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lara [203]
3 years ago
7

a boy stands 80m in front of a tall cliff and produces sound by clapping two planks together. A girl standing next to him time t

he return of the echo at 0.5. how fast does sound wave travel in this place​
Physics
1 answer:
harina [27]3 years ago
4 0

Answer:

320 m/s

Explanation:

Speed = Distance / Time

= 160m/0.5s

= <u>3</u><u>2</u><u>0</u><u> </u><u>m</u><u>/</u><u>s</u>

<em>[</em><em>N</em><em>o</em><em>t</em><em>e</em><em>:</em><em> </em><em>T</em><em>h</em><em>e</em><em> </em><em>d</em><em>i</em><em>s</em><em>t</em><em>a</em><em>n</em><em>c</em><em>e</em><em> </em><em>i</em><em>s</em><em> </em><em>1</em><em>6</em><em>0</em><em>m</em><em> </em><em>a</em><em>s</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>s</em><em>o</em><em>u</em><em>n</em><em>d</em><em> </em><em>t</em><em>r</em><em>a</em><em>v</em><em>e</em><em>l</em><em>l</em><em>e</em><em>d</em><em> </em><em>t</em><em>o</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>c</em><em>l</em><em>i</em><em>f</em><em>f</em><em> </em><em>b</em><em>e</em><em>f</em><em>o</em><em>r</em><em>e</em><em> </em><em>r</em><em>e</em><em>f</em><em>l</em><em>e</em><em>c</em><em>t</em><em>i</em><em>n</em><em>g</em><em> </em><em>b</em><em>a</em><em>c</em><em>k</em><em>.</em><em> </em><em>H</em><em>e</em><em>n</em><em>c</em><em>e</em><em> </em><em>i</em><em>t</em><em> </em><em>t</em><em>r</em><em>a</em><em>v</em><em>e</em><em>l</em><em>l</em><em>e</em><em>d</em><em> </em><em>8</em><em>0</em><em> </em><em>+</em><em> </em><em>8</em><em>0</em><em> </em><em>=</em><em> </em><em>1</em><em>6</em><em>0</em><em> </em><em>m</em><em>]</em>

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A net force of –8750N is used to stop of 1250.kg car travelling 25m/s. What braking distance is needed to bring the car to a hal
Karolina [17]

Answer:

d = 44.64 m

Explanation:

Given that,

Net force acting on the car, F = -8750 N

The mass of the car, m = 1250 kg

Initial speed of the car, u = 25 m/s

Final speed, v = 0 (it stops)

The formula for the net force is :

F = ma

a is acceleration of the car

a=\dfrac{F}{m}\\\\a=\dfrac{-8750}{1250}\\\\a=-7\ m/s^2

Let d be the breaking distance. It can be calculated using third equation of motion as :

v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0^2-(25)^2}{2\times (-7)}\\\\d=44.64\ m

So, the required distance covered by the car is 44.64 m.

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Minchanka [31]

Answer:

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Explanation:

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Question 9
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Answer

Can the wave travel between the Sun and Earth.

Explanation:

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Where else could Snell’s law be useful in determining the path of a light ray in your everyday life?
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Contact glasses.

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