2. The length of a day is based on the amount of time that

(a) Write an equation describing a sinusoidal transverse wave traveling on a cord in the positive of a y axis with an angular wa

Vedmedyk [2.9K]

Missing data: the wave number

$k=60 cm^{-1}$

(a) $z = 0.003 sin (6000y-31.4 t)$

For a transverse wave travelling in the positive y-direction and with vibration along the z-direction, the equation of the wave is

$z = A sin (ky-\omega t)$

where

A is the amplitude of the wave

k is the wave number

$\omega$ is the angular frequency

t is the time

In this situation:

A = 3.0 mm = 0.003 m is the amplitude

$k = 60 cm^{-1} = 6000 m^{-1}$ is the wave number

$T = 0.20 s$ is the period, so the angular frequency is

$\omega=\frac{2\pi}{T}=\frac{2\pi}{0.20}=31.4 rad/s$

So, the wave equation (in meters) is

$z = 0.003 sin (6000y-31.4 t)$

(b) 0.094 m/s

For a transverse wave, the transverse speed is equal to the derivative of the displacement of the wave, so in this case:

$v_t = z' = -A \omega cos (ky-\omega t)$

So the maximum transverse wave occurs when the cosine term is equal to 1, therefore the maximum transverse speed must be

$v_{t}_{max} =\omega A$

where

$\omega = 31.4 rad/s\\A = 0.003 m$

Substituting,

$v_{t}_{max}=(31.4)(0.003)=0.094 m/s$

(c) 5.24 mm/s

The wave speed is given by

$v=f \lambda$

where

f is the frequency of the wave

$\lambda$ is the wavelength

The frequency can be found from the angular frequency:

$f=\frac{\omega}{2\pi}=\frac{31.4}{2\pi}=5 Hz$

While the wavelength can be found from the wave number:

$\lambda = \frac{2\pi}{k}=\frac{2\pi}{6000}=1.05\cdot 10^{-3} m$

Therefore, the wave speed is

$v=(5)(1.05\cdot 10^{-3} )=5.24 \cdot 10^{-3} m/s = 5.24 mm/s$

7 0

3 years ago

A child with a weight of 230 N swings on a playground swing attached to 1.90 m long chains. What is the gravitational potential

nlexa [21]

Answer:

437 J

Explanation:

Parameters given:

Weight of child, W = 230 N

Height of swing, h = 1.9 m

Gravitational Potential Energy is given as:

P. E. = m*g*h = W*h

m = mass

h = height above the ground

W = weight

P. E. = 230 * 1.9

P. E. = 437 J

7 0

2 years ago

Read 2 more answers

An alkane has molecular weight 86. Write its molecular formula. What will be its physical state?

Brums [2.3K]

C6H14

Gaseous state

it's unsaturated hence gaseous

8 0

2 years ago

A circular ring with area 4.45 cm2 is carrying a current of 13.5 A. The ring, initially at rest, is immersed in a region of unif

Gwar [14]

Answer:

a) ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ ) N.m

b) ΔU = -0.000747871 J

c) w = 47.97 rad / s

Explanation:

Given:-

- The area of the circular ring, A = 4.45 cm^2

- The current carried by circular ring, I = 13.5 Amps

- The magnetic field strength, vec ( B ) = (1.05×10−2T).(12i^+3j^−4k^)

- The magnetic moment initial orientation, vec ( μi ) = μ.(−0.8i^+0.6j^)

- The magnetic moment final orientation, vec ( μf ) = -μ k^

- The inertia of ring, T = 6.50×10^−7 kg⋅m2

Solution:-

- First we will determine the magnitude of magnetic moment ( μ ) from the following relation:

μ = N*I*A

Where,

N: The number of turns

I : Current in coil

A: the cross sectional area of coil

- Use the given values and determine the magnitude ( μ ) for a single coil i.e ( N = 1 ):

μ = 1*( 13.5 ) * ( 4.45 / 100^2 )

μ = 0.0060075 A-m^2

- From definition the torque on the ring is the determined from cross product of the magnetic moment vec ( μ ) and magnetic field strength vec ( B ). The torque on the ring in initial position:

vec ( τi ) = vec ( μi ) x vec ( B )

= 0.0060075*( -0.8 i^ + 0.6 j^ ) x 0.0105*( 12 i^ + 3 j^ -4 k^ )

= ( -0.004806 i^ + 0.0036045 j^ ) x ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

- Perform cross product:

$\left[\begin{array}{ccc}i&j&k\\-0.004806&0.0036045&0\\0.126&0.0315&-0.042\end{array}\right] = \left[\begin{array}{ccc}-0.00015139\\-0.00020185\\-0.00060556\end{array}\right] \\\\$

- The initial torque ( τi ) is written as follows:

vec ( τi ) = ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ )

- The magnetic potential energy ( U ) is the dot product of magnetic moment vec ( μ ) and magnetic field strength vec ( B ):

- The initial potential energy stored in the circular ring ( Ui ) is:

Ui = - vec ( μi ) . vec ( B )

Ui =- ( -0.004806 i^ + 0.0036045 j^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

Ui = -[( -0.004806*0.126 ) + ( 0.0036045*0.0315 ) + ( 0*-0.042 )]

Ui = - [(-0.000605556 + 0.00011)]

Ui = 0.000495556 J

- The final potential energy stored in the circular ring ( Uf ) is determined in the similar manner after the ring is rotated by 90 degrees with a new magnetic moment orientation ( μf ) :

Uf = - vec ( μf ) . vec ( B )

Uf = - ( -0.0060075 k^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

Uf = - [( 0*0.126 ) + ( 0*0.0315 ) + ( -0.0060075*-0.042 ) ]

Uf = -0.000252315 J

- The decrease in magnetic potential energy of the ring is arithmetically determined:

ΔU = Uf - Ui

ΔU = -0.000252315 - 0.000495556

ΔU = -0.000747871 J

Answer: There was a decrease of ΔU = -0.000747871 J of potential energy stored in the ring.

- We will consider the system to be isolated from any fictitious forces and gravitational effects are negligible on the current carrying ring.

- The conservation of magnetic potential ( U ) energy in the form of Kinetic energy ( Ek ) is valid for the given application:

Ui + Eki = Uf + Ekf

Where,

Eki : The initial kinetic energy ( initially at rest ) = 0

Ekf : The final kinetic energy at second position

- The loss in potential energy stored is due to the conversion of potential energy into rotational kinetic energy of current carrying ring.

-ΔU = Ekf

0.5*T*w^2 = -ΔU

w^2 = -ΔU*2 / T

Where,

w: The angular speed at second position

w = √(0.000747871*2 / 6.50×10^−7)

w = 47.97 rad / s

6 0

3 years ago

A mass of 1.00×10−6 μkg is the same as

tankabanditka [31]

I'm not sure but I know u is 10^6

6 0

2 years ago