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PilotLPTM [1.2K]
2 years ago
6

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?
Physics
1 answer:
umka21 [38]2 years ago
3 0

Answer:

(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

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<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

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Forwad force acting on the boat,v₁ = 1,800-N

Resistive force acting on the boat,v₂ = 1,200-N

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The velocity at the end of that time is found as;

\rm  v_f = v_ 0 + at \\\\ v_f =- 0+ at \\\\ v_f = 8.5 m/sec

Hence the acceleration of the boat, the distance through which the boat moves in 20.0 s, and velocity at the end of that time will be 0.425 m/sec²,85 m, and 8.5 m/sec.

To learn more about the velocity, refer to the link: brainly.com/question/862972

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