Answer:
distance is 13 m for 100 dB
distance is 409 km for 10 dB
Explanation:
Given data
distance r = 2.30 m
source β = 115 dB
to find out
distance at sound level 100 dB and 10 dB
solution
first we calculate here power and intensity and with this power and intensity we will find distance
we know sound level β = 10 log(I/
) ......................a
put here value (I/) = 10^−12 W/m² and β = 115
115 = 10 log(I/10^−12)
so
I = 0.316228 W/m²
and we know power = intensity × 4π r² ...............b
power = 0.316228 × 4π (2.30)²
power = 21.021604 W
we know at 100 dB intensity is 0.01 W/m²
so by equation b
power = intensity × 4π r²
21.021604 = 0.01 × 4π r²
so by solving r
r = 12.933855 m = 13 m
distance is 13 m
and
at 10 dB intensity is 1 × 10^–11 W/m²
so by equation b
power = intensity × 4π r²
21.021604 = 1 × 10^–11 × 4π r²
by solving r we get
r = 409004.412465 m = 409 km
The answer is
C. periodic fluctuations in the intensity if sound waves.
Answer:magnitude -5; angle 160°
Explanation:
Vector A is described as having magnitude 5 and angle -20°.
To get an equivalent vector, we either leave the magnitude at 5 and add 360° to the angle, or we reverse the magnitude to -5 and add 180° to the angle.
5 @ -20° = 5 @ 340°
5 @ -20° = -5 @ 160°
The third one is the answer.
<span>Acceleration is the rate of
change of the velocity of an object that is moving. This value is a result of
all the forces that is acting on an object which is described by Newton's
second law of motion. Calculations of such is straightforward, if we are given
the final velocity, the initial velocity and the total time interval. However, we are not given these values. We are only left by using the kinematic equation expressed as:
d = v0t + at^2/2
We cancel the term with v0 since it is initially at rest,
d = at^2/2
44 = a(6.2)^2/2
a = 2.3 m/s^2
</span>
Answer:
When the liquid moves through the hydrosphere, the water collects into a cloud. When it falls to the earth, turning into snow and sleet collecting in rivers and lakes.
Explanation:
Hope that helps
