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Sladkaya [172]
1 year ago
8

An open tank contain oil of specific gravity 0.75 on top of

Engineering
1 answer:
n200080 [17]1 year ago
8 0

Let's not take into account atmospheric pressure.

Pressure P = Pressure of oil column + Pressure of water column

= h1 . d1.g + h2.d2. g

=( 0.5 x 800 x 9.8) + ( 1x 1000x 9.8) = 3920+ 9800 = 13720

Hence guage pressure at bottom is

P= 13720 N/ m^2

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Explanation:

White Board Activity: Practice: A sound has a frequency of 800 Hz. What is the period of the wave? The wave repeats 800 times in 1 second and the period of the function is 1/800 or 0.00125.

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if both the ram air input and drain hole of the pitot system become blocked, the indicated airspeed will
egoroff_w [7]

If both the ram air input and drain hole of the pitot system become blocked, the indicated airspeed will: a) increase during a climb.

<h3>What is a ram air input?</h3>

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This ultimately implies that, a ram air input allows a greater mass-flow of air through the engine of an automobile, thereby, increasing the engine's power.

In conclusion, indicated airspeed will increase during a climb when both the ram air input and drain hole of the pitot system become blocked.

Read more on pilots here: brainly.com/question/10381526

#SPJ1

Complete Question:

If both the ram air input and drain hole of the pitot system become blocked, the indicated airspeed will

a) increase during a climb

b) decrease during a climb

c) remain constant regardless of altitude change

6 0
2 years ago
What types of issues MAY occur to slow or prevent the best outcome?
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Which goal incorporates most of the criteria required for a SMART goal?
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8 0
3 years ago
A rigid, well-insulated tank of volume 0.9 m is initially evacuated. At time t = 0, air from the surroundings at 1 bar, 27°C beg
Eva8 [605]

Answer:

\dot{w}= -0.303 KW

Explanation:

This is the case of unsteady flow process because properties are changing with time.

From first law of thermodynamics for unsteady flow process

\dfrac{dU}{dt}=\dot{m_i}h_i+\dot{Q}-\dot{m_e}h_i+\dot{w}

Given that tank is insulated so\dot{Q}=0 and no mass is leaving so

\dot{m_e}=0

\int dU=\int \dot{m_i}h_i\ dt-\int \dot{w}\ dt

m_2u_2-m_1u_1=(m_2-m_1)h_i- \dot{w}\Delta t

Mass conservation m_2-m_1=m_e-m_i

m_1,m_2 is the initial and final mass in the system respectively.

Initially tank is evacuated so m_1=0

We know that for air u=C_vT ,h=C_p T,P_2v_2=m_2RT_2

m_2=0.42 kg

So now putting values

0.42 \times 0.71 \times 730=0.42\times 1.005\times 300- \dot{w} \times 300

\dot{w}= -0.303 KW

3 0
3 years ago
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