Describe three possible careers in physical science

A sinusoidal wave is traveling along a rope. The oscillator that generates the wave completes 40.0 vibrations in 30.0s . A given

Phantasy [73]

From the solution that I have done, the wavelength in the question that we have is 31.88 cm

<h3>How to solve for the wavelength</h3>

The frequency in the question is given as 40/30 = 1.33 hz

Next we have to solve for V

= 425/10

= 42.5 cm/s

v = frequency * wavelength

we have to put in the values in the formula. This would be

42.5 = 1.33 x wavelength

we have to divide through by 1.33 to get the wavelength. This would be

42.5/1.333 = wavelength

31.88 cm = wavelength

Hence we can say that the wavelength in the question that we have here is 31.88 cm

Read more on wavelength here:

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8 0

1 year ago

A traveling wave on a string can be described by the equation : y = (5.26 ~\text{m}) \cdot \sin \big( (1.65 ~\frac{\text{rad}}{\

zloy xaker [14]

Answer:

t = 1.77 s

Explanation:

The equation of a traveling wave is

y = A sin [2π (x /λ -t /T)]

where A is the oscillation amplitude, λ the wavelength and T the period

the speed of the wave is constant and is given by

v = λ f

Where the frequency and period are related

f = 1 / T

we substitute

v = λ / T

let's develop the initial equation

y = A sin [(2π / λ) x - (2π / T) t +Ф]

where Ф is a phase constant given by the initial conditions

the equation given in the problem is

y = 5.26 sin (1.65 x - 4.64 t + 1.33)

if we compare the terms of the two equations

2π /λ = 1.65

λ = 2π / 1.65

λ = 3.81 m

2π / T = 4.64

T = 2π / 4.64

T = 1.35 s

we seek the speed of the wave

v = 3.81 / 1.35

v = 2.82 m / s

Since this speed is constant, we use the uniformly moving ratios

v = d / t

t = d / v

t = 5 / 2.82

t = 1.77 s

3 0

3 years ago

The site from which an airplane takes off is the origin. The X axis points east, the y axis points straight up. The position and

Contact [7]

Answer:

d = 3.5*10^4 m

Explanation:

In order to calculate the displacement of the airplane you need only the information about the initial position and final position of the airplane. THe initial position is at the origin (0,0,0) and the final position is given by the following vector:

$\vec{r}=(1.21*10^3\hat{i}+3.45*10^4\hat{j})m$

The displacement of the airplane is obtained by using the general form of the Pythagoras theorem:

$d=\sqrt{(x-x_o)^2+(y-y_o)^2}$ (1)

where x any are the coordinates of the final position of the airplane and xo and yo the coordinates of the initial position. You replace the values of all variables in the equation (1):

$d=\sqrt{(1.12*10^3-0)^2+(3.45*10^4-0)^2}=3.45*10^4m$