He produced the first orderly arrangement of known elements, he used patterns to predict undiscovered elements
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 length of a 2 sec pendulum is 1 m.
Given that, initial length of the simple pendulum L₁ = 1 m
Initial time period T₁ = 2 sec
We need to find the length of the pendulum whose time period is 2 sec
T₂ = 2 sec
L₂ = ?
We know that the time period of the simple pendulum is given by the formula,
T = 2π√(L/g)
From the above relation, we can write T ∝ √L
T₁ / T₂ = √(L₁/L₂)
Making L₂ from the above relation, we have,
L₂ = (T₂² * L₁)/ T₁² = 2² * 1/ 2² = 1 m
Thus, the length of a 2 sec pendulum is 1 m.
To know more about time period:
brainly.com/question/17350379
#SPJ4
If each side is 0.1 feet extra,
The volume will be 5.1*2.1*1.1= about 11.781.
Perhaps this helps.