Water in a tank is pressurized by air and pressure measured using a multi-fluid manometer. Determine the gage pressure of air in
1 answer:
Answer:
The gauge pressure of air is 110 kpa
Explanation:
Atmospheric pressure, = 101 Kpa
where;
ρw is the density of water = 1000 kg/m³
ρo is the density of oil = 800 kg/m³
ρHg is the density of mercury = 13,600 kg/m³
g is acceleration due to gravity = 9.8 m/s²
Therefore, the gauge pressure of air is 110 kpa
You might be interested in
Answer:
E=12.2V/m
Explanation:
To solve this problem we must address the concepts of drift velocity. A drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.
The equation is given by,
Where,
V= Drift Velocity
I= Flow of current
n= number of electrons
q = charge of electron
A = cross-section area.
For this problem we know that there is a rate of 1.8*10^{18} electrons per second, that is
Mobility
We can find the drift velocity replacing,
The electric field is given by,
Explanation:
The internal heat sources for Jupiter and Saturn derive from primordial heat resulting from the initial gravitational contraction of each planet. Jupiter also generates heat by slow contraction, which liberates substantial gravitational energy. A significant part of Saturn’s heat comes from the release of gravitational energy from helium separating from the lighter hydrogen and sinking to its core. What one considers to be a star is a matter of definition, as we discuss in more detail in the chapter on The Birth of Stars and the Discovery of Planets outside the Solar System. While both Jupiter and Saturn generate much of their energy internally, they are not large enough (by a significant factor) to support nuclear reactions in their interiors, and so are not considered to be stars.
