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3 years ago

Wha is the frequency of a wave having a period equal to 18 seconds?

Physics

1 answer:

Rainbow [258]3 years ago

5 0

Answer:

5.5 × 10-2 hertz

Explanation:

The time taken by a wave crest to travel a distance equal to the length of wave is known as wave period.

= 0.055 per second (1 cycle per second = 1 Hertz)

Thus, we can conclude that the frequency of the wave is 5.5 X 10^{-2} hertz.

Hopes this helps, love <3

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A ball is thrown upward from the top of a building at an angle of 30.0° to the horizontal and with

Ray Of Light [21]

Answer:

See below

Explanation:

Vertical position = 45 + 20 sin (30) t - 4.9 t^2

when it hits ground this = 0

0 = -4.9t^2 + 20 sin (30 ) t + 45

0 = -4.9t^2 + 10 t +45 = 0 solve for t =4.22 sec

max height is at t= - b/2a = 10/9.8 =1.02

use this value of 't' in the equation to calculate max height = 50.1 m

it has 4.22 - 1.02 to free fall = 3.2 seconds free fall

v = at = 9.81 * 3.2 = 31.39 m/s VERTICAL

it will <u>also</u> still have horizontal velocity = 20 cos 30 = 17.32 m/s

total velocity will be sqrt ( 31.39^2 + 17.32^2) = 35.85 m/s

Horizontal range = 20 cos 30 * t = 20 * cos 30 * 4.22 = 73.1 m

8 0

1 year ago

A delivery truck leaves a warehouse and travels 3.20 km east. The truck makes a right turn and travels 2.45 km south to arrive a

boyakko [2]

Explanation:

It is given that,

Displacement of the delivery truck, $d_1=3.2\ km$ (due east)

Then the truck moves, $d_2=2.45\ km$ (due south)

Let d is the magnitude of the truck’s displacement from the warehouse. The net displacement is given by :

$d=\sqrt{d_1^2+d_2^2}$

$d=\sqrt{3.2^2+2.45^2}$

d = 4.03 km

Let $\theta$ is the direction of the truck’s displacement from the warehouse from south of east.

$\theta=tan^{-1}(\dfrac{2.45}{3.2})$

$\theta=37.43^{\circ}$

So, the magnitude and direction of the truck’s displacement from the warehouse is 4.03 km, 37.4° south of east.

8 0

3 years ago

An electron emitted in the beta decay of bismuth-210 has a mean kinetic energy of 390 keV. (a) Find the de Broglie wavelength of

Sauron [17]

Explanation:

Given that,

The mean kinetic energy of the emitted electron, $E=390\ keV=390\times 10^3\ eV$

(a) The relation between the kinetic energy and the De Broglie wavelength is given by :

$\lambda=\dfrac{h}{\sqrt{2meE}}$

$\lambda=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 390\times 10^3}}$

$\lambda=1.96\times 10^{-12}\ m$

(b) According to Bragg's law,

$n\lambda=2d\ sin\theta$

n = 1

For nickel, $d=0.092\times 10^{-9}\ m$

$\theta=sin^{-1}(\dfrac{\lambda}{2d})$

$\theta=sin^{-1}(\dfrac{1.96\times 10^{-12}}{2\times 0.092\times 10^{-9}})$

$\theta=0.010^{\circ}$

As the angle made is very small, so such an electron is not useful in a Davisson-Germer type scattering experiment.

4 0

3 years ago

On what criteria is the Fujita scale for Tornadoes based?

svp [43]

B. Tornado destruction
It is based on the amount of damage

7 0

3 years ago

Ian and Dane are twins. Ian loves art and music and work after school. Dane loves sports and would do anything to avoid having a

Brilliant_brown [7]

Answer:

Explanation:

Pliability

5 0

3 years ago