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attashe74 [19]
2 years ago
14

A dancer moves around a path like the one

Physics
1 answer:
mash [69]2 years ago
7 0

Given what we know, despite not having the figure attached to the question, we can still confirm that the magnitude for the acceleration of the dancer will be zero.

<h3>Why is the dancer's acceleration equal to zero?</h3>

This has to do with how the question clarifies the speed of the dancer. Though it does not give us an exact value, we are told that the speed is constant. This is an indicator that the acceleration is zero because with any other value for acceleration the speed <u>cannot remain</u> constant.

Therefore, given that any value for acceleration will increase or decrease the speed of the dancer, but we are told that the dancer's speed is constant throughout the trip, we can confirm that the magnitude for the acceleration of the dancer is zero.

To learn more about acceleration visit;

brainly.com/question/12134554?referrer=searchResults

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when asked about weight, it would be 1000N. but in mass , same 600kg.

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4 years ago
By what factor is the intensity of sound at a rock concert louder than that of a whisper when the two intensity levels are 120 d
Kipish [7]

Answer:

The intensity of sound at a rock concert is louder than that of a whisper by a factor of 1 x 10¹⁰

Explanation:

Given;

rock concert sound intensity level, β₁ = 120 dB

whisper sound intensity level, β₂ = 20 dB

The sound intensity level is given as;

\beta = 10Log(\frac{I}{I_o} )\\\\

where;

I₀ is the threshold sound intensity of hearing = 10⁻¹² W/m²

I is the sound intensity

Intensity of sound at rock concert ;

120 =  10Log(\frac{I}{10^{-12}} )\\\\12 =  Log(\frac{I}{10^{-12}} )\\\\10^{12} = \frac{I}{10^{-12}}\\\\I = 10^{12}  * 10^{-12}\\\\I = 10^0\\\\I = 1 \ W/m^2

The intensity of sound of a whisper;

20 =  10Log(\frac{I}{10^{-12}} )\\\\2 =  Log(\frac{I}{10^{-12}} )\\\\10^{2} = \frac{I}{10^{-12}}\\\\I = 10^{2}  * 10^{-12}\\\\I = 10^{-10} \ W/m^2\\\\

Determine the factor by which the intensity of sound at a rock concert louder than that of a whisper

\frac{I_{Concert}}{I_{whisper}} = \frac{1}{10^{-10}} \\\\\frac{I_{Concert}}{I_{whisper}} = 1 * 10^{10}\\\\I_{Concert} = 1 * 10^{10}*I_{whisper}

Therefore, the intensity of sound at a rock concert is louder than that of a whisper by a factor of 1 x 10¹⁰

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3 years ago
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