Answer:
3 times louder
Explanation:
The Loudness in decibel Db L = 10㏒(I/I₀) where I = sound intensity level and I₀ = threshold of hearing = 10⁻¹² W/m².
Now, for Jessica, I₁ = sound intensity level of Jessica's music = 10⁻⁹
and I₂ = sound intensity level of Braylee's music = 10⁻³
So, substituting the variables into the equation, we have
L₁ = 10㏒(I₁/I₀)
L₁ = 10㏒(10⁻⁹/10⁻¹²)
L₁ = 10㏒(10³)
L₁ = 3 × 10㏒10
L₁ = 30㏒10
L₁ = 30 dB
Now, for Braylee, I₂ = sound intensity level of Braylee's music = 10⁻³
So, substituting the variables into the equation, we have
L₂ = 10㏒(I₁/I₀)
L₂ = 10㏒(10⁻³/10⁻¹²)
L₂ = 10㏒(10⁹)
L₂ = 9 × 10㏒10
L₂ =90㏒10
L₂ = 90 dB
So, the number of times Braylee's music is louder than Jessica's music is L₂/L₁ = 90 dB/30 dB = 3
So, Braylee's music is 3 times louder than Jessica's music
iIn this case the mass of a body cannot be considered to be concentrated at the centre of mass of the body for the purpose of computing the rotational motion
Therefore the answer is False
Pressure at a given surface is given as ratio of normal force and area
so here force due to heel of the shoes is given as 80 N
and the area of the heel is given as 16 cm^2
so we can say

here we have
F = 80 N



so pressure at the surface due to its heel will be 5 * 10^4 N/m^2
Answer:
U = 11.67 W/m² °C
Explanation:
Inner diameter, d = 2.5 cm = 0.025 m
Thickness of the wall, t = 2 mm = 0.002 m
thus, the outer diameter, D = d + 2t = 0.025 + 2 x 0.002 = 0.029 m


Now, based on outside convection heat transfer
we have

on rearranging, we get

where,
are the perimeter respective to inner and and outer diameter
on substituting the values, we get

or
U = 11.67 W/m² °C