Answer:38.66 m
Explanation:
Given
launch angle
launch velocity
center of mass continue to travel its original Path so it center of mass will be at a distance of
Center of mass will be at
(b)if one of the piece will be at x=26 m then other will be at
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.