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Physics

1 answer:

Aleonysh [2.5K]3 years ago

3 0

I think the answer is x

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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

77julia77 [94]

Answer:

The intensity of sound at rock concert is 10¹⁰ greater than that of a whisper.

Explanation:

The intensity of sound is given by;

$I(dB) = 10Log(\frac{I}{I_o} )$

where;

I is the intensity of the sound

I₀ is the threshold of sound intensity = 1 x 10⁻¹² W/m²

The intensity of sound at a rock concert

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

The intensity of sound of a whisper

$20 = 10Log(\frac{I}{1*10^{-12}} )\\\\2 = Log(\frac{I}{1*10^{-12}} )\\\\\frac{I}{1*10^{-12}} = 10^{2}\\\\I = 1*10^{-12} *10^{2}\\\\I = 1*10^{-10}\\\\I =10^{-10} \ W/m^2$

Thus, the intensity of sound at rock concert is 10¹⁰ greater than that of a whisper.

4 0

3 years ago

How are frequency and period related to each other? A. They are the same for any given wave B. They have the same magnitude but

BabaBlast [244]

Either of them is. 1/(the other one). That's 'C' .

3 0

3 years ago

Juno created a diagram to describe characteristics of freezing and deposition. Which labels belong in the areas marked X and Y?

DIA [1.3K]

the correct answer to this question is A

7 0

3 years ago

Read 2 more answers

A 19 kg solid disk of radius0.44 m is rotated about an

Vlad1618 [11]

Answer:

0.915 Nm

Explanation:

1 revolution = 2π rad

We can use the following equation of motion to find out the acceleration acting on the disk

$\omega^2 - \omega_0^2 = 2\alpha\Delta \theta$

where $\omega v = 2.5 rad/s is the final angular velocity of the disk, [tex]\omega_0$ = 0 rad/s is the initial velocity of the can when it starts from rest, $\Delta \theta$ is the angular distance traveled, $\alpha$ is the angular acceleration of the disk, which we care looking for: