Based on the direction of propagation compared to direction of vibration, waves are classified into:
1- Transverse waves: The direction of propagation of the wave is perpendicular to the direction of vibration of the medium particles.
2- Longitudinal waves: The direction of propagation of the wave is the same as the direction of vibration of the medium particles.
For the question we have here, since the direction of the wave is the same as the direction of vibration of particles, therefore, this wave is a longitudinal wave
Answer:
The time rate of change in air density during expiration is 0.01003kg/m³-s
Explanation:
Given that,
Lung total capacity V = 6000mL = 6 × 10⁻³m³
Air density p = 1.225kg/m³
diameter of the trachea is 18mm = 0.018m
Velocity v = 20cm/s = 0.20m/s
dv /dt = -100mL/s (volume rate decrease)
= 10⁻⁴m³/s
Area for trachea =

0 - p × Area for trachea =



⇒

ds/dt = 0.01003kg/m³-s
Thus, the time rate of change in air density during expiration is 0.01003kg/m³-s
Answer: The power is 156 watt
Explanation:
is in the attachment
Answer:
0.00001266 m
Explanation:
D = Distance from source to screen
m = Order
d = Slit separation
The distance from a point on the screen to the center line

At m = 0


At m = 1

The slit separation is 0.00001266 m
Answer: The final volume V₂ of the container is 0.039 m³.
Explanation:
Since the temperature is constant, the gas would expand isothermally.
For isothermal expansion,
P₁V₁=P₂V₂
Where, P₁ and P₂ are the initial and final pressure and V₁ and V₂ are initial and final volume.
It is given that:
V₁ = 0.0250 m³
P₁ = 1.5 × 10⁶ Pa
P₂ = 0.950 × 10⁶ Pa
V₂ = ?
⇒ 1.5 × 10⁶ Pa × 0.0250 m³ = 0.950 × 10⁶ Pa × V₂
⇒V₂ = 0.039 m³
Hence, the final volume V₂ of the container is 0.039 m³.