Which of the following changes would decrease the force between two charged particles by a factor of 9? A. Decreasing the distan
1 answer:
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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
Answer:
-24.28571 rad/s²
29.57239 revolutions
3.91176 seconds
52.026478 m
Explanation:
= Tangential acceleration = -6.8 m/s²
r = Radius of wheel = 0.28
= Initial angular velocity = 95 rad/s
= Angle of rotation
= Final angular velocity
t = Time taken
Angular acceleration is given by
The angular acceleration is -24.28571 rad/s²
The number of revolutions is 29.57239
The time it takes for the car to stop is 3.91176 seconds
Linear distance
The distance the car travels is 52.026478 m