Answer:
pressure=height × density×acc due to gravity
so
pressire is directly proportional to height hence it decreses with decrease in height
here air column height is measured upside down so decreases witn increment
Answer:
f = 614.28 Hz
Explanation:
Given that, the length of the air column in the test tube is 14.0 cm. It can be assumed that the speed of sound in air is 344 m/s. The test tube is a kind of tube which has a closed end. The frequency in of standing wave in a closed end tube is given by :


f = 614.28 Hz
So, the frequency of the this standing wave is 614.28 Hz. Hence, this is the required solution.
Answer:
Explanation:
Here the rope is 10m long and making an angle of θ with the direction of canal or direction of flow of water in the canal .
sinθ = 2 / 10 = .2
θ = 11.5⁰
a )
Component of 500 N along the line perpendicular to the canal
= 500 sinθ
= 500 sin 11.5
= 100 N
b )
component of 500 N along the canal
= 500 cos 11.5
= 490 N approx .
Answer:
volume of the bubble just before it reaches the surface is 5.71 cm³
Explanation:
given data
depth h = 36 m
volume v2 = 1.22 cm³ = 1.22 ×
m³
temperature bottom t2 = 5.9°C = 278.9 K
temperature top t1 = 16.0°C = 289 K
to find out
what is the volume of the bubble just before it reaches the surface
solution
we know at top atmospheric pressure is about P1 =
Pa
so pressure at bottom P2 = pressure at top + ρ×g×h
here ρ is density and h is height and g is 9.8 m/s²
so
pressure at bottom P2 =
+ 1000 × 9.8 ×36
pressure at bottom P2 =4.52 ×
Pa
so from gas law

here p is pressure and v is volume and t is temperature
so put here value and find v1

V1 = 5.71 cm³
volume of the bubble just before it reaches the surface is 5.71 cm³