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3 years ago

A wave is traveling at a speed of 15 m/s and it's wavelength is 5 m. Calculate the waves frequency

Physics

1 answer:

Anvisha [2.4K]3 years ago

3 0

Answer:

The frequency of this wave is $3\; \rm Hz$ .

Explanation:

The frequency $f$ of a wave is the number of wavelengths that this wave covers in unit time (typically a second.)

The wave in this question travels at $v = 15\; \rm m \cdot s^{-1}$ . In other words, this wave covers $15\; \rm m$ in unit time (a second.) How many wavelengths $\lambda$ would that $15\; \rm m\;$ correspond to?

The question states that the wavelength of this wave is $\lambda = 5\; \rm m$ . Therefore, there would be $15 / 5 = 3$ wavelengths in the $15\; \rm m$ span that this wave covered in the unit time of one second ( $1\; \rm s$ .) Hence, the frequency of this wave would be $3\; \rm s^{-1}$ (three per second,) which is equivalent to $3\; \rm Hz$ (three Hertzs.)

In general, the frequency $f$ of a wave with speed $v$ and wavelength $\lambda$ would be:

$\displaystyle f = \frac{v}{\lambda}$ .

For the wave in this question:

$\begin{aligned}f &= \frac{v}{\lambda} \\ &= \frac{15\; \rm m \cdot s^{-1}}{3\; \rm s} = 3\; \rm s^{-1} = 3\; \rm Hz\end{aligned}$ .

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Answer:

Check the explanation

Explanation:

This is the step by step explanation to the above question:

$v_i = v [ f_L *(v - v_b) - f_s*(v + v_b)] / [f_L * (v - v_b) + f_s*(v +v_b)]$

= v * (83.1 * (v-4.3) - 80.7 ( v+4.3))/ [83.1 *(v - 4.3) + 80.7*(v + 4.3)]

v = 344 m/s

vi = 344 * ( 83.1* (344-4.3) - 80.7*(344+4.3) ) / (83.1 *(344 - 4.3) + 80.7*(344 + 4.3))

= 0.74 m/s

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3 years ago

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3 years ago

Read 2 more answers

Monochromatic coherent light shines through a pair of slits. If the wavelength of the light is decreased, which of the following

Law Incorporation [45]

Answer:

he correct answers are a, b

Explanation:

In the two-slit interference phenomenon, the expression for interference is

d sin θ= m λ constructive interference

d sin θ = (m + ½) λ destructive interference

in general this phenomenon occurs for small angles, for which we can write

tanθ = y / L

tan te = sin tea / cos tea = sin tea

sin θ = y / La

derestimate the first two equations.

Let's do the calculation for constructive interference

d y / L = m λ

the distance between maximum clos is and

y = (me / d) λ

this is the position of each maximum, the distance between two consecutive maximums

y₂-y₁ = (L 2/d) λ - (L 1 / d) λ₁ y₂ -y₁ = L / d λ

examining this equation if the wavelength decreases the value of y also decreases

the same calculation for destructive interference

d y / L = (m + ½) κ

y = [(m + ½) L / d] λ

again when it decreases the decrease the distance

the correct answers are a, b

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3 years ago

A major-league pitcher can throw a ball in excess of 40.1 m/s. If a ball is thrown horizontally at this speed, how much will it

mote1985 [20]

Answer:

The ball will drop 0.881 m by the time it reaches the catcher.

Explanation:

The position of the ball at time "t" is described by the position vector "r":

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

Where:

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

When the ball reaches the catcher, the position vector will be "r final" (see attached figure).

The x-component of the vector "r final", "rx final", will be 17.0 m. We have to find the y-component.

Using the equation of the x-component of the position vector, we can calculate the time it takes the ball to reach the catcher (notice that the frame of reference is located at the throwing point so that x0 and y0 = 0):

x = x0 + v0x · t

17.0 m = 0 m + 40.1 m/s · t

t = 17.0 m/ 40. 1 m/s = 0.424 s

With this time, we can calculate the y-component of the vector "r final", the drop of the ball:

y = y0 + v0y · t + 1/2 · g · t²

Initially, there is no vertical velocity, then, v0y = 0.

y = 1/2 · g · t²

y = -1/2 · 9.8 m/s² · (0.424 s)²

y = -0.881 m

The ball will drop 0.881 m by the time it reaches the catcher.

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3 years ago

At a place where an object is thrown vertically downward with a speed of while a different object is thrown vertically upward wi

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Answer:

Both objects will undergo the same change in velocity

Explanation:

m = Mass of the Earth = 5.972 × 10²⁴ kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Earth = 6371000 m

m = Mass of object

Any object which is falling has only the acceleration due to gravity.

$ma=\dfrac{GMm}{r^2}\\\Rightarrow a=\dfrac{GM}{r^2}\\\Rightarrow a=\dfrac{6.67\times 10^{-11}\times 5.972\times 10^{24}}{(6.371\times 10^6)^2}\\\Rightarrow a=9.81364\ m/s^2$

The acceleration due to gravity on Earth is 9.81364 m/s²

So, the speeds of the objects will change at an equal rate of 9.81364 m/s² but the change will be negative when an object is thrown up.

Hence, both objects will undergo the same change in velocity.

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3 years ago