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Helga [31]
2 years ago
12

On a day in which the local atmospheric pressure is 99.5 kPa, answer each of the following: (a) Calculate the column height of m

ercury in a mercury barometer in units of meters, feet, and inches. (b) Francis is concerned about mercury poisoning, so he builds a water barometer to replace the mercury barometer. Calculate the column height of water in the water barometer in units of meters, feet, and inches. (c) Explain why a water barometer is not very practical. (d) Ignoring the practicality issue, which of the two (mercury or water) would be more precise
Engineering
1 answer:
horrorfan [7]2 years ago
8 0

Answer:

C . . . . . . . . . . . . . . . . . . . . . .

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The European Space Agency launched a probe called Rosetta in March 2004. In August​ 2014, Rosetta reached its​ destination: a co
Art [367]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The of the lander would be 75.66 pounds

Explanation:

From the question we are given that the

Weight of the Philae on Earth  is = 220 pound-mass

gravity of Mars is = 3.71 m/s^{2}

Weight of the Philae on Mars is = ?

Now Mass is quantity of matter in an object so it is always constant for a particular object

Generally Weight = Mass × Gravity :

                      \frac{Weight Of Earth}{Gravity Of Earth} = \frac{Weight Of Mars}{Gravity Of Mars}

Hence  

         Weight of the Philae on Mars is = \frac{Weight Of Earth *Gravity Of Mars}{Gravity Of Earth}

                                                              =\frac{200 *3.71}{9.81}

                                                              = 75.66 pounds

6 0
3 years ago
Why is tubing sometimes coiled when installed in a car or vechile
andriy [413]

Answer:

Coiled tubing is often used to carry out operations similar to wire lining.

8 0
3 years ago
A hemispherical shell with an external diameter of 500 mm and a thickness of 20 mm is going to be made by casting, located entir
Olenka [21]

Solution :

Given :

External diameter of the hemispherical shell, D = 500 mm

Thickness, t = 20 mm

Internal diameter, d = D - 2t

                                 = 500 - 2(20)

                                 = 460 mm

So, internal radius, r = 230 mm

                                 = 0.23 m

Density of molten metal, ρ = $7.2 \ g/cm^3$

                                                  = $7200 \ kg/m^3$

The height of pouring cavity above parting surface is h = 300 mm

                                                                                                  = 0.3 m

So, the metallostatic thrust on the upper mold at the end of casting is :

$F=\rho g A h$

Area, A $=2 \pi r^2$

            $=2 \pi (0.23)^2$

            $=0.3324 \ m^2$

$F=\rho g A h$

   $=7200 \times 9.81 \times 0.3324 \times 0.3$

     = 7043.42 N

3 0
3 years ago
Water at a pressure of 3 bars enters a short horizontal convergent channel at 3.5 m/s. The upstream and downstream diameters of
earnstyle [38]

Answer:

The pressure reduces to 2.588 bars.

Explanation:

According to Bernoulli's theorem for ideal flow we have

\frac{P}{\gamma _{w}}+\frac{V^{2}}{2g}+z=constant

Since the losses are neglected thus applying this theorm between upper and lower porion we have

\frac{P_{u}}{\gamma _{w}}+\frac{V-{u}^{2}}{2g}+z_{u}=\frac{P_{L}}{\gamma _{w}}+\frac{V{L}^{2}}{2g}+z_{L}

Now by continuity equation we have

A_{u}v_{u}=A_{L}v_{L}\\\\\therefore v_{L}=\frac{A_{u}}{A_{L}}\times v_{u}\\\\v_{L}=\frac{d^{2}_{u}}{d^{2}_{L}}\times v_{u}\\\\\therefore v_{L}=\frac{2500}{900}\times 3.5\\\\\therefore v_{L}=9.72m/s

Applying the values in the Bernoulli's equation we get

\frac{P_{L}}{\gamma _{w}}=\frac{300000}{\gamma _{w}}+\frac{3.5^{2}}{2g}-\frac{9.72^{2}}{2g}(\because z_{L}=z_{u})\\\\\frac{P_{L}}{\gamma _{w}}=26.38m\\\\\therefore P_{L}=258885.8Pa\\\\\therefore P_{L}=2.588bars

6 0
3 years ago
An amplifier with 40 dB of small-signal, open-circuit voltage gain, an input resistance of 1 MO, and an output resistance of 100
Vanyuwa [196]

convert 40db to standard gain

AL=10^40/20=100

calculate total voltage gain

=AL×RL/RL+Ri

=83.33

38.41 DB

calculate power

Pi=Vi^2/Ri Po=Vo^2/RL

power gain= Po/Pi

=13.90×10^6

3 0
3 years ago
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