Answer:
y₀ = 1020.3 m
Explanation:
This is a projectile launching exercise, in this case as the package is released its initial vertical velocity is zero.
y = y₀ +
t - ½ g t²
when it reaches the ground its height is zero
0 = y₀ + 0 - ½ g t²
y₀ = ½ g t²
let's calculate
y₀ = ½ 9.8 14.43²
y₀ = 1020.3 m
Answer:
The increase in potential energy of the ball is 115.82 J
Explanation:
Conceptual analysis
Potential Energy (U) is the energy of a body located at a certain height (h) above the ground and is calculated as follows:
U = m × g × h
U: Potential Energy in Joules (J)
m: mass in kg
g: acceleration due to gravity in m/s²
h: height in m
Equivalences
1 kg = 1000 g
1 ft = 0.3048 m
1 N = 1 (kg×m)/s²
1 J = N × m
Known data




Problem development
ΔU: Potential energy change
ΔU = U₂ - U₁
U₂ - U₁ = mₓgₓh₂ - mₓgₓh₁
U₂ - U₁ = mₓg(h₂ - h₁)

The increase in potential energy of the ball is 115.82 J
Answer:
The specific gravity of the unkown liquid is 15.
Explanation:
Gauge pressure, at the bottom of the tank in this case, can be calculated from

where
and
are the height of the column of oil and the unkown liquid, respectively. Writing for
, we have

Relative to water, the unknow liquid specific weight is 15 times bigger, therefore this is its specific gravity as well.
Explanation:
that the people closer too the head of the table will feel more vibrations than the people at the end of the table. since the vibrations will slow down as they travel farther down the table
Hope this helps!!