The second one is correct not sure about the first one sorry
Answer:
The value of the power is 
Explanation:
From the question we are told that
The power rating 
The frequency is 
The frequency at which the sound intensity decreases 
The decrease in intensity is by 
Generally the initial intensity of the speaker is mathematically represented as
![\beta_1 = 10 log_{10} [\frac{P_b}{P_a} ]](https://tex.z-dn.net/?f=%5Cbeta_1%20%3D%20%2010%20log_%7B10%7D%20%5B%5Cfrac%7BP_b%7D%7BP_a%7D%20%5D)
Generally the intensity of the speaker after it has been decreased is
![\beta_2 = 10 log_{10} [\frac{P_c}{P_a} ]](https://tex.z-dn.net/?f=%5Cbeta_2%20%3D%20%2010%20log_%7B10%7D%20%5B%5Cfrac%7BP_c%7D%7BP_a%7D%20%5D)
So
![\beta_1-\beta_2 = 10 log_{10} [\frac{P_c}{P_a} ]- 10 log_{10} [\frac{P_b}{P_a} ]](https://tex.z-dn.net/?f=%5Cbeta_1-%5Cbeta_2%20%3D%20%2010%20log_%7B10%7D%20%5B%5Cfrac%7BP_c%7D%7BP_a%7D%20%5D-%2010%20log_%7B10%7D%20%5B%5Cfrac%7BP_b%7D%7BP_a%7D%20%5D)
=> ![\beta = 10 log_{10} [\frac{P_c}{P_a} ]- 10 log_{10} [\frac{P_b}{P_a} ]= 1.3](https://tex.z-dn.net/?f=%5Cbeta%20%3D%20%2010%20log_%7B10%7D%20%5B%5Cfrac%7BP_c%7D%7BP_a%7D%20%5D-%2010%20log_%7B10%7D%20%5B%5Cfrac%7BP_b%7D%7BP_a%7D%20%5D%3D%201.3)
=> ![\beta =10log_{10} [\frac{\frac{P_b}{P_a}}{\frac{P_c}{P_a}} ] = 1.3](https://tex.z-dn.net/?f=%5Cbeta%20%3D10log_%7B10%7D%20%5B%5Cfrac%7B%5Cfrac%7BP_b%7D%7BP_a%7D%7D%7B%5Cfrac%7BP_c%7D%7BP_a%7D%7D%20%5D%20%3D%201.3)
=> ![\beta =10log_{10} [\frac{P_b}{P_c} ] = 1.3](https://tex.z-dn.net/?f=%5Cbeta%20%3D10log_%7B10%7D%20%5B%5Cfrac%7BP_b%7D%7BP_c%7D%20%5D%20%3D%201.3)
=> ![10log_{10} [\frac{P_b}{P_c} ] = 1.3](https://tex.z-dn.net/?f=10log_%7B10%7D%20%5B%5Cfrac%7BP_b%7D%7BP_c%7D%20%5D%20%3D%201.3)
=> ![log_{10} [\frac{P_b}{P_c} ] = 0.13](https://tex.z-dn.net/?f=log_%7B10%7D%20%5B%5Cfrac%7BP_b%7D%7BP_c%7D%20%5D%20%3D%200.13)
taking atilog of both sides
=>
=> 
=> 
It's angle of reflection must be 41 degrees
we know, by the first law of reflection that angle of incidence is always equal to angle of reflection..........
Compute the ball's angular speed <em>v</em> :
<em>v</em> = (1 rev) / (2.3 s) • (2<em>π</em> • 180 cm/rev) • (1/100 m/cm) ≈ 4.917 m/s
Use this to find the magnitude of the radial acceleration <em>a</em> :
<em>a</em> = <em>v </em>²/<em>R</em>
where <em>R</em> is the radius of the circular path. We get
<em>a</em> = <em>v</em> ² / (180 cm) = <em>v</em> ² / (1.8 m) ≈ 13.43 m/s²
The only force acting on the ball in the plane parallel to the circular path is the tension force. By Newton's second law, the net force acting on the ball has magnitude
∑ <em>F</em> = <em>m</em> <em>a</em>
where <em>m</em> is the mass of the ball. So, if <em>t</em> denotes the magnitude of the tension force, then
<em>t</em> = (1.6 kg) (13.43 m/s²) ≈ 21 N