Ask question

Ask question

All categories

2 years ago

15

20. GP A sinusoidal wave traveling in the negative x direction (to the left) has an amplitude of 20.0cm, a wavelength of 35.0cm,

and a frequency of 12.0Hz. The transverse position of an element of the medium at t=0, x=0 is y= 3.00 cm, and the element has a positive velocity here. We wish to find an expression for the wave function describing this wave. the wave Find (d) the angular frequency omega and

1 answer:

umka21 [38]2 years ago

7 0

The angular frequency of the wave is determined as 75.4 rad/s.

<h3>What is wave function?</h3>

A wave function is a mathematical equation for the motion of the wave.

y(x, t) = A sin(kx + ωt + Φ)

where;

ω is angular speed
k is angular wavenumber
Φ is phase angle

<h3>What is angular frequency?</h3>

The angular frequency is the angular displacement of any wave element per unit of time or the rate of change of the waveform phase.

<h3>Angular frequency</h3>

ω = 2πf

ω = 2π(12)

ω = 75.4 rad/s

Thus, the angular frequency of the wave is determined as 75.4 rad/s.

Learn more about angular frequency here: brainly.com/question/3654452

#SPJ4

You might be interested in

An electric motor rotating a workshop grinding wheel at a rate of 151 rev/min is switched off. Assume constant angular decelerat

tamaranim1 [39]

Answer:

Explanation:

Initial angular velocity ω₀ = 151 x 2π / 60

= 15.8 rad /s

final velocity = 0

Angular deceleration α = 2.23 rad / s

ω² = ω₀² - 2 α θ

0 = 15.8² - 2 x 2.23 θ

= 55.99 rad

one revolution = 2π radian

55.99 radian = 55.99 / 2 π no of terns

= 9 approx .

8 0

3 years ago

A luggage handler pulls a suitcase of mass 19.6 kg up a ramp inclined at an angle 24.0 ∘ above the horizontal by a force F⃗ of m

Dvinal [7]

(a) 638.4 J

The work done by a force is given by

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the direction of the force and the displacement

Here we want to calculate the work done by the force F, of magnitude

F = 152 N

The displacement of the suitcase is

d = 4.20 m along the ramp

And the force is parallel to the displacement, so $\theta=0^{\circ}$ . Therefore, the work done by this force is

$W_F=(152)(4.2)(cos 0)=638.4 J$

b) -328.2 J

The magnitude of the gravitational force is

W = mg

where

m = 19.6 kg is the mass of the suitcase

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting,

$W=(19.6)(9.8)=192.1 N$

Again, the displacement is

d = 4.20 m

The gravitational force acts vertically downward, so the angle between the displacement and the force is

$\theta= 90^{\circ} - \alpha = 90+24=114^{\circ}$

Where $\alpha = 24^{\circ}$ is the angle between the incline and the horizontal.

Therefore, the work done by gravity is

$W_g=(192.1)(4.20)(cos 114^{\circ})=-328.2 J$

c) 0

The magnitude of the normal force is equal to the component of the weight perpendicular to the ramp, therefore:

$R=mg cos \alpha$

And substituting

m = 19.6 kg

g = 9.8 m/s^2

$\alpha=24^{\circ}$

We find

$R=(19.6)(9.8)(cos 24)=175.5 N$

Now: the angle between the direction of the normal force and the displacement of the suitcase is 90 degrees:

$\theta=90^{\circ}$

Therefore, the work done by the normal force is

$W_R=R d cos \theta =(175.4)(4.20)(cos 90)=0$

d) -194.5 J

The magnitude of the force of friction is

$F_f = \mu R$

where

$\mu = 0.264$ is the coefficient of kinetic friction

R = 175.5 N is the normal force

Substituting,

$F_f = (0.264)(175.5)=46.3 N$

The displacement is still

d = 4.20 m

And the friction force points down along the slope, so the angle between the friction and the displacement is

$\theta=180^{\circ}$

Therefore, the work done by friction is

$W_f = F_f d cos \theta =(46.3)(4.20)(cos 180)=-194.5 J$

e) 115.7 J

The total work done on the suitcase is simply equal to the sum of the work done by each force,therefore:

$W=W_F + W_g + W_R +W_f = 638.4 +(-328.2)+0+(-194.5)=115.7 J$

f) 3.3 m/s

First of all, we have to find the work done by each force on the suitcase while it has travelled a distance of

d = 3.80 m

Using the same procedure as in part a-d, we find:

$W_F=(152)(3.80)(cos 0)=577.6 J$

$W_g=(192.1)(3.80)(cos 114^{\circ})=-296.9 J$

$W_R=(175.4)(3.80)(cos 90)=0$

$W_f =(46.3)(3.80)(cos 180)=-175.9 J$

So the total work done is

$W=577.6+(-296.9)+0+(-175.9)=104.8 J$

Now we can use the work-energy theorem to find the final speed of the suitcase: in fact, the total work done is equal to the gain in kinetic energy of the suitcase, therefore

$W=\Delta K = K_f - K_i\\W=\frac{1}{2}mv^2\\v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(104.8)}{19.6}}=3.3 m/s$

6 0

3 years ago

Jimena is stopped at a red light. When the light turns green, she

Ludmilka [50]

First one accelerated

8 0

3 years ago

In what layer of earth does the temperature rise the fastest <br><br> This is for earth science!!!

CaHeK987 [17]

Answer:

thermosphere

Explanation:

i think

6 0

3 years ago

Question 7 Points 3 5 A vehicle is traveling with the velocity of 120 km/h. How much distance will it cover in 30 s? d

ludmilkaskok [199]

Answer:

Velocity v = 120 km/h

Time t = 30 s

Let Distance be d

First, we need to convert 120 km/h into m/s

1 km/h = (1000m)/(3600 s) = 5/18 m/s

So, 120 km/h = 120×5/18 m/s

So, v = 200/6 m/s

Now, velocity = distance/time

So d = v×t

So, d = (200/6) × 30

So, d = 1000 m

So, d = 1 km

Thus, distance travelled is 1 km

Explanation:

8 0

3 years ago

Read 2 more answers