a) Potential energy: 147 m [J]
The gravitational potential energy of an object is given by
where
m is its mass
is the acceleration of gravity
h is the height of the object above the ground
In this problem,
h = 15 m
We call 'm' the mass of the ball, since we don't know it
So, the potential energy of the ball at the top of the hill is
(J)
b) Velocity of the ball at the bottom of the hill: 17.1 m/s
According to the law of conservation of energy, in absence of friction all the potential energy of the ball is converted into kinetic energy as the ball reaches the bottom of the hill. Therefore we can write:
where
v is the final velocity of the ball
We know from part a) that
U = 147 m
Substituting into the equation above,
And re-arranging for v, we find the velocity:
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.
Answer:
Hipparchus was an ancient Greek who classified stars based on the brightness in 129 B.C. He grouped the brightest stars and ranked them 1 (first magnitude) and dimmest stars as 6 (sixth magnitude). Thus, the smaller numbers indicated brighter stars. Now, the scale extends in negative axis as well. More the negative number, brighter is the star. For example, Sun has magnitude -26.74.
This the apparent magnitude which means the classification is based on the brightness of the star as it appears from the Earth.
