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bekas [8.4K]
3 years ago
13

What is the pressure of the gas in this mercury manometer if h = 89 mm and atmospheric pressure is 775 mmhg?

Physics
2 answers:
ludmilkaskok [199]3 years ago
6 0
The pressure of the gas is 686 mmhg.
If h = 89 mm
and atmospheric pressure = 775 mmhg
Pressure of the gas = ?
We can find the pressure of the gas by finding the difference between both values.
pressure of the gas = 775 mmhg - 89 mm = 686 mmhg
aleksandr82 [10.1K]3 years ago
3 0

The pressure of the gas in this mercury manometer ( Figure 1 ) is about 864 mmHg

\texttt{ }

<h3>Further explanation</h3>

The basic formula of pressure that needs to be recalled is:

Pressure = Force / Cross-sectional Area

or symbolized:

\large {\boxed {P = F \div A} }

<em>P = Pressure (Pa)</em>

<em>F = Force (N)</em>

<em>A = Cross-sectional Area (m²)</em>

Let us now tackle the problem !

\texttt{ }

In this problem , we will use Ideal Gas Law as follows:

<u>Given:</u>

height of mercury column = h = 89 mm

atmospheric pressure = Po = 775 mmHg

<u>Asked:</u>

the pressure of the gas = P = ?

<u>Solution:</u>

We will use Hydrostatic Pressure formula to solve this problem as follows:

P = Po + \rho g h

P = 775 \texttt{ mmHg} + 89 \texttt{ mmHg}

P = 864 \texttt{ mmHg}

\texttt{ }

<h3>Conclusion :</h3>

The pressure of the gas in this mercury manometer ( Figure 1 ) is about 864 mmHg

\texttt{ }

<h3>Learn more</h3>
  • Minimum Coefficient of Static Friction : brainly.com/question/5884009
  • The Pressure In A Sealed Plastic Container : brainly.com/question/10209135
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Pressure

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From the window of a house that is placed 15 m
kow [346]

Answer:

a) 52.915 m

b) The vertical velocity is approximately 21.092 m/s

The resultant velocity is approximately 26.5 m/s

Explanation:

a) The height of the window in the house from which the water was thrown = 15 m

The speed of the stream of water thrown = 20 m/s

The angle at which the water was thrown = 37° over the horizontal

The acceleration due to gravity, g = 10 m/s²

a) The distance from the base of the house at which the water will fall is given as follows;

y = y₀ + u·t·sin(θ) + 1/2·g·t²

Where;

y = The vertical height reached    

u = The initial velocity

t = Time of flight

From the point the steam of water is thrown, we get;

y₀ = 15 m

Therefore;

y = 15 + 20 × t × sin(37°) - 1/2 × 10 × t²

y = 15 + 20 × t × sin(37°) - 5 × t²

When y = 0, Ground level, we get

0 = 15 + 20 × t × sin(37°) - 5 × t²

5·t² - 20×sin(37°)×t -15 = 0

∴ t = (20 ×sin(37°) ± √((-20 × ·sin(37°))² - 4 × (5) × (-15)))/(2 × 5)

t ≈ 3.3128302, or t ≈ 0.906

Therefore, the time of flight of the water, t ≈ 3.3128302 seconds

The distance from the base of the house at which the water will fall = The horizontal distance travelled by the water, x

x = u·cos(θ)×t

∴ x = 20 × cos(37°) × 3.3128302 ≈ 52.915

The distance from the base of the house at which the water will fall = x ≈ 52.915 m

b) The velocity at which the water will reach the ground, 'v', is given as follows;

The vertical velocity, v_y = u·sin(θ)·t - g·t

At the ground, t ≈ 3.3128302 seconds

∴ v_y = 20 × sin(37) - 10 × 3.3128302 ≈ -21.092

The vertical velocity at which the water will reach the ground, v_y ≈ 21.092 m/s (downwards)

The resultant velocity, v = √(v_y² + vₓ²)

∴ v = √(21.092² + (0 × cos(37°))²) ≈ 26.5

The resultant velocity at which the water will reach the ground, v ≈ 26.5 m/s.

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