Answer:
The amplitude of the eardrum's oscillation is 6.65×10^-13 m.
Explanation:
Given data:
The sound has a frequency of 262 Hz
The sound level is 84 dB
The air density is 1.21 kg/m^3
The speed of sound is 346 m/s
Solution:
As, Intensity of sound is given by,
I = Io×10^(s/10 db)
I = 2×π^2×ρ×v×f^2×Sm^2
Thus,
Sm = √(Io×10^(s/10 db)) / √( 2×π^2×ρ×v×f^2)
Now, put the values,
Sm = √( 10^-12 × 10^(84/10) ) / √( 2×(3.14)^2×1.21×346×(262)^2 )
Sm = √(2.51×10^-4 / 5.66×10^8)
Sm = √0.443×10^-12
Sm = 6.65×10^-13 m.
Answer:
frequency of a wave can be measured by the number of crests in a second or more
Answer:
option is option d)
Explanation:
the correct option is option d)
the characteristic of thunderstorm can be seen by the following scenario
in thunderstorm there is lot of movement of air the lighter i.e.warmer air moves up and the colder air moves down creating the condition of thunderstorm.
heavy rain , strong wind , potentially large hail stones is the situation of the thunderstorm.
during thunderstorm there is rain and snow at intermediate.
hence only option left is option d.
Answer:
Peak current= 84.86 A
Area of each turn = 0.029 m^2
Explanation:
The peak value of current can be obtained from Irms= 0.707Io. Where Io is the peak current.
Hence;
Irms= 60.0A
Io= Irms/0.707
Io = 60.0/0.707
Io= 84.86 A
Vrms= 0.707Vo
Vo= Vrms/0.707= 170/0.707 = 240.45 V
From;
V0 = NABω
Where;
Vo= peak voltage
N= number of turns
B= magnetic field
A= area of each coil
ω= angular velocity
But ω= 2πf = 2×π×95= 596.9 rads-1
Substituting values;
A= Vo/NBω
A= 240.45/550×0.025×596.9
A= 0.029 m^2
