Answer:
a) w₂ = 5.16 rev / s
, b) K₀ = 29.55 J, K₂ = 190.88 J
Explanation:
The skater forms an isolated system whereby its angular momentum is preserved
Initial. With outstretched arms
L₀ = I₀ w₀
Final. With arms stuck
L₂ = I₂W₂
L₀ = L₂
I₀ w₀ = I₂ w₂
w₂ = I₀ / I₂ w₀
Let's calculate
w₂ = 0.800 2.34 / 0.363
w₂ = 5.16 rev / s
b) The kinetic energy is
K = ½ I w²
Let's reduce to the SI system
w₀ = 0.800 rev / s (2π rad / 1 rev) = 5.026 rad / s
w₂ = 5.16 rev / s (2π rad / 1 rev) = 32.42 rad / s
The initial kinetic energy
K₀ = ½ I₀ w₀²
K₀ = ½ 2.34 5,026²
K₀ = 29.55 J
The final kinetic energy
K₂ = ½ I₂ w₂²
K₂ = ½ 0.363 32.42²
K₂ = 190.88 J
Answer:
Newtons Second Law of Motion
Explanation:
Answer:
The value of the power is ![P_c = 38.55 \ W](https://tex.z-dn.net/?f=P_c%20%20%3D%20%2038.55%20%5C%20%20W%20)
Explanation:
From the question we are told that
The power rating ![P_{1000} =P_b= 52 \ W](https://tex.z-dn.net/?f=P_%7B1000%7D%20%3DP_b%3D%20%2052%20%5C%20%20W)
The frequency is ![f = 1000 \ Hz](https://tex.z-dn.net/?f=f%20%3D%201000%20%5C%20%20Hz)
The frequency at which the sound intensity decreases ![f_k = 20 \ Hz](https://tex.z-dn.net/?f=f_k%20%20%3D%20%2020%20%5C%20%20Hz)
The decrease in intensity is by ![\beta = 1.3 dB](https://tex.z-dn.net/?f=%5Cbeta%20%20%3D%20%201.3%20dB)
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
=>
=> ![P_c = \frac{52}{1.34896}](https://tex.z-dn.net/?f=P_c%20%20%3D%20%20%5Cfrac%7B52%7D%7B1.34896%7D)
=> ![P_c = 38.55 \ W](https://tex.z-dn.net/?f=P_c%20%20%3D%20%2038.55%20%5C%20%20W%20)