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
Rutherford's experiment<span> utilized positively charged alpha particles (He with a +2 charge) which were deflected by the dense inner mass (nucleus). The conclusion that could be formed from this result was that </span>atoms<span> had an inner core which contained most of the mass of an </span>atom<span> and was positively charged.</span>
The combustion of fossil fuels is releasing more co2 into the atmosphere then what would occur naturally
Answer:
12 nC
Explanation:
Capacity of the parallel plate capacitor
C = ε₀ A/d
ε₀ is constant having value of 8.85 x 10⁻¹² , A area of plate , d is distance between plate
Area of plate = π r²
= 3.14 x (0.8x 10⁻²)²
= 2 x 10⁻⁴
C = ( 8.85 X 10⁻¹² x 2 x 10⁻⁴ ) / 2.8 x 10⁻³
= 7.08 x 10⁻¹³
Potential difference between plate = field strength x distance between plate
= 6 x 10⁶ x 2.8 x 10⁻³
= 16.8 x 10³ V
Charge on plate = CV
=7.08 x 10⁻¹³ X 16.8 X 10³
11.9 X 10⁻⁹ C
12 nC .
