A combination of longitudinal and transverse. :) yw
Answer:
The intensity level in the room is 63 dB
Explanation:
To calculate the intensity of sound in the room, we use the equation of definition of decibels
β = 10 log (I / Io) (1)
With “I” the sound intensity and “Io” the threshold intensity 1.0 10⁻⁻¹² W/m²
To calculate the intensity we will use the initial data and remember the power of the emitted sound is constant, in addition that the sound propagates in three-dimensional form or on a spherical surface
I = P/A ⇒ P = I A
The area of a sphere is 4 π r², where I can calculate of 1
β/10 = log (I/Io)
I / Io = 
I = Io 
I = 1 10⁻¹² 10⁽¹⁰⁰/¹⁰⁾ = 1 10⁻¹² 10¹⁰
I = 1.0 10⁻² W
With this we can calculate the intensity for a distance of 20 m
I = 1.0 10⁻² / ( 4π 20²)
I = 2.0 10⁻⁶ W/m²
We have already found the intensity at the point of interest, so we can calculate the intensity in decibels at this point with equation 1
β = 10 log(2.0 10⁻⁶ / 1.0 10⁻¹²)
β = 10 log ( 2 10⁶) = 10 6.3
β = 63 dB
The intensity level in the room is 63 dB
We can look at all the ages of the earth since it’s a big crack is reveals many layers of the earth and we can know about chemicals and metals that were in earth and diffrent times
+2 electron charges = 2x1.6x10^-19Coulombs
Answer:
The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.
Explanation:
Given that,
Initial velocity u= 128 ft/sec
Equation of height
....(I)
(a). We need to calculate the maximum height
Firstly we need to calculate the time

From equation (I)




Now, for maximum height
Put the value of t in equation (I)


(b). The number of seconds it takes the object to hit the ground.
We know that, when the object reaches ground the height becomes zero




Hence, The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.