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sashaice [31]
3 years ago
13

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

met called​ 67P/Churyumov-Gerasimenko. Rosetta is the first spacecraft to rendezvous with a comet. It contained a lander​ craft, called Philae.The Philae has a launch mass on the Earth of 220​pounds-mass ​[lbm​].If the Philae was deployed to Mars instead of Comet​ 67P, what would the weight of the lander be after landing on the

Engineering
1 answer:
Art [367]3 years ago
6 0

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

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A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28
Delicious77 [7]

Answer:

5984.67N

Explanation:

A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28 psi and a pressure drop of 2 psi occurs through the contraction if the upstream velocity is 4.0 ft/sec. What is the magnitude of the resultant force (lbs) needed to hold the pipe in place?

from continuity equation

v1A1=v2A2

equation of continuity

v1=4ft /s=1.21m/s

d1=14 inch=.35m

d2=14-2=0.304m

A1=pi*d^2/4

0.096m^2

a2=0.0706m^2

from continuity once again

1.21*0.096=v2(0.07)

v2=1.65

force on the pipe

(p1A1- p2A2) + m(v2 – v1)

from bernoulli

p1 + ρv1^2/2 = p2 + ρv2^2/2

difference in pressure or pressure drop

p1-p2=2psi

13.789N/m^2=rho(1.65^2-1.21^2)/2

rho=21.91kg/m^3

since the pipe is cylindrical

pressure is egh

13.789=21.91*9.81*h

length of the pipe is

0.064m

AH=volume of the pipe(area *h)

the mass =rho*A*H

0.064*0.07*21.91

m=0.098kg

(193053*0.096- 179263.6* 0.07) + 0.098(1.65 – 1.21)

force =5984.67N

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2) For P2 greater than P1 and V2 also is greater than V1. Since P2 is directly proportional to V2, hence, when P2 increases in pressure, P1 reduces in pressure. Similarly, velocity, V2 also increases and V1 reduces.

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