A mercury manometer is connected to a container of gas. (a) The height of the mercury column on the side connected to the gas is 22.0 cm (measured from the bottom of the manometer). What is the height of the mercury column on the open side if the gauge pressure is measured to be 13.3 kPa? (b) If the gauge pressure of the gas doubles, what are the new heights of the two columns?

(a) The height of the mercury on the open side if the gauge pressure is measured to be 13.3 kPa is 9.97395 cm

(b) When the gauge pressure is doubled;

The height of the mercury on the open side becomes 19.9479 cm

The height of the mercury on the side connected to the gas becomes 12.02605 cm

Explanation:

(a) The given parameters are;

The height of the mercury column on the side connected to the gas = 22.0 cm

The gauge pressure = 13.3 kPa

The gauge pressure is the pressure in the gas container relative to the atmospheric pressure, therefore, we have;

ΔP = P - P₀ = ρ × g × h

Where;

ΔP = The gauge pressure = P - P₀ = 13.3 kPa = 13,300 Pa

P₀ = The atmospheric pressure

P = The pressure of the gas in the container

ρ = The density of the mercury = 13593 kg/m³

g = The acceleration due to gravity = 9.81 m/s²

ΔP = ρ × g × h, by substitution gives

13,300 = 13593 × 9.81 × h

h = 13,300/(13593 × 9.81) ≈ 0.0997395 m = 9.97395 cm

The height of the mercury on the open side if the gauge pressure is measured to be 13.3 kPa is 9.97395 cm

(b) Given that the gauge pressure for the gas doubles, we have;

13,300 × 2 = 13593 × 9.81 × h

Therefore, h = 2 × 9.97395 = 19.9479 cm

The height of the mercury on the open side, h = 19.9479 cm

Therefore, the height of the mercury on the side connected to the gas becomes, 22 - 9.97395 = 12.02605 cm

falling water hydro = water .. spectcular places these plants are

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft