Answer:
121.3 cm^3
Explanation:
P1 = Po + 70 m water pressure (at a depth)
P2 = Po (at the surface)
T1 = 4°C = 273 + 4 = 277 K
V1 = 14 cm^3
T2 = 23 °C = 273 + 23 = 300 K
Let the volume of bubble at the surface of the lake is V2.
Density of water, d = 1000 kg/m^3
Po = atmospheric pressure = 10^5 N/m^2
P1 = 10^5 + 70 x 1000 x 10 = 8 x 10^5 N/m^2
Use the ideal gas equation
By substituting the values, we get
V2 = 121.3 cm^3
Thus, the volume of bubble at the surface of lake is 121.3 cm^3.
Answer:
0.37 m
Explanation:
The angular frequency, ω, of a loaded spring is related to the period, T, by
The maximum velocity of the oscillation occurs at the equilibrium point and is given by
A is the amplitude or maximum displacement from the equilibrium.
From the the question, T = 0.58 and A = 25 cm = 0.25 m. Taking π as 3.142,
To determine the height we reached, we consider the beginning of the vertical motion as the equilibrium point with velocity, v. Since it is against gravity, acceleration of gravity is negative. At maximum height, the final velocity is 0 m/s. We use the equation
is the final velocity,
is the initial velocity (same as v above), a is acceleration of gravity and h is the height.