Answer:
x = 2,864 m
, Ra = 32.1 m
Explanation:
Let's solve this problem in parts, let's start by finding the intensity of the sound in each observer
observer A β = 64 db
β = 10 log Iₐ / I₀
where I₀ = 1 10⁻¹² W / m²
Iₐ = I₀ 10 (β/ 10)
let's calculate
Iₐ = 1 10⁻¹² (64/10)
Iₐ = 2.51 10⁻⁶ W / m²
Observer B β = 85 db
I_b = 1 10-12 10 (85/10)
I_b = 3.16 10⁻⁴ W / m²
now we use that the emitted power that is constant is the intensity over the area of the sphere where the sound is distributed
P = I A
therefore for the two observers
P = Ia Aa = Ib Ab
the area of a sphere is
A = 4π R²
we substitute
Ia 4pi Ra2 = Ib 4pi Rb2
Ia Ra2 = Ib Rb2
Let us call the distance from the observer be to the haughty R = ax, so the distance from the observer A to the haughty is R = 35 ax; we substitute
Ia (35 -x) 2 = Ib x2
we develop and solve
35-x = Ra (Ib / Ia) x
35 = [Ra (Ib / Ia) +1] x
x (11.22 +1) = 35
x = 35 / 12.22
x = 2,864 m
This is the distance of observer B
The distance from observer A
Ra = 35 - x
Ra = 35 - 2,864
Ra = 32.1 m
Out of the following choices given, if you insert a piece of rubber between the wires, the electrons do not flow. The rubber material is an insulator. The correct answer is D.
Igneous rocks are formed from molten rock cooling. Rift zones are a common place for them to form because there is two tectonic plates connecting which is a common place for volcanoes to form.
Answer:
correct option is b. 31.3 m/s
Explanation:
given data
artificial gravity a1 = 1 g
artificial gravity a2 = 2 g
diameter = 100 m
radius r= 50 m
speed v1 = 22.1 m/s
solution
As acceleration is ∝ v²
so we can say
.....................1
put here value
solve it
v2 = 
× 22.1
v2 = 31.25 m/s
so correct option is b. 31.3 m/s
