Answer:
P₁- P₂ = 91.1 10³ Pa
Explanation:
For this exercise we will use Bernoulli's equation, where point 1 is at the bottom of the house and point 2 on the second floor
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
P1-P2 = ½ ρ (v₂² - v₁²) + ρ g (y₂-y₁)
In the exercise they give us the speeds and the height of the turbid, so we can calculate the pressure difference
For heights let's set a reference system on the ground floor of the house, so we have 5m for the second floor and an entrance at -2m
P₁-P₂ = ½ 1.0 10³ (7² - 2²) + 1.0 10³ 9.8 (5 + 2)
P₁-P₂ = 22.5 10³ + 68.6 10³
P₁- P₂ = 91.1 10³ Pa
Answer:
The intensity of sound at a rock concert is louder than that of a whisper by a factor of 1 x 10¹⁰
Explanation:
Given;
rock concert sound intensity level, β₁ = 120 dB
whisper sound intensity level, β₂ = 20 dB
The sound intensity level is given as;
where;
I₀ is the threshold sound intensity of hearing = 10⁻¹² W/m²
I is the sound intensity
Intensity of sound at rock concert ;
The intensity of sound of a whisper;
Determine the factor by which the intensity of sound at a rock concert louder than that of a whisper
Therefore, the intensity of sound at a rock concert is louder than that of a whisper by a factor of 1 x 10¹⁰
According to Newton's third law, the foot and the soccer ball<span> exert equal and opposite </span>forces<span> on each other. An </span>action force<span> and the </span>reaction force<span> that results are called a </span>force<span> pair. Newton's third law states that the </span>forces<span> in a </span>force<span> pair are equal in size, but opposite in direction.
hope it helped</span>
