The loudness of the sound at the rock concert, where the intensity of the sound is1 x 10⁻¹ Wm⁻² is 110 dB.
Here we are dealing with loudness which is the perception of the Intensity of the sound.
The formula to refer to in order to find the value of the loudness of a sound is ,
db= 10log(I/I₀)
As we are provided with the current intensity which is 1 x 10⁻¹ Wm⁻². and the initial intensity which is 1 x 10⁻¹² Wm⁻².
So, by substituting the required values in the formula we get
db= 10 * log( 1 x 10⁻¹ /1 x 10⁻¹²)
= 10 * 11 log(10)
= 110
So, the result is 110 dB.
To know more about the intensity of sound refer to the link brainly.com/question/9323731?referrer=searchResults.
To know more about questions related to loudness refer to the link brainly.com/question/21094511?referrer=searchResults.
#SPJ4
The AREA of the shaded region is the moving object's displacement.
Answer:
<h2>compression</h2>
Explanation:
Such type of the longitudinal waves that causes compression during the travelling through the medium is called as compression. Generally, displacement of the medium takes place in this wave travelling conditions. There are many waves that are considered as longitudinal waves such as sound waves, seismic p-waves and some other. Waves have different types of properties that are changed due to different medium.
Answer:
0.0133 A
Explanation:
The time at which B=1.33 T is given by
1.33 = 0.38*t^3
t = (1.33/0.38)^(1/3) = 1.52 s
Using Faraday's Law, we have
emf = - dΦ/dt = - A dB/dt = - A d/dt ( 0.380 t^3 )
Area A = pi * r² = 3.141 *(0.025 *0.025) = 0.00196 m²
emf = - A*(3*0.38)*t^2
thus, the emf at t=1.52 s is
emf = - 0.00196*(3*0.38)*(1.52)^2 = -0.0052 V
if the resistance is 0.390 ohms, then the current is given by
I = V/R = 0.0052/0.390 = 0.0133 A
Answer:
Explanation:
Given that,
Mass of the thin hoop
M = 2kg
Radius of the hoop
R = 0.6m
Moment of inertial of a hoop is
I = MR²
I = 2 × 0.6²
I = 0.72 kgm²
Period of a physical pendulum of small amplitude is given by
T = 2π √(I / Mgd)
Where,
T is the period in seconds
I is the moment of inertia in kgm²
I = 0.72 kgm²
M is the mass of the hoop
M = 2kg
g is the acceleration due to gravity
g = 9.8m/s²
d is the distance from rotational axis to center of of gravity
Therefore, d = r = 0.6m
Then, applying the formula
T = 2π √ (I / MgR)
T = 2π √ (0.72 / (2 × 9.8× 0.6)
T = 2π √ ( 0.72 / 11.76)
T = 2π √0.06122
T = 2π × 0.2474
T = 1.5547 seconds
T ≈ 1.55 seconds to 2d•p
Then, the period of oscillation is 1.55seconds
