Answer:
![\mathbf{\beta = 123.75 \ dB}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Cbeta%20%3D%20123.75%20%5C%20dB%7D)
Explanation:
From the question, using the expression:
![125 \ dB = 10 \ log (\dfrac{I}{I_o})](https://tex.z-dn.net/?f=125%20%5C%20dB%20%3D%2010%20%5C%20log%20%28%5Cdfrac%7BI%7D%7BI_o%7D%29)
where;
![I_o = 10^{-12} \ W/m^2](https://tex.z-dn.net/?f=I_o%20%3D%2010%5E%7B-12%7D%20%5C%20W%2Fm%5E2)
![I = 10^{12.5} \times 10^{-12} \ W/m^2](https://tex.z-dn.net/?f=I%20%3D%2010%5E%7B12.5%7D%20%5Ctimes%2010%5E%7B-12%7D%20%5C%20W%2Fm%5E2)
![I = 3.162 \ W/m^2](https://tex.z-dn.net/?f=I%20%3D%203.162%20%5C%20W%2Fm%5E2)
This is a combined intensity of 4 speakers.
Thus, the intensity of 3 speakers = ![\dfrac{3.162\times 3}{4}](https://tex.z-dn.net/?f=%5Cdfrac%7B3.162%5Ctimes%203%7D%7B4%7D)
= 2.372 W/m²
Thus;
![\beta = 10 \ log ( \dfrac{2.372}{10^{-12}} ) \ W/m^2](https://tex.z-dn.net/?f=%5Cbeta%20%3D%2010%20%5C%20%20log%20%28%20%5Cdfrac%7B2.372%7D%7B10%5E%7B-12%7D%7D%20%29%20%5C%20W%2Fm%5E2)
![\mathbf{\beta = 123.75 \ dB}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Cbeta%20%3D%20123.75%20%5C%20dB%7D)
Answer:
I think u have to do 580N times 40KG
M1 v1 = (m1 + m2)v2.
All of the exponents should be lowered to the bottom right of the letters.
Answer:
a) ![s \approx 6.676\,m](https://tex.z-dn.net/?f=s%20%5Capprox%206.676%5C%2Cm)
Explanation:
a) Let assume that the ground is not inclined, since the bottom of the playground slide is tangent to ground. Then, the length of given by the definition of a circular arc:
![s = \frac{\pi}{4}\cdot R](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B%5Cpi%7D%7B4%7D%5Ccdot%20R)
![s=\frac{\pi}{4}\cdot (8.5\,m)](https://tex.z-dn.net/?f=s%3D%5Cfrac%7B%5Cpi%7D%7B4%7D%5Ccdot%20%288.5%5C%2Cm%29)
![s \approx 6.676\,m](https://tex.z-dn.net/?f=s%20%5Capprox%206.676%5C%2Cm)
The bottom of the slide has a height of zero. The physical phenomenon around Dr. Ritchey's daughter is modelled after Principle of Energy Conservation. The child begins at rest:
![U_{g,A} = K_{B} + W_{fr}](https://tex.z-dn.net/?f=U_%7Bg%2CA%7D%20%3D%20K_%7BB%7D%20%2B%20W_%7Bfr%7D)
![m\cdot g \cdot h_{A} = \frac{1}{2}\cdot m \cdot v_{B}^{2} + f\cdot s](https://tex.z-dn.net/?f=m%5Ccdot%20g%20%5Ccdot%20h_%7BA%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20m%20%5Ccdot%20v_%7BB%7D%5E%7B2%7D%20%2B%20f%5Ccdot%20s)
The average frictional force is cleared within the expression:
![f = \frac{m\cdot (g\cdot h_{A}-\frac{1}{2}\cdot v_{B}^{2})}{s}](https://tex.z-dn.net/?f=f%20%3D%20%5Cfrac%7Bm%5Ccdot%20%28g%5Ccdot%20h_%7BA%7D-%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20v_%7BB%7D%5E%7B2%7D%29%7D%7Bs%7D)
![f = \frac{(12\,kg)\cdot [(9.807\,\frac{m}{s^{2}} )\cdot (3\,m)-\frac{1}{2}\cdot (4.5\,\frac{m}{s} )^{2} ]}{6.676\,m}](https://tex.z-dn.net/?f=f%20%3D%20%5Cfrac%7B%2812%5C%2Ckg%29%5Ccdot%20%5B%289.807%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%20%29%5Ccdot%20%283%5C%2Cm%29-%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20%284.5%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%29%5E%7B2%7D%20%5D%7D%7B6.676%5C%2Cm%7D)
![f = 34.684\,N](https://tex.z-dn.net/?f=f%20%3D%2034.684%5C%2CN)
false. all obejects in motion have friction