Answer:
h = 13.06 m
Explanation:
Given:
- Specific gravity of gasoline S.G = 0.739
- Density of water p_w = 997 kg/m^3
- The atmosphere pressure P_o = 101.325 KPa
- The change in height of the liquid is h m
Find:
How high would the level be in a gasoline barometer at normal atmospheric pressure?
Solution:
- When we consider a barometer setup. We dip the open mouth of an inverted test tube into a pool of fluid. Due to the pressure acting on the free surface of the pool, the fluid starts to rise into the test-tube to a height h.
- The relation with the pressure acting on the free surface and the height to which the fluid travels depends on the density of the fluid and gravitational acceleration as follows:
P = S.G*p_w*g*h
Where, h = P / S.G*p_w*g
- Input the values given:
h = 101.325 KPa / 0.739*9.81*997
h = 13.06 m
- Hence, the gasoline will rise up to the height of 13.06 m under normal atmospheric conditions at sea level.
Speed is the magnitude of velocity.
Velocity depends on both magnitude as well as direction.
Since speed is constant, direction of the object must be changed to make velocity not constant or to make velocity variable.
Hope this helps, have a great day ahead!
The temperature of the water and the and the salinity of water
The answer to the first question is that, the sumo wrestler loses energy equal to the absorbed energy of the trampoline plus the air friction drag loss. With the second question, the gymnast should maintain energy equal to the Potential energy gymnast has at the peak of bounce height by exerting little force to compensate for the energy spent on the internal friction of the trampolines action.
To hit the target the crew drop the crate before the plane is directly over the target. It is because <span>because the cargo has forward velocity and therefore before it reaches the ground it travels some distance. The answer is A. Hope it helps. </span>