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2 years ago

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On a given day, a barometer at the base of the Washington Monument reads 29.97 in. of mercury. What would the barometer reading

be when you carry it up to the observation deck 500 ft above the base of the monument?

1 answer:

Umnica [9.8K]2 years ago

8 0

Answer:

The barometer reading will be 29.43 in

Explanation:

Using the formula of pressure variation

p2 - p1 = -yair * H

= 7.65 * $10^{-2} \frac{lb}{ft^{3} } * 500 ft\\$

= 38.5 $\frac{lb}{ft^{2} }$

According to the relationship between the pressure and the height of the mercury column

p = yHg * h --> where yHg and h is the barometer reading

yHg $(\frac{29.97}{12} ft)$ - yHg * h1 = 38.5 $\frac{lb}{ft^{2} }$

h1 = ( $\frac{29.97}{12} ft$ ) - $\frac{38.5 \frac{lb}{ft^{2} } }{847 \frac{lb}{ft^{3} } }$

$[(\frac{29.97}{12} ft) - 0.0455 ft] - 12 \frac{in}{ft} \\\\h1 = 29.43 in$

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Steam enters a turbine operating at steady state at 1 MPa, 200 °C and exits at 40 °C with a quality of 83%. Stray heat transfer

Andrei [34K]

Answer:

(a) Work out put=692.83 $\frac{KJ}{Kg}$

(b) Change in specific entropy=0.0044 $\frac{KJ}{Kg-K}$

Explanation:

Properties of steam at 1 MPa and 200°C

$h_1=2827.4\frac{KJ}{Kg},s_1=6.69\frac{KJ}{Kg-K}$

We know that if we know only one property in side the dome then we will find the other property by using steam property table.

Given that dryness or quality of steam at the exit of turbine is 0.83 and temperature T=40°C.So from steam table we can find pressure corresponding to saturation temperature 40°C.

Properties of saturated steam at 40°C

$h_f= 167.5\frac{KJ}{Kg} ,h_g= 2537.4\frac{KJ}{Kg}$

$s_f= 0.57\frac{KJ}{Kg-K} ,s_g= 8.25\frac{KJ}{Kg-K}$

So the enthalpy of steam at the exit of turbine

$h_2=h_f+x(h_g-h_f)\frac{KJ}{Kg}$

$h_2=167.5+0.83(2537.4-167.5)\frac{KJ}{Kg}$

$h_2=2134.57\frac{KJ}{Kg}$

$s_2=s_f+x(s_g-s_f)\frac{KJ}{Kg-K}$

$s_2=0.57+0.83(8.25-0.57)\frac{KJ}{Kg-K}$

$s_2=6.6944\frac{KJ}{Kg-K}$

(a)

Work out put = $h_1-h_2$

=2827.4-2134.57 $\frac{KJ}{Kg}$

Work out put =692.83 $\frac{KJ}{Kg}$

(b) Change in specific entropy

$s_2-s_1=6.6944-6.69\frac{KJ}{Kg-K}$

Change in specific entropy =0.0044 $\frac{KJ}{Kg-K}$

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2 years ago

The small washer is sliding down the cord OA. When it is at the midpoint, its speed is 28 m/s and its acceleration is 7 m/s 2 .

Neporo4naja [7]

Answer:

Velocity components

$V_r = -16.28 m/s$

$V_z = -22.8 m/s$

$V_q = 0 m/s$

For Acceleration components;

$a_r = -4.07m/s^2$

$a_z = -5.70m/s^2$

$a_q = 0m/s^2$

Explanation:

We are given:

$Speed v_o = 28 m/s$

$Acceleration a_o= 7 m/s^2$

We first need to find the radial position r of washer in x-y plane.

Therefore

$r = \sqrt{300^2 + 400^2}$

r = 500 mm

To find length along direction OA we have:

$L = \sqrt{500^2 + 700^2}L = 860 mm$

Therefore, the radial and vertical components of velocity will be given as:

$V_r = V_o*cos(Q)$

$V_z = V_o*sin(Q)$

Where Q is the angle between OA and vector r.

Therefore,

$V_r = 28 * \frac{r}{L} = > 28 * \frac{500}{860}$

$V_r = -16.28 m/s$

• $V_z = 28 * \frac{700}{860} = -22.8$

• $V_q = 0 m/s$

The radial and vertical components of acceleration will be:

$a_r = a_o*cos(Q)$

$a_z = a_o*sin(Q)$

Therefore we have:

• $a_r = 7* \frac{500}{860} = -4.07m/s^2$

• $a_z = 7 * \frac{700}{860} = -5.70 m/s^2$

• $a_q = 0 m/s^2$

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2 years ago

Michelle is the general manager of a power plant. This morning, she will meet with city officials to discuss environmental issue

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Answer:

interpersonal.

Explanation:

Out of all the activities performed by Michelle, three activities involves the interpersonal skills.

1. Meeting with city officials

2. Meeting with section managers

3. Handling the complaint filed by an employee

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Answer:

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Explanation:

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The hydro static force is given by $Force=Pressure\times Area$

We know that the hydrostatic pressure on a rectangular surface in vertical position is given by $Pressure=\rho \times g\times h_{c.g}$

For the given rectangular surface we have $h_{c.g}=\frac{h}{2}=\frac{1.5}{2}=0.75m$

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$Force=1000\times 9.81\times 0.75\times 1.5\times 2.5=27.59kN$

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Thus applying the calculated values we get

$Moment=27.59\times \frac{1.5}{3}=13.8kN.m$

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