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Fittoniya [83]
3 years ago
6

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?
Engineering
1 answer:
Umnica [9.8K]3 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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Explanation:

Given;

angular frequency, ω = 400 rpm = (2π /60) x (400) = 41.893 rad/s

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The peak emf is given by;

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B = (24) / (300 x 0.003055 x 41.893)

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Answer:

The following query is used to display TripName. Reservationld, FirstName. LastName and TotalCost of trip by adding the trip price plus other fees and multiplying the result by the number of persons which have number of persons >4.

Query:

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3 years ago
A satellite orbits the Earth every 2 hours at an average distance from the Earth's centre of 8000km. (i) What is the average ang
AlexFokin [52]

Answer:

i)ω=3600 rad/s

ii)V=7059.44 m/s

iii)F=1245.8 N

Explanation:

i)

We know that angular speed given as

\omega =\dfrac{d\theta}{dt}

We know that for one revolution

θ=2π

Given that time t= 2 hr

So

ω=θ/t

ω=2π/2 = π rad/hr

ω=3600 rad/s

ii)

Average speed V

V=\sqrt{\dfrac{GM}{R}}

Where M is the mass of earth.

R is the distance

G is the constant.

Now by putting the values

V=\sqrt{\dfrac{GM}{R}}

V=\sqrt{\dfrac{6.667\times 10^{-11}\times 5.98\times 10^{24}}{8000\times 10^3}}

V=7059.44 m/s

iii)

We know that centripetal fore given as

F=\dfrac{mV^2}{R}

Here given that m= 200 kg

R= 8000 km

so now by putting the values

F=\dfrac{mV^2}{R}

F=\dfrac{200\times 7059.44^2}{8000\times 10^3}

F=1245.8 N

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