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jonny [76]
3 years ago
13

A student is on a balcony 15.6 m above the street. She throws a ball vertically downward at 10.7 m/s.

Physics
1 answer:
andrew11 [14]3 years ago
3 0

Answer:

20.65 m/s

Explanation:

You are gonna wanna use a calculator for this one

so we will write all the known variables:

h = 15.6 m

initial velocity (u) = 10.7 m/s

acceleration (a) = 10 m/s/s (gravity)

From the third equation of motion:

<em>v² - u² = 2ah</em>

<em>v² - (10.7)² = 2(10)(15.5)</em>

<em>v² - (10.7)² = 312</em>

<em>v² = 312 + 114.49</em>

<em>v² = 426.49</em>

v = 20.65 m/s

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A sound source is moving at 80 m/s toward a stationary listener that is standing in still air (a) Find the wavelength of the sou
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Answer:

a. wavelength of the sound, \vartheta = 1.315\vartheta_{o}

b. observed frequecy, \lambda = 0.7604\lambda_{o}

Given:

speed of sound source, v_{s} = 80 m/s

speed of sound in air or vacuum, v_{a} = 343 m/s

speed of sound observed, v_{o} = 0 m/s

Solution:

From the relation:

v = \vartheta \lambda        (1)

where

v = velocity of sound

\vartheta = observed frequency of sound

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(a) The wavelength of the sound between source and the listener is given by:

\lambda = \frac{v_{a}}{\vartheta }         (2)

(b) The observed frequency is given by:

\vartheta = \frac{v_{a}}{v_{a} - v_{s}}\vartheta_{o}

\vartheta = \frac{334}{334 - 80}\vartheta_{o}

\vartheta = 1.315\vartheta_{o}                (3)

Using eqn (2) and (3):

\lambda = \frac{334}{1.315} = \frac{1}{1.315}\frac{v_{a}}{\vartheta_{o}}

\lambda = 0.7604\lambda_{o}

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