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.
Alpha particles are composite particles consisting of two protons and two neutrons tightly bound together.
(not much explanation, I hope this is what you were looking for!)
Answer:
The temperature reported by a thermometer is never precisely the same as its surroundings
Explanation:
In this experiment to determine the specific heat of a material the theory explains that when a heat interchange takes place between two bodies that were having different temperatures at the start, the quantity of heat the warmer body looses is equal to that gained by the cooler body to reach the equilibrium temperature. <u>This is true only if no heat is lost or gained from the surrounding.</u> If heat is gained or lost from the surrounding environment, the temperature readings by the thermometer will be incorrect. The experimenter should therefore keep in mind that for accurate results, the temperature recorded by the thermometer is similar to that of the surrounding at the start of the experiment and if it differs then note that there is either heat gained or lost to the environment.
Explanation:
33. The 1.5kg owl is now soaring at 20m/s. What is the owl’s KE?
a. Step 1: Formula <u>½mv²</u>
b. Step 2: Data m = <u>1</u><u>.</u><u>5</u><u> </u><u>kg</u>, v = <u>2</u><u>0</u><u> </u><u>m</u><u>/</u><u>s</u>
c. Step 3: Solve
KE = (1/2)(<u>1</u><u>.</u><u>5</u>)(<u>2</u><u>0</u>)² = <u>3</u><u>0</u><u>0</u><u> </u><u>J</u>