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
Explanation:
Given:
u = 20 m/s
a = 5 m/s^2
v = 30 m/s
t = ?
Use the first kinematic equation of motion:
v = u + at
t = (v - u)/a = 10/5 = 2 seconds
Answer:
answer is 2 option because more force is applied
Answer:
– 2.5 m/s²
Explanation:
We have,
• Initial velocity, u = 180 km/h = 50 m/s
• Final velocity, v = 0 m/s (it stops)
• Time taken, t = 20 seconds
We have to find acceleration, a.
a = (v ― u)/t
a = (0 – 50)/20 m/s²
a = –50/20 m/s²
a = – 5/2 m/s²
a = – 2.5 m/s² (Velocity is decreasing) [Answer]
5.5 s
Explanation:
The time it takes for the ball to reach its maximum height can be calculated using

since
at the top of its trajectory. Plugging in the numbers,
