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aniked [119]
2 years ago
8

A person drops a pebble of mass m1 from a height h, and it hits the floor with kinetic energy KE. The person drops another pebbl

e of mass m2 from a height of 4h, and it hits the floor with the same kinetic energy KE. How do the masses of the pebbles compare
Physics
1 answer:
Leona [35]2 years ago
8 0
<h2>Answer:</h2>

<em>Hello, </em>

<h3><u>QUESTION)</u></h3>

<em>✔ We have: KE = PE (potential energy) </em>

<em>PE = m x g x h </em>

The potential energy that the pebble of mass 1 has is called PE1 and the potential energy that the pebble of mass 2 has is called PE2  

PE1 = PE2 ⇔ PE1/PE2 = 1

\frac{m_1\times g\times h}{m_2\times g\times 4h} = 1 \\ \\  \frac{m_1}{m_2\times 4} = 1 \\ \\  \frac{m_1}{m_2} = 4

The mass m1 is therefore 4 times greater than that of the stone of mass m2.

 

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Two loudspeakers in a 20c room emit 686 hz sound waves along the x-axis.a. if the speakers are in phase, what is the smallest di
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<u>Answer :</u>

(a) d = 0.25 m

(b) d = 0.5 m

<u>Explanation :</u>

It is given that,

Frequency of sound waves, f = 686 Hz

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(1) Perfectly destructive interference occurs when the path difference is half integral multiple of wavelength i.e.

d=\dfrac{\lambda}{2}........(1)

Velocity of sound wave is given by :

v=f\times \lambda

d=\dfrac{v}{2f}

d=\dfrac{343\ m/s}{2\times 686\ Hz}

d=0.25\ m

Hence, when the speakers are in phase the smallest distance between the speakers for which the interference of the sound waves is perfectly destructive is 0.25 m.

(2) For constructive interference, the path difference is integral multiple of wavelengths i.e.

d=n\lambda  ( n = integers )

Let n = 1

So, d=\dfrac{v}{f}

d=\dfrac{343\ m/s}{686\ Hz}

d=0.5\ m

Hence, the smallest distance between the speakers for which the interference of the sound waves is maximum constructive is 0.5 m.

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