Answer:
The intensity level of the sound wave due to the ambulance is 153.5 dB.
Explanation:
The intensity level of the sound wave due to the ambulance can be calculated using the following equation:
<u>Where</u>:
I: is the intensity of the sound wave from a siren = 111.2 W/m²
I₀: is the reference intensity = 1.0x10⁻¹² W/m²
Now, since the second sound wave from a nearby ambulance has an intensity level 13 dB we have:
Therefore, the intensity level of the sound wave due to the ambulance is 153.5 dB.
I hope it helps you!
Answer:
work done = 117 kJ
Explanation:
given data
mass m = 3 kg
constant pressure P = 200 kPa
temperature T = 200°C
solution
we know that work done by steam is express as
work done = pressure × ΔVolume ....................1
and here ΔVolume = final volume - initial volume
we use here steam table and get at pressure 200 kPa
final specific volume = 1.08052 m³/kg
and
initial specific volume = 0.885735 m³/kg
so here
ΔV = 3 × (1.08052 - 0.885735)
ΔV = 0.584 m³
so put value in equation 1 we get
work done by steam = 200 × 0.584
work done = 117 kJ
<span>The speed of a wave, V, is f *lambda. Where f is the frequency and lambda is the distance. If a new crest reaches the end every 4 secs; it takes 8s to cover the distance. Hence, f, which is the number of oscillations covered is 8s. So we have V = 8 * 5 = 40 ms^1.</span>
At a point near the rim of the disk, it will have a<span> non-zero radial acceleration and a zero tangential acceleration. Also known as centripetal acceleration, radial acceleration takes place along the radius of the disk. On the other hand, the tangential acceleration is along the path of disk's motion.</span>
Look first for the relation between deBroglie wavelength (λ) and kinetic energy (K):
K = ½mv²
v = √(2K/m)
λ = h/(mv)
= h/(m√(2K/m))
= h/√(2Km)
So λ is proportional to 1/√K.
in the potential well the potential energy is zero, so completely the electron's energy is in the shape of kinetic energy:
K = 6U₀
Outer the potential well the potential energy is U₀, so
K = 5U₀
(because kinetic and potential energies add up to 6U₀)
Therefore, the ratio of the de Broglie wavelength of the electron in the region x>L (outside the well) to the wavelength for 0<x<L (inside the well) is:
1/√(5U₀) : 1/√(6U₀)
= √6 : √5
