + 205

1200 m / 410 minut

Minuty do sekund

1200 / (410 * 60)

1200/24600

= 0,0487804

= 0,049 m / s

8 0
2 years ago
How do the chemical properties of the halogens compare to those of the noble gases?
serg [7]
Halogens<span> are extremely reactive elements because they need one more electron to gain a full octet of valence electrons, whereas the </span>noble gases<span>are extremely unstable because they already have their full octet.</span>
8 0
3 years ago
A wire of radius R has a current I uniformly distributed across its cross-sectional area. Ampere's law is used with a concentric
MrMuchimi

Answer:

Please refer to the figure.

Explanation:

The crucial point here is to calculate the enclosed current. If the current I is flowing through the whole cross-sectional area of the wire, the current density is

J = \frac{I}{\pi R^2}

The current density is constant for different parts of the wire. This idea is similar to that of the density of a glass of water is equal to the density of a whole bucket of water.

So,

J = \frac{I}{\pi R^2} = \frac{I_{enc}}{\pi r^2}\\I_{enc} = \frac{Ir^2}{R^2}

This enclosed current is now to be used in Ampere’s Law.

\mu_o I_{enc} = \int {B} \, dl

Here, \int \, dl represents the circular path of radius r. So we can replace the integral with the circumference of the path, 2\pi r.

As a result, the magnetic field is

B = \frac{\mu_0}{2\pi}\frac{Ir}{R^2}

5 0
3 years ago
How much elastic potential energy is stored in a bungee cord with a spring constant of 10.0 N/m when the cord is stretched 2.00
Aleonysh [2.5K]

Answer:

<em> The elastic potential energy stored in the bungee cord = 20 J</em>

Explanation:

potential energy: This is the energy possessed by a body due to its position. The S.I unit of energy is Joules. The mathematical expression for elastic potential energy is given below

E = 1/2ke²................ Equation 1

Where E = elastic potential energy of the spring, k = force constant of the spring, e = extension

<em>Given: K = 10 N/m, e = 2.00 m</em>

<em>Substituting these values into Equation 1</em>

<em>E = 1/2(10)(2)²</em>

<em>E = 5×4</em>

<em>E = 20 Joules.</em>

<em>Therefore the elastic potential energy stored in the bungee cord = 20 J</em>

<em></em>

6 0
3 years ago
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